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Physics LibreTexts

12.3: Examples of Static Equilibrium

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Learning Objectives

  • Identify and analyze static equilibrium situations
  • Set up a free-body diagram for an extended object in static equilibrium
  • Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations

All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Equation 12.2.9 to Equation 12.2.11 . We introduced a problem-solving strategy in Example 12.1 to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps.

Problem-Solving Strategy: Static Equilibrium

  • Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
  • Set up a free-body diagram for the object. (a) Choose the xy-reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x- and y-directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign (+) means that the working direction is the actual direction. A minus sign (−) means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis.
  • Set up the equations of equilibrium for the object. (a) Use the free-body diagram to write a correct equilibrium condition Equation 12.2.9 for force components in the x-direction. (b) Use the free-body diagram to write a correct equilibrium condition Equation 12.2.13 for force components in the y-direction. (c) Use the free-body diagram to write a correct equilibrium condition Equation 12.2.11 for torques along the axis of rotation. Use Equation 12.2.12 to evaluate torque magnitudes and senses.
  • Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved.
  • Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in Example 12.1 .

Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.

Example 12.3: The Torque Balance

Three masses are attached to a uniform meter stick, as shown in Figure \(\PageIndex{1}\). The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are m 1 = 50.0 g and m 2 = 75.0 g. Find the mass m3 that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

Figure is a schematic drawing of a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Mass 3 is 30 cm to the right of S. Mass 2 is 40 cm to the left of S. Mass 1 is 30 cm to the left of Mass 2.

For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:

  • w 1 = m 1 g is the weight of mass m 1 ;
  • w 2 = m 2 g is the weight of mass m 2 ;
  • w = mg is the weight of the entire meter stick;
  • w 3 = m 3 g is the weight of unknown mass m 3 ;
  • F S is the normal reaction force at the support point S.

We choose a frame of reference where the direction of the y-axis is the direction of gravity, the direction of the xaxis is along the meter stick, and the axis of rotation (the z-axis) is perpendicular to the x-axis and passes through the support point S. In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot Figure \(\PageIndex{2}\). At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w, at the 50-cm mark.

Figure is a schematic drawing of a force distribution for a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Force Fs at the point S is pointing upward. Force w3, to the right of point S and separated by distance r3 is pointing downward. Forces w, w2, and w1 are to the left of point S and are pointing downward. They are separated by distance r, r2, and r1, respectively.

With Figure \(\PageIndex{1}\) and Figure \(\PageIndex{2}\) for reference, we begin by finding the lever arms of the five forces acting on the stick:

\[\begin{split} r_{1} & = 30.0\; cm + 40.0\; cm = 70.0\; cm \\ r_{2} & = 40.0\; cm \\ r & = 50.0\; cm - 30.0\; cm = 20.0\; cm \\ r_{S} & = 0.0\; cm\; (because\; F_{S}\; is\; attached\; at\; the\; pivot) \\ r_{3} & = 30.0\; cm \ldotp \end{split}\]

Now we can find the five torques with respect to the chosen pivot:

\[\begin{split} \tau_{1} & = +r_{1} w_{1} \sin 90^{o} = +r_{1} m_{1} g \quad (counterclockwise\; rotation,\; positive\; sense) \\ \tau_{2} & = +r_{2} w_{2} \sin 90^{o} = +r_{2} m_{2} g \quad (counterclockwise\; rotation,\; positive\; sense) \\ \tau & = +rw \sin 90^{o} = +rmg \quad \quad \quad (gravitational\; torque) \\ \tau_{S} & = r_{S} F_{S} \sin \theta_{S} = 0 \quad \quad \quad \quad \quad (because\; r_{S} = 0\; cm) \\ \tau_{3} & = -r_{3} w_{3} \sin 90^{o} = -r_{3} m_{3} g \quad (counterclockwise\; rotation,\; negative\; sense) \end{split}\]

The second equilibrium condition (equation for the torques) for the meter stick is

\[\tau_{1} + \tau_{2} + \tau + \tau_{S} + \tau_{3} = 0 \ldotp\]

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is

\[+r_{1} m_{1} g + r_{2} m_{2} g + rmg - r_{3} m_{3} g = 0 \ldotp \label{12.17}\]

Selecting the +y-direction to be parallel to \(\vec{F}_{S}\), the first equilibrium condition for the stick is

\[-w_{1} - w_{2} - w + F_{S} - w_{3} = 0 \ldotp\]

Substituting the forces, the first equilibrium condition becomes

\[-m_{1} g - m_{2} g - mg + F_{S} - m_{3} g = 0 \ldotp \label{12.18}\]

We solve these equations simultaneously for the unknown values m 3 and F S . In Equation \ref{12.17}, we cancel the g factor and rearrange the terms to obtain

\[r_{3} m_{3} = r_{1} m_{1} + r_{2} m_{2} + rm \ldotp\]

To obtain m 3 we divide both sides by r 3 , so we have

\[\begin{split} m_{3} & = \frac{r_{1}}{r_{3}} m_{1} + \frac{r_{2}}{r_{3}} m_{2} + \frac{r}{r_{3}} m \\ & = \frac{70}{30} (50.0\; g) + \frac{40}{30} (75.0\; g) + \frac{20}{30} (150.0\; g) = 315.0 \left(\dfrac{2}{3}\right)\; g \simeq 317\; g \ldotp \end{split} \label{12.19}\]

To find the normal reaction force, we rearrange the terms in Equation \ref{12.18}, converting grams to kilograms:

\[\begin{split} F_{S} & = (m_{1} + m_{2} + m + m_{3}) g \\ & = (50.0 + 75.0 + 150.0 + 316.7) \times (10^{-3}\; kg) \times (9.8\; m/s^{2}) = 5.8\; N \ldotp \end{split} \label{12.20}\]

Significance

Notice that Equation \ref{12.17} is independent of the value of g. The torque balance may therefore be used to measure mass, since variations in g-values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force.

Exercise 12.3

Repeat Example 12.3 using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick.

In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by Equation 12.2.9 and Equation 12.2.10 . We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics.

Example 12.4: Forces in the Forearm

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in Figure \(\PageIndex{3}\). His forearm is positioned at \(\beta\) = 60° with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.

Figure is a schematic drawing of a forearm rotated around the elbow. A 50 pound ball is held in the palm. The distance between the elbow and the ball is 13 inches. The distance between the elbow and the biceps muscle, which causes a torque around the elbow, is 1.5 inches. Forearm forms a 60 degree angle with the upper arm.

We identify three forces acting on the forearm: the unknown force \(\vec{F}\) at the elbow; the unknown tension \(\vec{T}_{M}\) in the muscle; and the weight \(\vec{w}\) with magnitude w = 50 lb. We adopt the frame of reference with the x-axis along the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The x-axis makes an angle \(\beta\) = 60° with the vertical. The y-axis is perpendicular to the x-axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle \(\beta\) and represent each force by its x- and y-components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in Figure \(\PageIndex{4}\). At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x- and y-components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm.

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force T is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Forces F and T form angle beta with the x axis. Force W forms an angle beta with line connecting it with its projection to the y axis.

Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y-components of the forces because the x-components of the forces are all parallel to their lever arms, so that for any of them we have sin \(\theta\) = 0 in Equation 12.2.12 . For the y-components we have \(\theta\) = ± 90° in Equation 12.2.12 . Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of T y and of w y .

We see from the free-body diagram that the x-component of the net force satisfies the equation

\[+F_{x} + T_{x} - w_{x} = 0 \label{12.21}\]

and the y-component of the net force satisfies

\[+F_{y} + T_{y} - w_{y} = 0 \ldotp \label{12.22}\]

Equation \ref{12.21} and Equation \ref{12.22} are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is

\[+r_{T} T_{y} - r_{w} w_{y} = 0 \ldotp \label{12.23}\]

Equation \ref{12.23} is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are r T = 1.5 in. and r w = 13.0 in. At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Equation 12.23, they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

\[\begin{split} F_{x} & = F \cos \beta = F \cos 60^{o} = \frac{F}{2} \\ T_{x} & = T \cos \beta = T \cos 60^{o} = \frac{T}{2} \\ w_{x} & = w \cos \beta = w \cos 60^{o} = \frac{w}{2} \\ F_{y} & = F \sin \beta = F \sin 60^{o} = \frac{F \sqrt{3}}{2} \\ T_{y} & = T \sin \beta = T \sin 60^{o} = \frac{T \sqrt{3}}{2} \\ w_{y} & = w \sin \beta = w \sin 60^{o} = \frac{w \sqrt{3}}{2} \ldotp \end{split}\]

We substitute these magnitudes into Equation \ref{12.21}, Equation \ref{12.22}, and Equation \ref{12.23} to obtain, respectively,

\[\begin{split} \frac{F}{2} + \frac{T}{2} - \frac{w}{2} & = 0 \\ \frac{F \sqrt{3}}{2} + \frac{T \sqrt{3}}{2} - \frac{w \sqrt{3}}{2} & = 0 \\ \frac{r_{T} T \sqrt{3}}{2} - \frac{r_{w} w \sqrt{3}}{2} & = 0 \ldotp \end{split}\]

When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T, because Equation \ref{12.21} for the x-component is equivalent to Equation \ref{12.22} for the y-component. In this way, we obtain the first equilibrium condition for forces

\[F + T - w = 0 \label{12.24}\]

and the second equilibrium condition for torques

\[r_{T} T - r_{w} w = 0 \ldotp \label{12.25}\]

The magnitude of tension in the muscle is obtained by solving Equation \ref{12.25}:

\[T = \frac{r_{w}}{r_{T}} w =frac{13.0}{1.5} (50\; lb) = 433 \frac{1}{3}\; lb \simeq 433.3\; lb \ldotp\]

The force at the elbow is obtained by solving Equation \ref{12.24}:

\[F = w - T = 50.0\; lb - 433.3\; lb = -383.3\; lb \ldotp\]

The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is

\[F = 383.3\; lb = 383.3(4.448\; N) = 1705\; N\; downward\]

\[T = 433.3\; lb = 433.3(4.448\; N) = 1927\; N\; upward \ldotp\]

Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction , and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule.

Suppose we adopt a reference frame with the direction of the y-axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only y-components, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in Figure \(\PageIndex{5}\), indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles \(\theta_{T}\) and \(\theta_{w}\) that the forces \(\vec{T}_{M}\) and \(\vec{w}\) (respectively) make with their lever arms. In the definition of torque given by Equation 12.2.12 , the angle \(\theta_{T}\) is the direction angle of the vector \(\vec{T}_{M}\), counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle \(\theta_{w}\) is measured counterclockwise from the radial direction of the lever arm to the vector \(\vec{w}\). Done this way, the non-zero torques are most easily computed by directly substituting into Equation 12.2.12 as follows:

\[\tau_{T} = r_{T} T \sin \theta_{T} = r_{T} T \sin \beta = r_{T} T \sin 60^{o} = + \frac{r_{T} T \sqrt{3}}{2}\]

\[\tau_{w} = r_{w} w \sin \theta_{w} = r_{w} w \sin (\beta + 180^{o}) = -r_{w} w \sin \beta = - \frac{r_{w} w \sqrt{3}}{2} \ldotp\]

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force Tm is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Force Tm forms and angle theta tau that is equal to beta with the direction of the lever arm. Force W forms an angle theta w that is equal to the sum of beta and Pi with the direction of the lever arm.

