Collection of Solved Problems in Physics

A leaning ladder, task number: 654, solution to hint 1.

The weight \(\vec{F}_g\) affects in the centre of mass downwards.

Forces \(\vec{F}_f\) and \(\vec{F}_w\) (by which the surface affects the ladder in the points where it stands or leans) affect perpendicular both the wall and the floor.

The friction forces \(\vec{F}_{ff}\) between the ladder and the floor or \(\vec{F}_{fw}\) between the ladder and the wall, which act against the direction of the slipping ladder, are tangetial to the surface.

Solution to Hint 2

Solution to hint 3.

The resultant force equals zero: \(\vec{F}_g + \vec{F}_f + \vec{F}_w + \vec{F}_{ff}+ \vec{F}_{fw} = \vec{0}\)

We choose the appropriate coordinate system according to the picture and then rewrite to the scalar form:

The resultant moment of the forces about the random point equals zero:

We will determine the moment of force (Torque ) for example about the centre of mass. It implies: \(\vec{M}_g=\vec{0}.\)

Let us remember the expression for the moment of force \(\vec{F}\) about the point O:

The following holds for the magnitude of the moment: \(M = rF\sin \varphi\) where \(\varphi\) is the angle which is formed by vectors \(\vec{r}\) and \(\vec{F}\).

The direction of the moment of force is perpendicular to the flat surface where the vectors \(\vec{r}\) and \(\vec{F}\) lie. The orientation is determined by the Right Hand Rule: we place the fingers of the right hand to direct from \(\vec{r}\) to \(\vec{F}\) the tight thumb points to the vector \(\vec{M}\).

We choose a clockwise direction for the negative direction of the rotation, we will mark the moment of the force directing backwards from the picture with the negative sign. Then

The following holds for the magnitude of the moment:

We divide L /2 and adjust:

Solution to Hint 4

In this case the static friction force reaches the maximum of the value.

Solution to Hint 5

Substitute into equation (3):

According to (4) and (5):

Substitute into equation (6):

The magnitude F f does not equal zero and therefore we can divide:

We add the members of functions sine and cosine:

We adjust the equation so that on one side there is an article with α and on the other side there is the rest. We divide both sides by the expression 2 μ sf cos α .

We get the resultant:

For the given numerical values we get:

The following holds for the found angle:

Numerically:

Complete solution

We draw a picture, mark affecting forces and choose the coordinate system.

The affecting forces keep the ladder in balance. It means that the resultant force affecting the ladder has to equal zero as well as the resultant moment of the forces has to equal zero, too (due to a random point).

For the limit angle α it holds that static friction force reaches the maximum value.

  • 12.2 Examples of 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
  • 12.1 Conditions for 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 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 12.7 to Equation 12.9 . 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 12.7 for force components in the x -direction. (b) Use the free-body diagram to write a correct equilibrium condition Equation 12.11 for force components in the y -direction. (c) Use the free-body diagram to write a correct equilibrium condition Equation 12.9 for torques along the axis of rotation. Use Equation 12.10 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.

w 1 = m 1 g w 1 = m 1 g is the weight of mass m 1 ; m 1 ; w 2 = m 2 g w 2 = m 2 g is the weight of mass m 2 ; m 2 ;

w = m g w = m g is the weight of the entire meter stick; w 3 = m 3 g w 3 = m 3 g is the weight of unknown mass m 3 ; m 3 ;

F S 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 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 12.10 . 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.

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 + y + y -direction to be parallel to F → S , F → S , 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 m 3 m 3 and F S . F S . In Equation 12.17 , we cancel the g factor and rearrange the terms to obtain

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

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

Significance

Check your understanding 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 12.7 and Equation 12.8 . 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.