The second equilibrium condition, \(\tau_{T}\) + \(\tau_{w}\) = 0, can be now written as

\[\frac{r_{T} T \sqrt{3}}{2} - \frac{r_{w} w \sqrt{3}}{2} = 0 \ldotp \label{12.26}\]

From the free-body diagram, the first equilibrium condition (for forces) is

\[-F + T - w = 0 \ldotp \label{12.27}\]

Equation \ref{12.26} is identical to Equation \ref{12.25} and gives the result T = 433.3 lb. Equation \ref{12.27} gives

\[F = T - w = 433.3\; lb - 50.0\; lb = 383.3\; lb\]

We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

Exercise 12.4

Repeat Example 12.4 assuming that the forearm is an object of uniform density that weighs 8.896 N.

Example 12.5: A Ladder Resting Against a Wall

A uniform ladder is L = 5.0 m long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in Figure \(\PageIndex{6}\). The inclination angle between the ladder and the rough floor is \(\beta\) = 53°. Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction \(\mu_{s}\) at the interface of the ladder with the floor that prevents the ladder from slipping.

Figure is a schematic drawing of a 5.0-m-long ladder resting against a wall. Ladder forms a 53 degree angle with the floor.

We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. The second force is the static friction force f = \(\mu_{s}\)N directed horizontally along the floor toward the wall—this force prevents the ladder from slipping. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends. The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall. There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y-axis in the vertical direction (parallel to the wall) and the x-axis in the horizontal direction (parallel to the floor). In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution. We select the pivot at the contact point with the floor. In the free-body diagram for the ladder, we indicate the pivot, all four forces and their lever arms, and the angles between lever arms and the forces, as shown in Figure \(\PageIndex{7}\). With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot.

Figure is a free-body diagram for a ladder that forms an angle beta with the floor and rests against a wall. Force N is applied at the point at the floor and is perpendicular to the floor. Force W is applied at the mid-point of the ladder. Force F is applied at the point resting at the wall and is perpendicular to the wall. Force W forms an angle theta w with the direction of the lever arm. Theta w is equal to the sum of Pi and half Pi with the beta subtracted. Force F forms an angle theta F with the direction of the lever arm. Theta F is equal to the Pi minus beta.

From the free-body diagram, the net force in the x-direction is

\[+f - F = 0 \label{12.28}\]

the net force in the y-direction is

\[+ N - w = 0 \label{12.29}\]

and the net torque along the rotation axis at the pivot point is

\[\tau_{w} + \tau_{F} = 0 \ldotp \label{12.30}\]

where \(\tau_{w}\) is the torque of the weight w and \(\tau_{F}\) is the torque of the reaction F. From the free-body diagram, we identify that the lever arm of the reaction at the wall is r F = L = 5.0 m and the lever arm of the weight is r w = \(\frac{L}{2}\) = 2.5 m. With the help of the free-body diagram, we identify the angles to be used in Equation 12.2.12 for torques: \(\theta_{F}\) = 180° − \(\beta\) for the torque from the reaction force with the wall, and \(\theta_{w}\) = 180° + (90° − \(\beta\)) for the torque due to the weight. Now we are ready to use Equation 12.2.12 to compute torques:

\[\tau_{w} = r_{w} w \sin \theta_{w} = r_{w} w \sin (180^{o} + 90^{o} - \beta) = - \frac{L}{2} w \sin (90^{o} - \beta) = - \frac{L}{2} w \cos \beta\]

\[tau_{F} = r_{F} F \sin \theta_{F} = r_{F} F \sin (180^{o} - \beta) = LF \sin \beta \ldotp\]

We substitute the torques into Equation \ref{12.30} and solve for F :

\[- \frac{L}{2} w \cos \beta + LF \sin \beta = 0 \label{12.31}\]

\[F = \frac{w}{2} \cot \beta = \frac{400.0\; N}{2} \cot 53^{o} = 150.7\; N\]

We obtain the normal reaction force with the floor by solving Equation \ref{12.29}: N = w = 400.0 N. The magnitude of friction is obtained by solving Equation \ref{12.28}: f = F = 150.7 N. The coefficient of static friction is \(\mu_{s}\) = \(\frac{f}{N}\) = \(\frac{150.7}{400.0}\) = 0.377.

The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces:

\[\vec{F}_{floor} = \vec{f} + \vec{N} = (150.7\; N)(- \hat{i}) + (400.0\; N)(+ \hat{j}) = (-150.7\; \hat{i} + 400.0\; \hat{j}) N \ldotp\]

Its magnitude is

\[F_{floor} = \sqrt{f^{2} + N^{2}} = \sqrt{150.7^{2} + 400.0^{2}}\; N = 427.4\; N\]

and its direction is

\[\varphi = \tan^{-1} \left(\dfrac{N}{f}\right) = \tan^{-1} \left(\dfrac{400.0}{150.7}\right) = 69.3^{o}\; above\; the\; floor \ldotp\]

We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use Equation 12.2.12 for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in Equation 12.2.12 is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, Equation 12.2.12 gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and Equation 12.2.12 expresses the rectangular component of this vector product along the axis of rotation.

This result is independent of the length of the ladder because L is canceled in the second equilibrium condition, Equation \ref{12.31}. No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is 53°, our results hold. But the ladder will slip if the net torque becomes negative in Equation \ref{12.31}. This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.

Exercise 12.5

For the situation described in Example 12.5, determine the values of the coefficient \(\mu_{s}\) of static friction for which the ladder starts slipping, given that β is the angle that the ladder makes with the floor.

Example 12.6: Forces on Door Hinges

A swinging door that weighs w = 400.0 N is supported by hinges A and B so that the door can swing about a vertical axis passing through the hinges Figure \(\PageIndex{8}\). The door has a width of b = 1.00 m, and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance a = 2.00 m. Find the forces on the hinges when the door rests half-open.

Figure is a schematic drawing of a swinging vertical door supported by two hinges attached at points A and B. The distance between points A and B is 2 meters. Door is one meter wide.

The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force \(\vec{A}\) from hinge A, an unknown force \(\vec{B}\) from hinge B, and the known weight \(\vec{w}\) attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the y-axis along the direction of gravity and the x-axis in the plane of the slab, as shown in panel (a) of Figure \(\PageIndex{9}\), and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force \(\vec{A}\) (A x and A y ), and two components of force \(\vec{B}\) (B x and B y ). In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns (A x , B x , A y , and B y ), we must set up four independent equations. One equation is the equilibrium condition for forces in the x-direction. The second equation is the equilibrium condition for forces in the y-direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation, A y = B y . To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of Figure \(\PageIndex{9}\). Finally, we solve the equations for the unknown force components and find the forces.

Figure A is a geometrical representation for a swinging vertical door supported by two hinges attached at points A and B. Forces A and B are applied at the points A and B. Projections of these forces to the x and y axes are shown. Force w is applied at the point CM. Point CM is lower than point A by half-a and to the right of point A by half-b. Line from point A to CM forms an angle beta with the edge of the wall. Figure B is a free-body diagram for a swinging vertical door is supported by two hinges attached at points A and B. Force Ay forms an angle beta with the line connecting points P and CM. Force By forms an angle beta with the line connecting points B and CM. Force W forms an angle beta with the line that is the continuation of the line connecting points P and CM. Distance between points P and CM is d.

From the free-body diagram for the door we have the first equilibrium condition for forces:

in the x-direction, $$-A_{x} + B_{x} = 0 \Rightarrow A_{x} + B_{x}$$in y-direction, $$+ A_{y} + B_{y} - w = 0 \Rightarrow A_{y} = B_{y} = \frac{w}{2} = \frac{400.0\; N}{2} = 200.0\; N \ldotp\]

We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P:

pivot at P: $$\tau_{w} + \tau_{Bx} + \tau_{By} = 0 \ldotp \label{12.32}\]

We use the free-body diagram to find all the terms in this equation:

\[\begin{split} \tau_{w} & = dw \sin (- \beta) = -dw \sin \beta = -dw \frac{\frac{b}{2}}{d} = -w \frac{b}{2} \\ \tau_{Bx} & = a B_{x} \sin 90^{o} = + a B_{x} \\ \tau_{By} & = a B_{y} \sin 180^{o} = 0 \ldotp \end{split}\]

In evaluating sin \(\beta\), we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Equation \ref{12.32} and compute B x :

pivot at P: $$-w \frac{b}{2} + a B_{x} = 0 \Rightarrow B_{x} = w \frac{b}{2a} = (400.0\; N) \frac{1}{2\; \cdotp 2} = 100.0\; N \ldotp\]

Therefore the magnitudes of the horizontal component forces are A x = B x = 100.0 N. The forces on the door are

at the upper hinge: $$\vec{F}_{A\; on\; door} = -100.0\; N\; \hat{i} + 200.0\; N\; \hat{j}$$at the lower hinge: $$\vec{F}_{B\; on\; door} = +100.0\; N\; \hat{i} + 200.0\; N\; \hat{j} \ldotp\]

The forces on the hinges are found from Newton’s third law as

on the upper hinge: $$\vec{F}_{door\; on\; A} = 100.0\; N\; \hat{i} - 200.0\; N\; \hat{j}$$on the lower hinge: $$\vec{F}_{door\; on\; B} = - 100.0\; N\; \hat{i} - 200.0\; N\; \hat{j} \ldotp\]

Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.

Exercise 12.6

Solve the problem in Example 12.6 by taking the pivot position at the center of mass.

Exercise 12.7

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.

Figure is a schematic drawing of a woman standing 1.5 m away from one end and 4.5 m away from another end of a scaffold.

Exercise 12.8

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut is 4.0 m long and is supported by a 5.0 m long cable tied to the wall at a point 3.0 m above the left end of the strut.

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12 Static Equilibrium and Elasticity

12.2 Examples of Static Equilibrium

Learning objectives.

By the end of this section, you will be able to:

  • Identify and analyze static equilibrium situations
  • Set up a free-body diagram for an extended object in static equilibrium
  • Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations

All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Figure to Figure . We introduced a problem-solving strategy in Figure to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps.

Problem-Solving Strategy: Static Equilibrium

  • Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
  • Set up a free-body diagram for the object. (a) Choose the xy -reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x – and y -directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign [latex](+)[/latex] means that the working direction is the actual direction. A minus sign [latex](-)[/latex] means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis.
  • Set up the equations of equilibrium for the object. (a) Use the free-body diagram to write a correct equilibrium condition Figure for force components in the x -direction. (b) Use the free-body diagram to write a correct equilibrium condition Figure for force components in the y -direction. (c) Use the free-body diagram to write a correct equilibrium condition Figure for torques along the axis of rotation. Use Figure to evaluate torque magnitudes and senses.
  • Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved.
  • Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in Figure .

Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.

The Torque Balance

Three masses are attached to a uniform meter stick, as shown in Figure . The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are [latex]{m}_{1}=50.0\,\text{g}[/latex] and [latex]{m}_{2}=75.0\,\text{g}.[/latex] Find the mass [latex]{m}_{3}[/latex] that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

Figure is a schematic drawing of a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Mass 3 is 30 cm to the right of S. Mass 2 is 40 cm to the left of S. Mass 1 is 30 cm to the left of Mass 2.

For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:

[latex]{w}_{1}={m}_{1}g[/latex] is the weight of mass [latex]{m}_{1};[/latex] [latex]{w}_{2}={m}_{2}g[/latex] is the weight of mass [latex]{m}_{2};[/latex]

[latex]w=mg[/latex] is the weight of the entire meter stick; [latex]{w}_{3}={m}_{3}g[/latex] is the weight of unknown mass [latex]{m}_{3};[/latex]

[latex]{F}_{S}[/latex] is the normal reaction force at the support point S .

We choose a frame of reference where the direction of the y -axis is the direction of gravity, the direction of the x -axis is along the meter stick, and the axis of rotation (the z -axis) is perpendicular to the x -axis and passes through the support point S . In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot Figure . At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w , at the 50-cm mark.

Figure is a schematic drawing of a force distribution for a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Force Fs at the point S is pointing upward. Force w3, to the right of point S and separated by distance r3 is pointing downward. Forces w, w2, and w1 are to the left of point S and are pointing downward. They are separated by distance r, r2, and r1, respectively.

With Figure and Figure for reference, we begin by finding the lever arms of the five forces acting on the stick:

Now we can find the five torques with respect to the chosen pivot:

The second equilibrium condition (equation for the torques) for the meter stick is

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is

Selecting the [latex]+y[/latex]-direction to be parallel to [latex]{\mathbf{\overset{\to }{F}}}_{S},[/latex] the first equilibrium condition for the stick is

Substituting the forces, the first equilibrium condition becomes

We solve these equations simultaneously for the unknown values [latex]{m}_{3}[/latex] and [latex]{F}_{S}.[/latex] In Figure , we cancel the g factor and rearrange the terms to obtain

To obtain [latex]{m}_{3}[/latex] we divide both sides by [latex]{r}_{3},[/latex] so we have

To find the normal reaction force, we rearrange the terms in Figure , converting grams to kilograms:

Significance

Notice that Figure is independent of the value of g . The torque balance may therefore be used to measure mass, since variations in g -values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force.

Check Your Understanding

Repeat Figure using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick.

316.7 g; 5.8 N

In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by Figure and Figure . We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics.

Forces in the Forearm

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in Figure . His forearm is positioned at [latex]\beta =60^\circ[/latex] with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.

Figure is a schematic drawing of a forearm rotated around the elbow. A 50 pound ball is held in the palm. The distance between the elbow and the ball is 13 inches. The distance between the elbow and the biceps muscle, which causes a torque around the elbow, is 1.5 inches. Forearm forms a 60 degree angle with the upper arm.

We identify three forces acting on the forearm: the unknown force [latex]\mathbf{\overset{\to }{F}}[/latex] at the elbow; the unknown tension [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}}[/latex] in the muscle; and the weight [latex]\mathbf{\overset{\to }{w}}[/latex] with magnitude [latex]w=50\,\text{lb}.[/latex] We adopt the frame of reference with the x -axis along the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The x -axis makes an angle [latex]\beta =60^\circ[/latex] with the vertical. The y -axis is perpendicular to the x -axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle [latex]\beta[/latex] and represent each force by its x – and y -components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in Figure . At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x – and y -components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm.

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force T is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Forces F and T form angle beta with the x axis. Force W forms an angle beta with line connecting it with its projection to the y axis.

Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y -components of the forces because the x -components of the forces are all parallel to their lever arms, so that for any of them we have [latex]\text{sin}\,\theta =0[/latex] in Figure . For the y -components we have [latex]\theta =\pm90^\circ[/latex] in Figure . Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of [latex]{T}_{y}[/latex] and of [latex]{w}_{y}.[/latex]

We see from the free-body diagram that the x -component of the net force satisfies the equation

and the y -component of the net force satisfies

Figure and Figure are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is

Figure is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are [latex]{r}_{T}=1.5\,\text{in}\text{.}[/latex] and [latex]{r}_{w}=13.0\,\text{in}\text{.}[/latex] At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Figure , they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

We substitute these magnitudes into Figure , Figure , and Figure to obtain, respectively,

When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T , because Figure for the x -component is equivalent to Figure for the y -component. In this way, we obtain the first equilibrium condition for forces

and the second equilibrium condition for torques

The magnitude of tension in the muscle is obtained by solving Figure :

The force at the elbow is obtained by solving Figure :

The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is

Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction , and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule.

Suppose we adopt a reference frame with the direction of the y -axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only y -components, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in Figure , indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles [latex]{\theta }_{T}[/latex] and [latex]{\theta }_{w}[/latex] that the forces [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}}[/latex] and [latex]\mathbf{\overset{\to }{w}}[/latex] (respectively) make with their lever arms. In the definition of torque given by Figure , the angle [latex]{\theta }_{T}[/latex] is the direction angle of the vector [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}},[/latex] counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle [latex]{\theta }_{w}[/latex] is measured counterclockwise from the radial direction of the lever arm to the vector [latex]\mathbf{\overset{\to }{w}}.[/latex] Done this way, the non-zero torques are most easily computed by directly substituting into Figure as follows:

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force Tm is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Force Tm forms and angle theta tau that is equal to beta with the direction of the lever arm. Force W forms an angle theta w that is equal to the sum of beta and Pi with the direction of the lever arm.

The second equilibrium condition, [latex]{\tau }_{T}+{\tau }_{w}=0,[/latex] can be now written as

From the free-body diagram, the first equilibrium condition (for forces) is

Figure is identical to Figure and gives the result [latex]T=433.3\,\text{lb}.[/latex] Figure gives

We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

Repeat Figure assuming that the forearm is an object of uniform density that weighs 8.896 N.

[latex]T=\text{1963 N};\,\text{F}=1732\,\text{N}[/latex]

A Ladder Resting Against a Wall

A uniform ladder is [latex]L=5.0\,\text{m}[/latex] long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in Figure . The inclination angle between the ladder and the rough floor is [latex]\beta =53^\circ.[/latex] Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction [latex]{\mu }_{\text{s}}[/latex] at the interface of the ladder with the floor that prevents the ladder from slipping.

Figure is a schematic drawing of a 5.0-m-long ladder resting against a wall. Ladder forms a 53 degree angle with the floor.

We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. The second force is the static friction force [latex]f={\mu }_{\text{s}}N[/latex] directed horizontally along the floor toward the wall—this force prevents the ladder from slipping. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends. The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall. There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y -axis in the vertical direction (parallel to the wall) and the x -axis in the horizontal direction (parallel to the floor). In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution. We select the pivot at the contact point with the floor. In the free-body diagram for the ladder, we indicate the pivot, all four forces and their lever arms, and the angles between lever arms and the forces, as shown in Figure . With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot.

Figure is a free-body diagram for a ladder that forms an angle beta with the floor and rests against a wall. Force N is applied at the point at the floor and is perpendicular to the floor. Force W is applied at the mid-point of the ladder. Force F is applied at the point resting at the wall and is perpendicular to the wall. Force W forms an angle theta w with the direction of the lever arm. Theta w is equal to the sum of Pi and half Pi with the beta subtracted. Force F forms an angle theta F with the direction of the lever arm. Theta F is equal to the Pi minus beta.

From the free-body diagram, the net force in the x -direction is

the net force in the y -direction is

and the net torque along the rotation axis at the pivot point is

where [latex]{\tau }_{w}[/latex] is the torque of the weight w and [latex]{\tau }_{F}[/latex] is the torque of the reaction F . From the free-body diagram, we identify that the lever arm of the reaction at the wall is [latex]{r}_{F}=L=5.0\,\text{m}[/latex] and the lever arm of the weight is [latex]{r}_{w}=L\,\text{/}\,2=2.5\,\text{m}.[/latex] With the help of the free-body diagram, we identify the angles to be used in Figure for torques: [latex]{\theta }_{F}=180^\circ-\beta[/latex] for the torque from the reaction force with the wall, and [latex]{\theta }_{w}=180^\circ+(90^\circ-\beta )[/latex] for the torque due to the weight. Now we are ready to use Figure to compute torques:

We substitute the torques into Figure and solve for [latex]F:[/latex]

We obtain the normal reaction force with the floor by solving Figure : [latex]N=w=400.0\,\text{N}.[/latex] The magnitude of friction is obtained by solving Figure : [latex]f=F=150.7\,\text{N}.[/latex] The coefficient of static friction is [latex]{\mu }_{\text{s}}=f\,\text{/}\,N=150.7\,\text{/}\,400.0=0.377.[/latex]

The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces:

Its magnitude is

and its direction is

We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use Figure for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in Figure is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, Figure gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and Figure expresses the rectangular component of this vector product along the axis of rotation.

This result is independent of the length of the ladder because L is cancelled in the second equilibrium condition, Figure . No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is [latex]53^\circ,[/latex] our results hold. But the ladder will slip if the net torque becomes negative in Figure . This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.

For the situation described in Figure , determine the values of the coefficient [latex]{\mu }_{\text{s}}[/latex] of static friction for which the ladder starts slipping, given that [latex]\beta[/latex] is the angle that the ladder makes with the floor.

[latex]{\mu }_{s} \lt 0.5\,\text{cot}\,\beta[/latex]

Forces on Door Hinges

A swinging door that weighs [latex]w=400.0\,\text{N}[/latex] is supported by hinges A and B so that the door can swing about a vertical axis passing through the hinges Figure . The door has a width of [latex]b=1.00\,\text{m},[/latex] and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance [latex]a=2.00\,\text{m}.[/latex] Find the forces on the hinges when the door rests half-open.

Figure is a schematic drawing of a swinging vertical door supported by two hinges attached at points A and B. The distance between points A and B is 2 meters. Door is one meter wide.