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 θ = 0 sin θ = 0 in Equation 12.10 . For the y -components we have θ = ± 90 ° θ = ± 90 ° in Equation 12.10 . 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 T y and of w y . w y .

and the y -component of the net force satisfies

Equation 12.21 and Equation 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

Equation 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 . r T = 1.5 in . and r w = 13.0 in . 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:

We substitute these magnitudes into Equation 12.21 , Equation 12.22 , and Equation 12.23 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 Equation 12.21 for the x -component is equivalent to Equation 12.22 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 Equation 12.25 :

The force at the elbow is obtained by solving Equation 12.24 :

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

The second equilibrium condition, τ T + τ w = 0 , τ T + τ w = 0 , can be now written as

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

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

Check Your Understanding 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.

the net force in the y -direction is

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

where τ w τ w is the torque of the weight w and τ F τ 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 r F = L = 5.0 m and the lever arm of the weight is r w = L / 2 = 2.5 m . r w = L / 2 = 2.5 m . With the help of the free-body diagram, we identify the angles to be used in Equation 12.10 for torques: θ F = 180 ° − β θ F = 180 ° − β for the torque from the reaction force with the wall, and θ w = 180 ° + ( 90 ° − β ) θ w = 180 ° + ( 90 ° − β ) for the torque due to the weight. Now we are ready to use Equation 12.10 to compute torques:

We substitute the torques into Equation 12.30 and solve for F : F :

We obtain the normal reaction force with the floor by solving Equation 12.29 : N = w = 400.0 N . N = w = 400.0 N . The magnitude of friction is obtained by solving Equation 12.28 : f = F = 150.7 N . f = F = 150.7 N . The coefficient of static friction is μ s = f / N = 150.7 / 400.0 = 0.377 . μ s = f / N = 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:

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 Equation 12.10 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 12.10 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 12.10 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 12.10 expresses the rectangular component of this vector product along the axis of rotation.

Check Your Understanding 12.5

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

Example 12.6

Forces on door hinges.

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 sin β , sin β , we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Equation 12.32 and compute B x : B x :

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

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

Check Your Understanding 12.6

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

Check Your Understanding 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.

Check Your Understanding 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.

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Physics LibreTexts

9.3: Virtual Work

  • Last updated
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  • Page ID 6984

  • Jeremy Tatum
  • University of Victoria

We have seen that a mechanical system subject to conservative forces is in equilibrium when the derivatives of the potential energy with respect to the coordinates are zero. A method of solving such problems, therefore, is to write down an expression for the potential energy and put the derivatives equal to zero.

A very similar method is to use the principle of virtual work. In this method, we imagine that we act upon the system in such a manner as to increase one of the coordinates. We imagine, for example, what would happen if we were to stretch one of the springs, or to increase the angle between two jointed rods, or the angle that the ladder makes as it leans against the wall. We ask ourselves how much work we have to do on the system in order to increase this coordinate by a small amount. If the system starts from equilibrium, this work will be very small, and, in the limit of an infinitesimally small displacement, this “virtual work” will be zero. This method is very little different from setting the derivative of the potential energy to zero. I mention it here, however, because the concept might be useful in Chapter 13 in describing Hamilton’s variational principle.

Let’s start by doing a simple ladder problem by the method of virtual work. The usual uniform ladder of high school physics, of length \( 2l\) and weight \( mg\), is leaning in limiting static equilibrium against the usual smooth vertical wall and the rough horizontal floor whose coefficient of limiting static friction is \( \mu\). What is the angle \(\theta \) that the ladder makes with the vertical wall?

I have drawn the four forces on the ladder, namely: its weight \( mg\); the normal reaction of the floor on the ladder, which must also be \( mg\); the frictional force, which is \( \mu mg\); and the normal (and only) reaction of the wall on the ladder, which must also be \( \mu mg\).

There are several ways of doing this, which will be familiar to many readers. The only small reminder that I will give is to point out that, if you wish to combine the two forces at the foot of the ladder into a single force acting upwards and somewhat to the left, so that there are then just three forces acting on the ladder, the three forces must act through a single point, which will be above the middle of the ladder and to the right of the point of contact with the wall. But we are interested now in solving this problem by the principle of virtual work.