The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force [latex]\mathbf{\overset{\to }{A}}[/latex] from hinge [latex]A,[/latex] an unknown force [latex]\mathbf{\overset{\to }{B}}[/latex] from hinge [latex]B,[/latex] and the known weight [latex]\mathbf{\overset{\to }{w}}[/latex] attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the y -axis along the direction of gravity and the x -axis in the plane of the slab, as shown in panel (a) of Figure , and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force [latex]\mathbf{\overset{\to }{A}}[/latex] [latex]({A}_{x}[/latex] and [latex]{A}_{y}),[/latex] and two components of force [latex]\mathbf{\overset{\to }{B}}[/latex] [latex]({B}_{x}[/latex] and [latex]{B}_{y}).[/latex] In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns [latex]({A}_{x},[/latex] [latex]{B}_{x},[/latex] [latex]{A}_{y},[/latex] and [latex]{B}_{y}),[/latex] we must set up four independent equations. One equation is the equilibrium condition for forces in the x -direction. The second equation is the equilibrium condition for forces in the y -direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation, [latex]{A}_{y}={B}_{y}.[/latex] To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of Figure . Finally, we solve the equations for the unknown force components and find the forces.

Figure A is a geometrical representation for a swinging vertical door supported by two hinges attached at points A and B. Forces A and B are applied at the points A and B. Projections of these forces to the x and y axes are shown. Force w is applied at the point CM. Point CM is lower than point A by half-a and to the right of point A by half-b. Line from point A to CM forms an angle beta with the edge of the wall. Figure B is a free-body diagram for a swinging vertical door is supported by two hinges attached at points A and B. Force Ay forms an angle beta with the line connecting points P and CM. Force By forms an angle beta with the line connecting points B and CM. Force W forms an angle beta with the line that is the continuation of the line connecting points P and CM. Distance between points P and CM is d.

From the free-body diagram for the door we have the first equilibrium condition for forces:

We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P :

We use the free-body diagram to find all the terms in this equation:

In evaluating [latex]\text{sin}\,\beta ,[/latex] we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Figure and compute [latex]{B}_{x}:[/latex]

Therefore the magnitudes of the horizontal component forces are [latex]{A}_{x}={B}_{x}=100.0\,\text{N}.[/latex] The forces on the door are

The forces on the hinges are found from Newton’s third law as

Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.

Solve the problem in Figure by taking the pivot position at the center of mass.

[latex]{\mathbf{\overset{\to }{F}}}_{\text{door on}\,A}=100.0\,\text{N}\mathbf{\hat{i}}-200.0\,\text{N}\mathbf{\hat{j}}\,\text{;}\,{\mathbf{\overset{\to }{F}}}_{\text{door on}\,B}=-100.0\,\text{N}\mathbf{\hat{i}}-200.0\,\text{N}\mathbf{\hat{j}}[/latex]

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.

Figure is a schematic drawing of a woman standing 1.5 m away from one end and 4.5 m away from another end of a scaffold.

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut is 4.0 m long and is supported by a 5.0 m long cable tied to the wall at a point 3.0 m above the left end of the strut.

  • A variety of engineering problems can be solved by applying equilibrium conditions for rigid bodies.
  • In applications, identify all forces that act on a rigid body and note their lever arms in rotation about a chosen rotation axis. Construct a free-body diagram for the body. Net external forces and torques can be clearly identified from a correctly constructed free-body diagram. In this way, you can set up the first equilibrium condition for forces and the second equilibrium condition for torques.
  • In setting up equilibrium conditions, we are free to adopt any inertial frame of reference and any position of the pivot point. All choices lead to one answer. However, some choices can make the process of finding the solution unduly complicated. We reach the same answer no matter what choices we make. The only way to master this skill is to practice.

Conceptual Questions

Is it possible to rest a ladder against a rough wall when the floor is frictionless?

Show how a spring scale and a simple fulcrum can be used to weigh an object whose weight is larger than the maximum reading on the scale.

A painter climbs a ladder. Is the ladder more likely to slip when the painter is near the bottom or near the top?

A uniform plank rests on a level surface as shown below. The plank has a mass of 30 kg and is 6.0 m long. How much mass can be placed at its right end before it tips? ( Hint: When the board is about to tip over, it makes contact with the surface only along the edge that becomes a momentary axis of rotation.)

Figure schematic drawing of uniform plank rests on a level surface. Part of the plank that is 4.2 meters long is supported by the plank. Part of the plank that is 1.8 meters long is hanging over it.

The uniform seesaw shown below is balanced on a fulcrum located 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has a mass 80 kg. What is the mass of the board?

Figure is a schematic drawing of two boys on the seesaw. One boy sits on the edge of the seesaw three meters from the center. Another boys sits at the opposite edge of the seesaw, five meters from the center.

In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the other end to a tree 15 m away, as shown below. He then pulls on the center of the rope with a force of 400 N, which causes its center to be displaced 0.30 m, as shown. What is the force of the rope on the car?

Figure is a schematic drawing that shows a rope tied to the front bumper and the other end to a tree 15 m away. A force of 400 N is applied to the center of the rope and causes it to get displaced 0.30 m.

A uniform 40.0-kg scaffold of length 6.0 m is supported by two light cables, as shown below. An 80.0-kg painter stands 1.0 m from the left end of the scaffold, and his painting equipment is 1.5 m from the right end. If the tension in the left cable is twice that in the right cable, find the tensions in the cables and the mass of the equipment.

Figure is a schematic drawing of a man standing at the left side and the bucket placed at the right side of a scaffold.

When the structure shown below is supported at point P , it is in equilibrium. Find the magnitude of force F and the force applied at P . The weight of the structure is negligible.

Figure shows the distribution of forces applied to point P. Force of 2000 N, two meters to the left of the point P, moves it downwards. Force F, two meters to the left and two meters above of the point P, moves it to the right. Force of 1000 N, two meters to the right and three meters below of the point P, moves it to the left.

To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2.00 m from the bottom. The person is standing 3.00 m from the bottom. Find the normal reaction and friction forces on the ladder at its base.

784 N, 376 N

A uniform horizontal strut weighs 400.0 N. One end of the strut is attached to a hinged support at the wall, and the other end of the strut is attached to a sign that weighs 200.0 N. The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the strut, find the tension in the cable and the force at the hinge of the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut forms a 30 degree angle with the cable tied to the wall above the left end of the strut.

The forearm shown below is positioned at an angle [latex]\theta[/latex] with respect to the upper arm, and a 5.0-kg mass is held in the hand. The total mass of the forearm and hand is 3.0 kg, and their center of mass is 15.0 cm from the elbow. (a) What is the magnitude of the force that the biceps muscle exerts on the forearm for [latex]\theta =60^\circ\text{?}[/latex] (b) What is the magnitude of the force on the elbow joint for the same angle? (c) How do these forces depend on the angle [latex]\theta ?[/latex]

Figure is a schematic drawing of a forearm rotated around the elbow. A 5 kilogram ball is held in the palm. The distance between the elbow and the ball is 35 centimeters. The distance between the elbow and the biceps muscle, which causes a torque around the elbow, is 4 centimeters. Forearm forms a theta angle with the upper arm.

The uniform boom shown below weighs 3000 N. It is supported by the horizontal guy wire and by the hinged support at point A . What are the forces on the boom due to the wire and due to the support at A ? Does the force at A act along the boom?

Figure is a schematic drawing of a 2000 N weight that is supported by the horizontal guy wire and by the hinged support at point A. Hinged support forms a 45 degree angle with the ground.

The uniform boom shown below weighs 700 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?

Figure is a schematic drawing of a 400 N weight that is by a cable and by a hinge at the wall. Hinge forms a 20 degree angle with the line perpendicular to the wall. Cable forms a 45 degree angle with the line perpendicular to the wall.

A 12.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 1000 kg. For the position shown, calculate tension T in the cable and the force at the axle A .

Figure is a schematic drawing of a crane lifting a 3000-kg load. Arm of a crane forms a 30 degree angle with the line parallel to the ground. Cable supporting load forms a 10 degree angle with the arm.

A uniform trapdoor shown below is 1.0 m by 1.5 m and weighs 300 N. It is supported by a single hinge (H), and by a light rope tied between the middle of the door and the floor. The door is held at the position shown, where its slab makes a [latex]30^\circ[/latex] angle with the horizontal floor and the rope makes a [latex]20^\circ[/latex] angle with the floor. Find the tension in the rope and the force at the hinge.

Figure is a schematic drawing of a trapdoor that is 1.0 m by 1.5 m. Door is supported by a single hinge labeled H, and by a light rope tied between the middle of the door and the floor. The door makes a 30 degree angle with the floor and the rope makes a 20 degree angle with the floor.

A 90-kg man walks on a sawhorse, as shown below. The sawhorse is 2.0 m long and 1.0 m high, and its mass is 25.0 kg. Calculate the normal reaction force on each leg at the contact point with the floor when the man is 0.5 m from the far end of the sawhorse. ( Hint: At each end, find the total reaction force first. This reaction force is the vector sum of two reaction forces, each acting along one leg. The normal reaction force at the contact point with the floor is the normal (with respect to the floor) component of this force.)

Figure is a schematic drawing of a man walks on a sawhorse. Each side of the sawhorse is supported by two connected legs. There are 60 degree angles between the legs.

12.2 Examples of Static Equilibrium Copyright © 2016 by OpenStax. All Rights Reserved.

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how to solve static equilibrium problems

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how to solve static equilibrium problems

If an object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to describe equilibrium situations. Thus, the net force is zero and the acceleration is 0 m/s/s. Objects at equilibrium must have an acceleration of 0 m/s/s. This extends from Newton's first law of motion . But having an acceleration of 0 m/s/s does not mean the object is at rest. An object at equilibrium is either ...

  • at rest and staying at rest, or
  • in motion and continuing in motion with the same speed and direction.

This too extends from Newton's first law of motion .

Analyzing a Static Equilibrium Situation

If an object is at rest and is in a state of equilibrium, then we would say that the object is at "static equilibrium." "Static" means stationary or at rest . A common physics lab is to hang an object by two or more strings and to measure the forces that are exerted at angles upon the object to support its weight. The state of the object is analyzed in terms of the forces acting upon the object. The object is a point on a string upon which three forces were acting. See diagram at right. If the object is at equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if all the forces are added together as vectors, then the resultant force (the vector sum) should be 0 Newton. (Recall that the net force is "the vector sum of all the forces" or the resultant of adding all the individual forces head-to-tail.) Thus, an accurately drawn vector addition diagram can be constructed to determine the resultant. Sample data for such a lab are shown below.

For most students, the resultant was 0 Newton (or at least very close to 0 N). This is what we expected - since the object was at equilibrium , the net force (vector sum of all the forces) should be 0 N.

Another way of determining the net force (vector sum of all the forces) involves using the trigonometric functions to resolve each force into its horizontal and vertical components. Once the components are known, they can be compared to see if the vertical forces are balanced and if the horizontal forces are balanced. The diagram below shows vectors A, B, and C and their respective components. For vectors A and B, the vertical components can be determined using the sine of the angle and the horizontal components can be analyzed using the cosine of the angle. The magnitude and direction of each component for the sample data are shown in the table below the diagram.

The data in the table above show that the forces nearly balance. An analysis of the horizontal components shows that the leftward component of A nearly balances the rightward component of B. An analysis of the vertical components show that the sum of the upward components of A + B nearly balance the downward component of C. The vector sum of all the forces is ( nearly ) equal to 0 Newton. But what about the 0.1 N difference between rightward and leftward forces and the 0.2 N difference between the upward and downward forces? Why do the components of force only nearly balance? The sample data used in this analysis are the result of measured data from an actual experimental setup. The difference between the actual results and the expected results is due to the error incurred when measuring force A and force B. We would have to conclude that this low margin of experimental error reflects an experiment with excellent results. We could say it's "close enough for government work."

Analyzing a Hanging Sign

The above analysis of the forces acting upon an object in equilibrium is commonly used to analyze situations involving objects at static equilibrium . The most common application involves the analysis of the forces acting upon a sign that is at rest. For example, consider the picture at the right that hangs on a wall. The picture is in a state of equilibrium, and thus all the forces acting upon the picture must be balanced. That is, all horizontal components must add to 0 Newton and all vertical components must add to 0 Newton. The leftward pull of cable A must balance the rightward pull of cable B and the sum of the upward pull of cable A and cable B must balance the weight of the sign.

Suppose the tension in both of the cables is measured to be 50 N and that the angle that each cable makes with the horizontal is known to be 30 degrees. What is the weight of the sign? This question can be answered by conducting a force analysis using trigonometric functions . The weight of the sign is equal to the sum of the upward components of the tension in the two cables. Thus, a trigonometric function can be used to determine this vertical component. A diagram and accompanying work is shown below.

Since each cable pulls upwards with a force of 25 N, the total upward pull of the sign is 50 N. Therefore, the force of gravity (also known as weight) is 50 N, down. The sign weighs 50 N.

In the above problem, the tension in the cable and the angle that the cable makes with the horizontal are used to determine the weight of the sign. The idea is that the tension, the angle, and the weight are related. If the any two of these three are known, then the third quantity can be determined using trigonometric functions.

Thinking Conceptually

There is an important principle that emanates from some of the trigonometric calculations performed above. The principle is that as the angle with the horizontal increases, the amount of tensional force required to hold the sign at equilibrium decreases. To illustrate this, consider a 10-Newton picture held by three different wire orientations as shown in the diagrams below. In each case, two wires are used to support the picture; each wire must support one-half of the sign's weight (5 N). The angle that the wires make with the horizontal is varied from 60 degrees to 15 degrees. Use this information and the diagram below to determine the tension in the wire for each orientation. When finished, click the button to view the answers.

At 60 degrees, the tension is 5.8 N. (5 N / sin 60 degrees).

At 45 degrees, the tension is 7.1 N. (5 N / sin 45 degrees).

At 15 degrees, the tension is 19.3 N (5 N / sin 15 degrees).

In conclusion, equilibrium is the state of an object in which all the forces acting upon it are balanced. In such cases, the net force is 0 Newton. Knowing the forces acting upon an object, trigonometric functions can be utilized to determine the horizontal and vertical components of each force. If at equilibrium, then all the vertical components must balance and all the horizontal components must balance.

   

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how to solve static equilibrium problems

Check Your Understanding

The following questions are meant to test your understanding of equilibrium situations. Click the button to view the answers to these questions.

1. The following picture is hanging on a wall. Use trigonometric functions to determine the weight of the picture.

The weight of the sign is 42.4 N .

The tension is 30.0 N and the angle is 45 degrees. Thus,

sine (45 degrees) = (F vert ) / (30.0 N).

The proper use of algebra leads to the equation:

F vert = (30.0 N) • sine (45 degrees) = 21.2 N

Each cable pulls upward with 21.2 N of force. Thus, the sign must weigh twice this - 42.4 N.

2. The sign below hangs outside the physics classroom, advertising the most important truth to be found inside. The sign is supported by a diagonal cable and a rigid horizontal bar. If the sign has a mass of 50 kg, then determine the tension in the diagonal cable that supports its weight.

The tension is 980 Newtons .

Since the mass is 50 kg, the weight is 490 N. Since there is only one "upward-pulling" cable, it must supply all the upward force. This cable pulls upwards with approximately 490 N of force. Thus,

sine (30 degrees) = (490 N ) / (F tens ).

Proper use of algebra leads to the equation

F tens = (490 N) / [ sine 30 (degrees) ] = 980 N.

3. The following sign can be found in Glenview. The sign has a mass of 50 kg. Determine the tension in the cables.

The tension is 346 Newtons .

Since the mass is 50.0 kg, the weight is 490 N. Each cable must pull upwards with 245 N of force.

Thus, sine (45 degrees) = (245 N ) / (F tens ).

F tens = (245 N) / [sine (45 degrees)] = 346 N.

4. After its most recent delivery, the infamous stork announces the good news. If the sign has a mass of 10 kg, then what is the tensional force in each cable? Use trigonometric functions and a sketch to assist in the solution.

The tension 56.6 Newtons .

Since the mass is 10.0 kg, the weight is 98.0 N. Each cable must pull upwards with 49.0 N of force. Thus,

sine 60 (degrees) = (49.0 N) / (F tens ).

F tens = (49.0 N) / [ sine 60 (degrees) ] = 56.6 N.

5. Suppose that a student pulls with two large forces (F 1 and F 2 ) in order to lift a 1-kg book by two cables. If the cables make a 1-degree angle with the horizontal, then what is the tension in the cable?

The tension 281 Newtons!

Since the mass is 1 kg, the weight is 9.8 N. Each cable must pull upwards with 4.9 N of force. Thus,

sine (1 degree) = (4.9 N) / (F tens ).

F tens = (4.9 N) / [ sine (1 degree) ] = 281 N.

  • 12.1 Conditions for Static Equilibrium
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Identify the physical conditions of static equilibrium.
  • Draw a free-body diagram for a rigid body acted on by forces.
  • Explain how the conditions for equilibrium allow us to solve statics problems.

We say that a rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. We say that a rigid body is in static equilibrium when it is at rest in our selected frame of reference . Notice that the distinction between the state of rest and a state of uniform motion is artificial—that is, an object may be at rest in our selected frame of reference, yet to an observer moving at constant velocity relative to our frame, the same object appears to be in uniform motion with constant velocity. Because the motion is relative , what is in static equilibrium to us is in dynamic equilibrium to the moving observer, and vice versa. Since the laws of physics are identical for all inertial reference frames, in an inertial frame of reference, there is no distinction between static equilibrium and equilibrium.

According to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or

Here, the sum is of all external forces acting on the body, where m is its mass and a → CM a → CM is the linear acceleration of its center of mass (a concept we discussed in Linear Momentum and Collisions on linear momentum and collisions). In equilibrium, the linear acceleration is zero. If we set the acceleration to zero in Equation 12.1 , we obtain the following equation:

First Equilibrium Condition

The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium:

The first equilibrium condition, Equation 12.2 , is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws.

This vector equation is equivalent to the following three scalar equations for the components of the net force:

Analogously to Equation 12.1 , we can state that the rotational acceleration α → α → of a rigid body about a fixed axis of rotation is caused by the net torque acting on the body, or

Here I I is the rotational inertia of the body in rotation about this axis and the summation is over all torques τ → k τ → k of external forces in Equation 12.2 . In equilibrium, the rotational acceleration is zero. By setting to zero the right-hand side of Equation 12.4 , we obtain the second equilibrium condition:

Second Equilibrium Condition

The second equilibrium condition for the static equilibrium of a rigid body expresses rotational equilibrium:

The second equilibrium condition, Equation 12.5 , is the equilibrium condition for torques that we encountered when we studied rotational dynamics. It is worth noting that this equation for equilibrium is generally valid for rotational equilibrium about any axis of rotation (fixed or otherwise). Again, this vector equation is equivalent to three scalar equations for the vector components of the net torque:

The second equilibrium condition means that in equilibrium, there is no net external torque to cause rotation about any axis.

The first and second equilibrium conditions are stated in a particular reference frame. The first condition involves only forces and is therefore independent of the origin of the reference frame. However, the second condition involves torque, which is defined as a cross product, τ → k = r → k × F → k , τ → k = r → k × F → k , where the position vector r → k r → k with respect to the axis of rotation of the point where the force is applied enters the equation. Therefore, torque depends on the location of the axis in the reference frame. However, when rotational and translational equilibrium conditions hold simultaneously in one frame of reference, then they also hold in any other inertial frame of reference, so that the net torque about any axis of rotation is still zero. The explanation for this is fairly straightforward.

Suppose vector R → R → is the position of the origin of a new inertial frame of reference S ′ S ′ in the old inertial frame of reference S . From our study of relative motion, we know that in the new frame of reference S ′ , S ′ , the position vector r → ′ k r → ′ k of the point where the force F → k F → k is applied is related to r → k r → k via the equation

Now, we can sum all torques τ → ′ k = r → ′ k × F → k τ → ′ k = r → ′ k × F → k of all external forces in a new reference frame, S ′ : S ′ :

In the final step in this chain of reasoning, we used the fact that in equilibrium in the old frame of reference, S , the first term vanishes because of Equation 12.5 and the second term vanishes because of Equation 12.2 . Hence, we see that the net torque in any inertial frame of reference S ′ S ′ is zero, provided that both conditions for equilibrium hold in an inertial frame of reference S .

The practical implication of this is that when applying equilibrium conditions for a rigid body, we are free to choose any point as the origin of the reference frame. Our choice of reference frame is dictated by the physical specifics of the problem we are solving. In one frame of reference, the mathematical form of the equilibrium conditions may be quite complicated, whereas in another frame, the same conditions may have a simpler mathematical form that is easy to solve. The origin of a selected frame of reference is called the pivot point .