Before starting, I should warn that it is important in using the principle of virtual work to be meticulously careful about signs, and in that respect I remind readers that in the differential calculus the symbols \( \delta\) and \( d\) in front of a scalar quantity \( x\) do not mean “a small change in” or “an infinitesimal change” in \( x\). Such language is vague. The symbols stand for “a small increase in” and “an infinitesimal increase in”.

alt

Let us take note of the following distances:

\[ CD = l \cos \theta \nonumber \]

\[ BE = 2l \sin \theta. \nonumber \]

If we were to increase \( \theta\) by \( \delta\theta\), keeping the ladder in contact with wall and floor, the increases in these distances would be

\[ \delta(CD)=-l\sin\theta d\theta. \nonumber \]

\[ \delta(BE)=2l\cos\theta d\theta. \nonumber \]

Further, if were to increase \( \theta\) by \( \delta\theta\), the work done by the force at C would be \( mg\) times the decrease of the distance CD, and the work done by the frictional force at E would be minus \( \mu mg\) times the increase of the distance BE. The other two forces do no work. Thus the “virtual work” done by the external forces on the ladder is

\[ mg. l\sin\theta\delta\theta-\mu mg.2l \cos\theta\delta\theta. \nonumber \]

On putting the expression for the virtual work to zero, we obtain

\[ \tan\theta=2\mu. \nonumber \]

You should verify that this is the same answer as you get from other methods – the easiest of which is probably to take moments about E.

There is something about virtual work which reminds me of thermodynamics. The first law of thermodynamics, for example is \( \Delta U=\Delta q+\Delta w\), where \( \Delta U\) is the increase of the internal energy of the system, \( \Delta q\) is the heat added to the system, and \( \Delta w\) is the work done on the system. Prepositions play an important part in thermodynamics. It is always mandatory to state clearly and without ambiguity whether work is done by the piston on the gas, or by the gas on the system; or whether heat is gained by the system or lost from it. Without these prepositions, all discussion is meaningless. Likewise in solving a problem by the principle of virtual work, it is always essential to say whether you are describing the work done by a force on what part of the system (on the ladder or on the floor?) and whether you are describing an increase or a decrease of some length or angle.

Let us move now to a slightly more difficult problem, which we’ll try by three different methods – including that of virtual work.

In Figure IX.5, a uniform rod AB of weight \( Mg\) and length \( 2a\) is freely hinged at A. The end B carries a smooth ring of negligible mass. A light inextensible string of length \( l\) has one end attached to a fixed point C at the same level as A and distant \( 2a\) from it. It passes through the ring and carries at its other end a weight \( \frac{1}{10}Mg\) hanging freely. (The “smooth” ring means that the tension in the string is the same on both sides of the ring.) Find the angle CAB when the system is in equilibrium.

I have marked in various angles and lengths, which can easily be determined from the geometry of the system, and I have also marked the four forces on the rod.

alt

Let us first try a very conventional method. We know rather little about the force R of the hinge on the rod (though see below), and therefore this is a good reason for taking moments about the point A. We immediately obtain

\[ Mga\cos\theta+\frac{1}{10}Mg.2a\cos\theta=\frac{1}{10}Mg.2a\cos\frac{1}{2}\theta. \nonumber \]

Divide by \( Mga\) and set \( \cos\theta=2c^{2}-1\), where \( c=\cos\frac{1}{2}\theta\). After a little algebra, we obtain \( 12c^{2}-c-6=0\) and hence we find for the equilibrium condition that \( \theta\) = 82 o 49' or 263 o 37'. The latter, by the way, is a physically valid solution – you might want to sketch it.

Now let’s try the same problem using energy conditions. We’ll take the zero of potential energy when the rod is horizontal – at which time the small mass is at a distance l below the level AC.