In the most general case, equilibrium conditions are expressed by the six scalar equations ( Equation 12.3 and Equation 12.6 ). For planar equilibrium problems with rotation about a fixed axis, which we consider in this chapter, we can reduce the number of equations to three. The standard procedure is to adopt a frame of reference where the z -axis is the axis of rotation. With this choice of axis, the net torque has only a z -component, all forces that have non-zero torques lie in the xy -plane, and therefore contributions to the net torque come from only the x - and y -components of external forces. Thus, for planar problems with the axis of rotation perpendicular to the xy -plane, we have the following three equilibrium conditions for forces and torques:

where the summation is over all N external forces acting on the body and over their torques. In Equation 12.9 , we simplified the notation by dropping the subscript z , but we understand here that the summation is over all contributions along the z -axis, which is the axis of rotation. In Equation 12.9 , the z -component of torque τ → k τ → k from the force F → k F → k is

where r k r k is the length of the lever arm of the force and F k F k is the magnitude of the force (as you saw in Fixed-Axis Rotation ). The angle θ θ is the angle between vectors r → k r → k and F → k , F → k , measuring from vector r → k r → k to vector F → k F → k in the counterclockwise direction ( Figure 12.2 ). When using Equation 12.10 , we often compute the magnitude of torque and assign its sense as either positive ( + ) ( + ) or negative ( − ) , ( − ) , depending on the direction of rotation caused by this torque alone. In Equation 12.9 , net torque is the sum of terms, with each term computed from Equation 12.10 , and each term must have the correct sense . Similarly, in Equation 12.7 , we assign the + + sign to force components in the + + x -direction and the − − sign to components in the − − x -direction. The same rule must be consistently followed in Equation 12.8 , when computing force components along the y -axis.

Interactive

View this demonstration to see two forces act on a rigid square in two dimensions. At all times, the static equilibrium conditions given by Equation 12.7 through Equation 12.9 are satisfied. You can vary magnitudes of the forces and their lever arms and observe the effect these changes have on the square.

In many equilibrium situations, one of the forces acting on the body is its weight. In free-body diagrams, the weight vector is attached to the center of gravity of the body. For all practical purposes, the center of gravity is identical to the center of mass, as you learned in Linear Momentum and Collisions on linear momentum and collisions. Only in situations where a body has a large spatial extension so that the gravitational field is nonuniform throughout its volume, are the center of gravity and the center of mass located at different points. In practical situations, however, even objects as large as buildings or cruise ships are located in a uniform gravitational field on Earth’s surface, where the acceleration due to gravity has a constant magnitude of g = 9.8 m/s 2 . g = 9.8 m/s 2 . In these situations, the center of gravity is identical to the center of mass. Therefore, throughout this chapter, we use the center of mass (CM) as the point where the weight vector is attached. Recall that the CM has a special physical meaning: When an external force is applied to a body at exactly its CM, the body as a whole undergoes translational motion and such a force does not cause rotation.

When the CM is located off the axis of rotation, a net gravitational torque occurs on an object. Gravitational torque is the torque caused by weight. This gravitational torque may rotate the object if there is no support present to balance it. The magnitude of the gravitational torque depends on how far away from the pivot the CM is located. For example, in the case of a tipping truck ( Figure 12.3 ), the pivot is located on the line where the tires make contact with the road’s surface. If the CM is located high above the road’s surface, the gravitational torque may be large enough to turn the truck over. Passenger cars with a low-lying CM, close to the pavement, are more resistant to tipping over than are trucks.

If you tilt a box so that one edge remains in contact with the table beneath it, then one edge of the base of support becomes a pivot. As long as the center of gravity of the box remains over the base of support, gravitational torque rotates the box back toward its original position of stable equilibrium. When the center of gravity moves outside of the base of support, gravitational torque rotates the box in the opposite direction, and the box rolls over. View this demonstration to experiment with stable and unstable positions of a box.

Example 12.1

Center of gravity of a car.

We are almost ready to write down equilibrium conditions Equation 12.7 through Equation 12.9 for the car, but first we must decide on the reference frame. Suppose we choose the x -axis along the length of the car, the y -axis vertical, and the z -axis perpendicular to this xy -plane. With this choice we only need to write Equation 12.7 and Equation 12.9 because all the y -components are identically zero. Now we need to decide on the location of the pivot point. We can choose any point as the location of the axis of rotation ( z -axis). Suppose we place the axis of rotation at CM, as indicated in the free-body diagram for the car. At this point, we are ready to write the equilibrium conditions for the car.

This condition is trivially satisfied because when we substitute the data, Equation 12.11 becomes + 0.52 w − w + 0.48 w = 0 . + 0.52 w − w + 0.48 w = 0 . The second equilibrium condition, Equation 12.9 , reads

where τ F τ F is the torque of force F F , τ w F F , τ w is the gravitational torque of force w , and τ R τ R is the torque of force F R . F R . When the pivot is located at CM, the gravitational torque is identically zero because the lever arm of the weight with respect to an axis that passes through CM is zero. The lines of action of both normal reaction forces are perpendicular to their lever arms, so in Equation 12.10 , we have | sin θ | = 1 | sin θ | = 1 for both forces. From the free-body diagram, we read that torque τ F τ F causes clockwise rotation about the pivot at CM, so its sense is negative; and torque τ R τ R causes counterclockwise rotation about the pivot at CM, so its sense is positive. With this information, we write the second equilibrium condition as

With the help of the free-body diagram, we identify the force magnitudes F R = 0.48 w F R = 0.48 w and F F = 0.52 w , F F = 0.52 w , and their corresponding lever arms r R = x r R = x and r F = d − x . r F = d − x . We can now write the second equilibrium condition, Equation 12.13 , explicitly in terms of the unknown distance x :

Here the weight w cancels and we can solve the equation for the unknown position x of the CM. The answer is x = 0.52 d = 0.52 ( 2.5 m ) = 1.3 m . x = 0.52 d = 0.52 ( 2.5 m ) = 1.3 m .

When we substitute the quantities indicated in the diagram, we obtain

The answer obtained by solving Equation 12.13 is, again, x = 0.52 d = 1.3 m . x = 0.52 d = 1.3 m .

Significance

Check your understanding 12.1.

Solve Example 12.1 by choosing the pivot at the location of the rear axle.

Check Your Understanding 12.2

Explain which one of the following situations satisfies both equilibrium conditions: (a) a tennis ball that does not spin as it travels in the air; (b) a pelican that is gliding in the air at a constant velocity at one altitude; or (c) a crankshaft in the engine of a parked car.

A special case of static equilibrium occurs when all external forces on an object act at or along the axis of rotation or when the spatial extension of the object can be disregarded. In such a case, the object can be effectively treated like a point mass. In this special case, we need not worry about the second equilibrium condition, Equation 12.9 , because all torques are identically zero and the first equilibrium condition (for forces) is the only condition to be satisfied. The free-body diagram and problem-solving strategy for this special case were outlined in Newton’s Laws of Motion and Applications of Newton’s Laws . You will see a typical equilibrium situation involving only the first equilibrium condition in the next example.

View this demonstration to see three weights that are connected by strings over pulleys and tied together in a knot. You can experiment with the weights to see how they affect the equilibrium position of the knot and, at the same time, see the vector-diagram representation of the first equilibrium condition at work.

Example 12.2

A breaking tension.

From the free-body diagram, the magnitudes of components in these equations are

We substitute these components into the equilibrium conditions and simplify. We then obtain two equilibrium equations for the tensions:

The equilibrium equation for the x -direction tells us that the tension T 1 T 1 in the 5.0-cm string is twice the tension T 2 T 2 in the 10.0-cm string. Therefore, the shorter string will snap. When we use the first equation to eliminate T 2 T 2 from the second equation, we obtain the relation between the mass m m on the pan and the tension T 1 T 1 in the shorter string:

The string breaks when the tension reaches the critical value of T 1 = 2.80 N . T 1 = 2.80 N . The preceding equation can be solved for the critical mass m that breaks the string:

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  • Authors: William Moebs, Samuel J. Ling, Jeff Sanny
  • Publisher/website: OpenStax
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12 Static Equilibrium and Elasticity

12.1 conditions for static equilibrium, learning objectives.

By the end of this section, you will be able to:

  • Identify the physical conditions of static equilibrium.
  • Draw a free-body diagram for a rigid body acted on by forces.
  • Explain how the conditions for equilibrium allow us to solve statics problems.

We say that a rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. We say that a rigid body is in static equilibrium when it is at rest in our selected frame of reference . Notice that the distinction between the state of rest and a state of uniform motion is artificial—that is, an object may be at rest in our selected frame of reference, yet to an observer moving at constant velocity relative to our frame, the same object appears to be in uniform motion with constant velocity. Because the motion is relative , what is in static equilibrium to us is in dynamic equilibrium to the moving observer, and vice versa. Since the laws of physics are identical for all inertial reference frames, in an inertial frame of reference, there is no distinction between static equilibrium and equilibrium.

According to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or

Here, the sum is of all external forces acting on the body, where m is its mass and [latex] {\overset{\to }{a}}_{\text{CM}} [/latex] is the linear acceleration of its center of mass (a concept we discussed in Linear Momentum and Collisions on linear momentum and collisions). In equilibrium, the linear acceleration is zero. If we set the acceleration to zero in (Figure) , we obtain the following equation:

First Equilibrium Condition

The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium:

The first equilibrium condition, (Figure) , is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws.

This vector equation is equivalent to the following three scalar equations for the components of the net force:

Analogously to (Figure) , we can state that the rotational acceleration [latex] \overset{\to }{\alpha } [/latex] of a rigid body about a fixed axis of rotation is caused by the net torque acting on the body, or

Here [latex] I [/latex] is the rotational inertia of the body in rotation about this axis and the summation is over all torques [latex] {\overset{\to }{\tau }}_{k} [/latex] of external forces in (Figure) . In equilibrium, the rotational acceleration is zero. By setting to zero the right-hand side of (Figure) , we obtain the second equilibrium condition:

Second Equilibrium Condition

The second equilibrium condition for the static equilibrium of a rigid body expresses rotational equilibrium:

The second equilibrium condition, (Figure) , is the equilibrium condition for torques that we encountered when we studied rotational dynamics. It is worth noting that this equation for equilibrium is generally valid for rotational equilibrium about any axis of rotation (fixed or otherwise). Again, this vector equation is equivalent to three scalar equations for the vector components of the net torque:

The second equilibrium condition means that in equilibrium, there is no net external torque to cause rotation about any axis.

The first and second equilibrium conditions are stated in a particular reference frame. The first condition involves only forces and is therefore independent of the origin of the reference frame. However, the second condition involves torque, which is defined as a cross product, [latex] {\overset{\to }{\tau }}_{k}={\overset{\to }{r}}_{k}\,×\,{\overset{\to }{F}}_{k}, [/latex] where the position vector [latex] {\overset{\to }{r}}_{k} [/latex] with respect to the axis of rotation of the point where the force is applied enters the equation. Therefore, torque depends on the location of the axis in the reference frame. However, when rotational and translational equilibrium conditions hold simultaneously in one frame of reference, then they also hold in any other inertial frame of reference, so that the net torque about any axis of rotation is still zero. The explanation for this is fairly straightforward.