When the angle CAB = \( \theta\), the distance of the centre of mass of the rod below AC is \( a\sin\theta\) and the distance of the small mass below AC is \( l-4a\sin\frac{1}{2}\theta+2a\sin\theta\) so that the potential energy is

\[ V=-Mga\sin\theta+\frac{1}{10}Mg[l-(l-4a\sin\frac{1}{2}\theta+2a\sin\theta)]=-\frac{2}{3}Mga(3\sin\theta-\sin\frac{1}{2}\theta) \nonumber \]

The derivative is

\[ \frac{dV}{d\theta}=-\frac{2}{3}Mga(3\cos\theta-\frac{1}{2}cos\frac{1}{2}\theta), \nonumber \]

and setting this to zero will produce the same results as before. Further differentiation (do it), or a graph of \( V\) : \( \theta\) (do it), will show that the 82 o 49' solution is stable and the 263 o 37' solution is unstable.

Now let’s try it by virtual work. We are going to increase \( \theta\) by \( \delta\theta\) and see how much work is done.

The distance of the centre of mass of the rod below AC is \( a\sin\theta\), and if \( \theta\) increases by \( \delta\theta\), this will increase by \( a\cos\theta \delta\theta\), and the work done by \( Mg\) will be \( Mga \cos \theta \delta\theta\).

The distance of the ring below AC is \( 2a\sin\theta\), and if \( \theta\) increases by \( \delta\theta\), this will increase by \( 2a\cos\theta\delta\theta\), and the work done by the downward force will be \( \frac{1}{10}Mg.2a\cos\theta \delta\theta\).

The distance BC is \( 4a\sin\frac{1}{2}\theta\), and if \( \theta\) increases by \( \delta\theta\), this will increase by \( 2a\cos\frac{1}{2}\theta\delta\theta\) and the work done by the sloping force will be MINUS \( \frac{1}{10}Mg.2a\cos\frac{1}{2}\theta\delta\theta\).

Thus the virtual work is

\[ Mg.a\cos\theta\delta\theta+\frac{1}{10}Mg.2a\cos\theta\delta\theta-\frac{1}{10}Mg.2a\cos\frac{1}{2}\theta\delta\theta. \nonumber \]

If we put this equal to zero, we obtain the same result as before.

Rhett Allain's Stuff

Confusion is the sweat of learning, the ladder problem.

I like to solve physics problems.  Here is one for you (I just made it up).

A 4 meter ladder leans against a frictionless wall at a 30 degree angle.  The mass of the ladder is 10 kg.  A human stands 1 meter up the ladder and has a mass of 70 kg.  What is the minimum coefficient of static friction between the floor and the ladder so that the ladder doesn’t slip?

Here is the solution—in video form.

But wait! There’s more.  Let’s do another problem.  

Suppose you have the same ladder and the same human.  However, in this case the coefficient of static friction between the ladder and the ground is 0.55.  How far up can the human move before the ladder slips?

I like this question because I don’t know how the answer will turn out.  That makes it fun.  So, let’s do it.

But what kind of problem is this?  I’ll make this a multiple choice question.  Here are your options.

  • A friction problem.
  • An equilibrium problem.
  • A ladder problem.
  • A work-energy problem.

Your answer?  Go ahead and answer.  It’s important to think about things like this if you want to become an expert problem solver.

Did you pick?  OK, I’ll tell you that there will be quite a few students that will say this is a friction problem because it has a coefficient of friction.  That’s not untrue—but it’s not a good way to classify the problem.  You could also say this is a “ladder problem”—again, not untrue but not helpful.

This is an equilibrium problem.  We are trying to find the point at which the ladder slips—the point it leaves equilibrium, but it’s still sort of in equilibrium.

For an object in equilibrium, there are two main ideas—especially for a rigid object.  First, it must have zero net force.  Second, it must have zero net torque about any point.  In two dimensions, I can write these two conditions as the following three equations.

F_\text{net-x}=0

Let’s talk about the torque stuff.  I don’t want to get into a whole thing about torque, so let me just say that torque is like a “rotational force” and it can be calculated as:

\tau = Fr\sin\theta

Oh, torques that would make an object rotate clockwise are negative.