Suppose vector [latex] \overset{\to }{R} [/latex] is the position of the origin of a new inertial frame of reference [latex] S\prime [/latex] in the old inertial frame of reference S . From our study of relative motion, we know that in the new frame of reference [latex] S\prime , [/latex] the position vector [latex] {{\overset{\to }{r}}^{\prime }}_{k} [/latex] of the point where the force [latex] {\overset{\to }{F}}_{k} [/latex] is applied is related to [latex] {\overset{\to }{r}}_{k} [/latex] via the equation

Now, we can sum all torques [latex] {{\overset{\to }{\tau }}^{\prime }}_{k}={{\overset{\to }{r}}^{\prime }}_{k}\,×\,{\overset{\to }{F}}_{k} [/latex] of all external forces in a new reference frame, [latex] S\prime : [/latex]

In the final step in this chain of reasoning, we used the fact that in equilibrium in the old frame of reference, S , the first term vanishes because of (Figure) and the second term vanishes because of (Figure) . Hence, we see that the net torque in any inertial frame of reference [latex] S\prime [/latex] is zero, provided that both conditions for equilibrium hold in an inertial frame of reference S .

The practical implication of this is that when applying equilibrium conditions for a rigid body, we are free to choose any point as the origin of the reference frame. Our choice of reference frame is dictated by the physical specifics of the problem we are solving. In one frame of reference, the mathematical form of the equilibrium conditions may be quite complicated, whereas in another frame, the same conditions may have a simpler mathematical form that is easy to solve. The origin of a selected frame of reference is called the pivot point .

In the most general case, equilibrium conditions are expressed by the six scalar equations ( (Figure) and (Figure) ). For planar equilibrium problems with rotation about a fixed axis, which we consider in this chapter, we can reduce the number of equations to three. The standard procedure is to adopt a frame of reference where the z -axis is the axis of rotation. With this choice of axis, the net torque has only a z -component, all forces that have non-zero torques lie in the xy -plane, and therefore contributions to the net torque come from only the x – and y -components of external forces. Thus, for planar problems with the axis of rotation perpendicular to the xy -plane, we have the following three equilibrium conditions for forces and torques:

where the summation is over all N external forces acting on the body and over their torques. In (Figure) , we simplified the notation by dropping the subscript z , but we understand here that the summation is over all contributions along the z -axis, which is the axis of rotation. In (Figure) , the z -component of torque [latex] {\overset{\to }{\tau }}_{k} [/latex] from the force [latex] {\overset{\to }{F}}_{k} [/latex] is

where [latex] {r}_{k} [/latex] is the length of the lever arm of the force and [latex] {F}_{k} [/latex] is the magnitude of the force (as you saw in Fixed-Axis Rotation ). The angle [latex] \theta [/latex] is the angle between vectors [latex] {\overset{\to }{r}}_{k} [/latex] and [latex] {\overset{\to }{F}}_{k}, [/latex] measuring from vector [latex] {\overset{\to }{r}}_{k} [/latex] to vector [latex] {\overset{\to }{F}}_{k} [/latex] in the counterclockwise direction ( (Figure) ). When using (Figure) , we often compute the magnitude of torque and assign its sense as either positive [latex] (+) [/latex] or negative [latex] (-), [/latex] depending on the direction of rotation caused by this torque alone. In (Figure) , net torque is the sum of terms, with each term computed from (Figure) , and each term must have the correct sense . Similarly, in (Figure) , we assign the [latex] + [/latex] sign to force components in the [latex] + [/latex] x -direction and the [latex] - [/latex] sign to components in the [latex] - [/latex] x -direction. The same rule must be consistently followed in (Figure) , when computing force components along the y -axis.

Figure A is the schematics of the torque of a force that causes counterclockwise rotation around the axis or rotation. Vector tau is aligned with Z axis and has a positive value. Angle theta that is formed by vectors F and r is bigger than zero. Figure B is the schematics of the torque of a force that causes clockwise rotation around the axis or rotation. Vector tau is aligned with Z axis and has a negative value. Angle theta that is formed by vectors F and r is smaller than zero.

Figure 12.2 Torque of a force: (a) When the torque of a force causes counterclockwise rotation about the axis of rotation, we say that its sense is positive, which means the torque vector is parallel to the axis of rotation. (b) When torque of a force causes clockwise rotation about the axis, we say that its sense is negative, which means the torque vector is antiparallel to the axis of rotation.

View this demonstration to see two forces act on a rigid square in two dimensions. At all times, the static equilibrium conditions given by (Figure) through (Figure) are satisfied. You can vary magnitudes of the forces and their lever arms and observe the effect these changes have on the square.

In many equilibrium situations, one of the forces acting on the body is its weight. In free-body diagrams, the weight vector is attached to the center of gravity of the body. For all practical purposes, the center of gravity is identical to the center of mass, as you learned in Linear Momentum and Collisions on linear momentum and collisions. Only in situations where a body has a large spatial extension so that the gravitational field is nonuniform throughout its volume, are the center of gravity and the center of mass located at different points. In practical situations, however, even objects as large as buildings or cruise ships are located in a uniform gravitational field on Earth’s surface, where the acceleration due to gravity has a constant magnitude of [latex] g=9.8\,{\text{m/s}}^{2}. [/latex] In these situations, the center of gravity is identical to the center of mass. Therefore, throughout this chapter, we use the center of mass (CM) as the point where the weight vector is attached. Recall that the CM has a special physical meaning: When an external force is applied to a body at exactly its CM, the body as a whole undergoes translational motion and such a force does not cause rotation.

When the CM is located off the axis of rotation, a net gravitational torque occurs on an object. Gravitational torque is the torque caused by weight. This gravitational torque may rotate the object if there is no support present to balance it. The magnitude of the gravitational torque depends on how far away from the pivot the CM is located. For example, in the case of a tipping truck ( (Figure) ), the pivot is located on the line where the tires make contact with the road’s surface. If the CM is located high above the road’s surface, the gravitational torque may be large enough to turn the truck over. Passenger cars with a low-lying CM, close to the pavement, are more resistant to tipping over than are trucks.

Figure A shows an evenly loaded truck with the center of gravity within the area of support. Figure B shows a truck with the center of gravity outside the area of support that is close to turning over. A car in equilibrium is shown next to it for the comparison. Figure C is the schematics that shows the position of the combined center of mass, a combination of load and truck centers of mass, between the two wheels that keep the vehicle stable. Figure D is the schematics that shows the position of the combined center of mass, a combination of load and truck centers of mass, outside the two wheels that make the vehicle unstable and can cause it to tip over.

Figure 12.3 The distribution of mass affects the position of the center of mass (CM), where the weight vector [latex] \overset{\to }{w} [/latex] is attached. If the center of gravity is within the area of support, the truck returns to its initial position after tipping [see the left panel in (b)]. But if the center of gravity lies outside the area of support, the truck turns over [see the right panel in (b)]. Both vehicles in (b) are out of equilibrium. Notice that the car in (a) is in equilibrium: The low location of its center of gravity makes it hard to tip over.

If you tilt a box so that one edge remains in contact with the table beneath it, then one edge of the base of support becomes a pivot. As long as the center of gravity of the box remains over the base of support, gravitational torque rotates the box back toward its original position of stable equilibrium. When the center of gravity moves outside of the base of support, gravitational torque rotates the box in the opposite direction, and the box rolls over. View this demonstration to experiment with stable and unstable positions of a box.

Center of Gravity of a Car

A passenger car with a 2.5-m wheelbase has 52% of its weight on the front wheels on level ground, as illustrated in (Figure) . Where is the CM of this car located with respect to the rear axle?

Picture shows a passenger car with a 2.5-m wheelbase that has 52% of its weight on the front wheels and 48% of its weight on the rear wheels on level ground. Distance between the rear axle and the center of mass is x.

Figure 12.4 The weight distribution between the axles of a car. Where is the center of gravity located?

We do not know the weight w of the car. All we know is that when the car rests on a level surface, 0.52 w pushes down on the surface at contact points of the front wheels and 0.48 w pushes down on the surface at contact points of the rear wheels. Also, the contact points are separated from each other by the distance [latex] d=2.5\,\text{m}. [/latex] At these contact points, the car experiences normal reaction forces with magnitudes [latex] {F}_{\text{F}}=0.52w [/latex] and [latex] {F}_{\text{R}}=0.48w [/latex] on the front and rear axles, respectively. We also know that the car is an example of a rigid body in equilibrium whose entire weight w acts at its CM. The CM is located somewhere between the points where the normal reaction forces act, somewhere at a distance x from the point where [latex] {F}_{R} [/latex] acts. Our task is to find x . Thus, we identify three forces acting on the body (the car), and we can draw a free-body diagram for the extended rigid body, as shown in (Figure) .

Figure are schematics that show the mass distribution for a passenger car with a wheelbase defined as d. Car has 52% of its weight on the front wheels (labeled as Ff) and 48% on the rear wheels (labeled Fr) and is on level ground. Distance between the rear axle and the center of mass (labeled rR) is x. Distance between the front axle and the center of mass (labeled rF) is d - x.

Figure 12.5 The free-body diagram for the car clearly indicates force vectors acting on the car and distances to the center of mass (CM). When CM is selected as the pivot point, these distances are lever arms of normal reaction forces. Notice that vector magnitudes and lever arms do not need to be drawn to scale, but all quantities of relevance must be clearly labeled.

We are almost ready to write down equilibrium conditions (Figure) through (Figure) for the car, but first we must decide on the reference frame. Suppose we choose the x -axis along the length of the car, the y -axis vertical, and the z -axis perpendicular to this xy -plane. With this choice we only need to write (Figure) and (Figure) because all the y -components are identically zero. Now we need to decide on the location of the pivot point. We can choose any point as the location of the axis of rotation ( z -axis). Suppose we place the axis of rotation at CM, as indicated in the free-body diagram for the car. At this point, we are ready to write the equilibrium conditions for the car.

Each equilibrium condition contains only three terms because there are [latex] N=3 [/latex] forces acting on the car. The first equilibrium condition, (Figure) , reads

This condition is trivially satisfied because when we substitute the data, (Figure) becomes [latex] +0.52w-w+0.48w=0. [/latex] The second equilibrium condition, (Figure) , reads

where [latex] {\tau }_{\text{F}} [/latex] is the torque of force [latex] {F}_{\text{F}},\,{\tau }_{w} [/latex] is the gravitational torque of force w , and [latex] {\tau }_{\text{R}} [/latex] is the torque of force [latex] {F}_{\text{R}}. [/latex] When the pivot is located at CM, the gravitational torque is identically zero because the lever arm of the weight with respect to an axis that passes through CM is zero. The lines of action of both normal reaction forces are perpendicular to their lever arms, so in (Figure) , we have [latex] |\,\text{sin}\,\theta |=1 [/latex] for both forces. From the free-body diagram, we read that torque [latex] {\tau }_{\text{F}} [/latex] causes clockwise rotation about the pivot at CM, so its sense is negative; and torque [latex] {\tau }_{\text{R}} [/latex] causes counterclockwise rotation about the pivot at CM, so its sense is positive. With this information, we write the second equilibrium condition as

With the help of the free-body diagram, we identify the force magnitudes [latex] {F}_{\text{R}}=0.48w [/latex] and [latex] {F}_{\text{F}}=0.52w, [/latex] and their corresponding lever arms [latex] {r}_{\text{R}}=x [/latex] and [latex] {r}_{\text{F}}=d-x. [/latex] We can now write the second equilibrium condition, (Figure) , explicitly in terms of the unknown distance x :

Here the weight w cancels and we can solve the equation for the unknown position x of the CM. The answer is [latex] x=0.52d=0.52(2.5\,\text{m})=1.3\,\text{m}\text{.} [/latex]

Choosing the pivot at the position of the front axle does not change the result. The free-body diagram for this pivot location is presented in (Figure) . For this choice of pivot point, the second equilibrium condition is

When we substitute the quantities indicated in the diagram, we obtain

The answer obtained by solving (Figure) is, again, [latex] x=0.52d=1.3\,\text{m}. [/latex]

Figure is the schematics that shows the mass distribution for a passenger car with a wheelbase defined as d. The car has 52% of its weight on its front wheels, now circled and labeled Pivot (Ff) and 48% of its weight on the rear wheels (Fr) on level ground. Distance between the rear axle and the center of mass is x. Distance between the front axle and the center of mass (rw) is d - x. The entire length of the whole axis is labeled with the equation rR=d.

Figure 12.6 The equivalent free-body diagram for the car; the pivot is clearly indicated.

Significance

This example shows that when solving static equilibrium problems, we are free to choose the pivot location. For different choices of the pivot point we have different sets of equilibrium conditions to solve. However, all choices lead to the same solution to the problem.

Check Your Understanding

Solve (Figure) by choosing the pivot at the location of the rear axle.

[latex] x=1.3\,\text{m} [/latex]

Explain which one of the following situations satisfies both equilibrium conditions: (a) a tennis ball that does not spin as it travels in the air; (b) a pelican that is gliding in the air at a constant velocity at one altitude; or (c) a crankshaft in the engine of a parked car.

A special case of static equilibrium occurs when all external forces on an object act at or along the axis of rotation or when the spatial extension of the object can be disregarded. In such a case, the object can be effectively treated like a point mass. In this special case, we need not worry about the second equilibrium condition, (Figure) , because all torques are identically zero and the first equilibrium condition (for forces) is the only condition to be satisfied. The free-body diagram and problem-solving strategy for this special case were outlined in Newton’s Laws of Motion and Applications of Newton’s Laws . You will see a typical equilibrium situation involving only the first equilibrium condition in the next example.

View this demonstration to see three weights that are connected by strings over pulleys and tied together in a knot. You can experiment with the weights to see how they affect the equilibrium position of the knot and, at the same time, see the vector-diagram representation of the first equilibrium condition at work.

A Breaking Tension

A small pan of mass 42.0 g is supported by two strings, as shown in (Figure) . The maximum tension that the string can support is 2.80 N. Mass is added gradually to the pan until one of the strings snaps. Which string is it? How much mass must be added for this to occur?

Figure shows small pan of mass supported by two strings intersecting at a 90 degree angle. The length of one string is 5 centimeters, the length of another string is 10 centimeters.

Figure 12.7 Mass is added gradually to the pan until one of the strings snaps.

This mechanical system consisting of strings, masses, and the pan is in static equilibrium. Specifically, the knot that ties the strings to the pan is in static equilibrium. The knot can be treated as a point; therefore, we need only the first equilibrium condition. The three forces pulling at the knot are the tension [latex] {\overset{\to }{T}}_{1} [/latex] in the 5.0-cm string, the tension [latex] {\overset{\to }{T}}_{2} [/latex] in the 10.0-cm string, and the weight [latex] \overset{\to }{w} [/latex] of the pan holding the masses. We adopt a rectangular coordinate system with the y -axis pointing opposite to the direction of gravity and draw the free-body diagram for the knot (see (Figure) ). To find the tension components, we must identify the direction angles [latex] {\alpha }_{1} [/latex] and [latex] {\alpha }_{2} [/latex] that the strings make with the horizontal direction that is the x -axis. As you can see in (Figure) , the strings make two sides of a right triangle. We can use the Pythagorean theorem to solve this triangle, shown in (Figure) , and find the sine and cosine of the angles [latex] {\alpha }_{1} [/latex] and [latex] {\alpha }_{2}. [/latex] Then we can resolve the tensions into their rectangular components, substitute in the first condition for equilibrium ( (Figure) and (Figure) ), and solve for the tensions in the strings. The string with a greater tension will break first.

Top figure shows the distribution of forces for the knot that ties the strings to the pan. T1 and T2 forces are pulling at the knot upward. Weight, a sum of M and m multiplied by g is pulling the knot downward. Projections of T1 and T2 at the x and y axes are shown. Bottom figure shows the representation of the knot that ties the strings to the pan as a right triangle. It has legs of the length a and 2a with a being equal 5 centimeters. Hypotenuse is a square root of five. Angle alpha 1 is formed by shorter leg and hypotenuse. Angle alpha 2 is formed by the longer leg and hypotenuse. Cosine of angle alpha 1 is equal to sine of angle alpha 2 and is equal to one divided by square root of five. Cosine of angle alpha 2 is equal to sine of angle alpha 1 and is equal to two divided by square root of five.

Figure 12.8 Free-body diagram for the knot in (Figure) .

The weight w pulling on the knot is due to the mass M of the pan and mass m added to the pan, or [latex] w=(M+m)g. [/latex] With the help of the free-body diagram in (Figure) , we can set up the equilibrium conditions for the knot:

From the free-body diagram, the magnitudes of components in these equations are

We substitute these components into the equilibrium conditions and simplify. We then obtain two equilibrium equations for the tensions:

The equilibrium equation for the x -direction tells us that the tension [latex] {T}_{1} [/latex] in the 5.0-cm string is twice the tension [latex] {T}_{2} [/latex] in the 10.0-cm string. Therefore, the shorter string will snap. When we use the first equation to eliminate [latex] {T}_{2} [/latex] from the second equation, we obtain the relation between the mass [latex] m [/latex] on the pan and the tension [latex] {T}_{1} [/latex] in the shorter string:

The string breaks when the tension reaches the critical value of [latex] {T}_{1}=2.80\,\text{N}. [/latex] The preceding equation can be solved for the critical mass m that breaks the string:

Suppose that the mechanical system considered in this example is attached to a ceiling inside an elevator going up. As long as the elevator moves up at a constant speed, the result stays the same because the weight [latex] w [/latex] does not change. If the elevator moves up with acceleration, the critical mass is smaller because the weight of [latex] M+m [/latex] becomes larger by an apparent weight due to the acceleration of the elevator. Still, in all cases the shorter string breaks first.

  • A body is in equilibrium when it remains either in uniform motion (both translational and rotational) or at rest. When a body in a selected inertial frame of reference neither rotates nor moves in translational motion, we say the body is in static equilibrium in this frame of reference.
  • Conditions for equilibrium require that the sum of all external forces acting on the body is zero (first condition of equilibrium), and the sum of all external torques from external forces is zero (second condition of equilibrium). These two conditions must be simultaneously satisfied in equilibrium. If one of them is not satisfied, the body is not in equilibrium.
  • The free-body diagram for a body is a useful tool that allows us to count correctly all contributions from all external forces and torques acting on the body. Free-body diagrams for the equilibrium of an extended rigid body must indicate a pivot point and lever arms of acting forces with respect to the pivot.

Conceptual Questions

What can you say about the velocity of a moving body that is in dynamic equilibrium?

Under what conditions can a rotating body be in equilibrium? Give an example.

What three factors affect the torque created by a force relative to a specific pivot point?

magnitude and direction of the force, and its lever arm

Mechanics sometimes put a length of pipe over the handle of a wrench when trying to remove a very tight bolt. How does this help?

For the next four problems, evaluate the statement as either true or false and explain your answer.

If there is only one external force (or torque) acting on an object, it cannot be in equilibrium.

True, as the sum of forces cannot be zero in this case unless the force itself is zero.

If an object is in equilibrium there must be an even number of forces acting on it.

If an odd number of forces act on an object, the object cannot be in equilibrium.

False, provided forces add to zero as vectors then equilibrium can be achieved.

A body moving in a circle with a constant speed is in rotational equilibrium.

What purpose is served by a long and flexible pole carried by wire-walkers?

It helps a wire-walker to maintain equilibrium.

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt. How much torque are you exerting relative to the center of the bolt?

When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850 m from the hinges. What torque are you exerting relative to the hinges?

[latex] 46.8\,\text{N}·\text{m} [/latex]

Find the magnitude of the tension in each supporting cable shown below. In each case, the weight of the suspended body is 100.0 N and the masses of the cables are negligible.

Figure A shows small pan of mass supported by string T3 that is tied to strings T1 and T2. Strings T1 and T2 are connected to two beams intersecting at a 90 degree angle. String T1 is perpendicular to the beam it is connected to. String T2 forms a 45 degree angle with the beam it is connected to. Figure B shows small pan of mass supported by string T2 that is tied to two identical strings T1. Strings T1 form 60 degree angles with the beam they are connected to. Figure C shows small pan of mass supported by string T3 that is tied to strings T1 and T2. String T1 and T2 form 60 and 45 degree angles, respectively, with the beam they are connected to. Figure D shows small pan of mass supported by string T4 that is tied to two strings T3 forming 6o degrees angle with the string T2. String T2 is connected to the beam by two strings T1. Strings T1 form 45 degree angles with the beam.

What force must be applied at point P to keep the structure shown in equilibrium? The weight of the structure is negligible.

Figure shows the distribution of forces applied to point P. Force of 2000 N, two meters to the left of the point P, moves it downwards. Force of 4000 N, two meters to the right and one meter above of the point P, moves it to the right.

Show Answer

Is it possible to apply a force at P to keep in equilibrium the structure shown? The weight of the structure is negligible.

Figure shows the distribution of forces applied to point P. Force of 2000 N, two meters to the left of the point P, moves it downwards. Force of 3000 N, two meters to the right of the point P, moves it upwards. Force of 5000 N, two meters to the right and one meter above of the point P, moves it to the right.

Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.

A small 1000-kg SUV has a wheel base of 3.0 m. If 60% if its weight rests on the front wheels, how far behind the front wheels is the wagon’s center of mass?

The uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of 40.0 kg. What is the mass of his friend?

Figure is a schematic drawing of two boys on the seesaw. One boy sits two meters from the edge of the seesaw and two meters from the center. Another boys sits at the opposite edge of the seesaw, four meters from the center.

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