One last thing about torques—especially the sum of the torques.  If an object is in rotational equilibrium about some point (point “o”) then it is also in rotational equilibrium about any other point.  When you set up the torque equation, you can pick whatever point you like to sum the torques—it’s your choice (but choose wisely).

Now we can start setting up some equations for equilibrium.  I’ll start with a force diagram for the ladder.

how to solve ladder problem physics

I know that’s a little busy—but it will have to do.  Here are some comments.

  • There are two normal forces.  One from the wall (labeled 1) and one from the floor.
  • There are two gravitational forces—and this is a cheat.  There is a gravitational force on the ladder and it is as though it is one force acting at the center of mass for the ladder. If the ladder has a uniform density, the center of mass is the center of the ladder.

m_2\vec{g}

Now for some equations.  First, this is the sum of the forces in the y-direction.

F_\text{net-y} = 0 = N_2 - m_1 g-m_2g

Just a quick reminder.  These are not vectors.  These are components of vectors in the y-direction.  That’s why the two gravitational forces have a negative sign.  I guess I can simplify this a little bit.

N_2 = (m_1 + m_2)g

Next is the sum of forces in the x-direction.

F_\text{net-x} =0=N_1 - F_f

Since we are at the point of maximum friction, I can include the expression for the frictional force in terms of the coefficient.

N_1 = \mu_s N_2

In order to write down the sum of the torques, I need to pick a point about which I calculate the torque.  I’m going to go with the bottom of the ladder.  At this point, there are two forces applied.  Since they are applied at the point about which the torque is calculated, they contribute zero torque and they won’t appear in the equation.  Winning.

Here is the sum of the torques about point O (at the bottom of the ladder).

\tau_\text{net-o} = 0 = m_2gs\cos\theta +m_1g\left(\frac{L}{2}\right)\cos\theta - N_1 L\sin\theta

Instead, let me make a plot of the frictional force as a function of human distance up the ladder.  That will be more fun, right?

Here’s what I get.

how to solve ladder problem physics

From this plot, the human can go up 1.16 meters before the required frictional force exceeds the maximum.  Oh, here is the code for that plot.

Now that you have the code, you can change stuff—like the angle of the ladder or the mass of the human or whatever.  

Share this:

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Thank you so much for this sir! Nice hypothesis on this video being helpful 2 yrs later! I really enjoyed this video.

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how to solve ladder problem physics

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Physics of a sliding ladder.

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Using results from related rate problems, some calculus books suggest that a ladder leaning against a wall and sliding under the influence of gravity will reach speeds that approach infinity. This Demonstration is built from the actual equations that govern the motion of the ladder as determined by the theory of rigid body mechanics. It shows that a sliding ladder never reaches very high speeds. The motion can be followed in two contrasting situations, with the top of the ladder either free to move away from the wall or constrained to be in contact with the wall. The forces are calculated for the falling ladder just before the top hits the floor.

Contributed by: Stelios Kapranidis and Reginald Koo   (March 2011) Open content licensed under CC BY-NC-SA

how to solve ladder problem physics

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Reginald Koo "Physics of a Sliding Ladder" http://demonstrations.wolfram.com/PhysicsOfASlidingLadder/ Wolfram Demonstrations Project Published: March 11 2011

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High School Physics

How to solve leaning ladder equilibrium numerical?

In this post, we will solve a numerical problem based on the equilibrium of a leaning ladder .

To do this you need to first understand the conditions of equilibrium . If you require some revision then you can quickly go through our physics tutorial on the equilibrium conditions first, before attempting this numerical.

Anyways, let’s begin solving the numerical.

Leaning ladder equilibrium numerical

Problem statement:

how to solve ladder problem physics

A uniform ladder of length l rests against a smooth, vertical wall (Fig. 1).

The mass of the ladder is m , and the coefficient of static friction between the ladder and the ground is μ s =0.40.

Find the minimum angle θ min at which the ladder does not slip.

How to solve this kind of problem:

In this post, we will understand this problem first and then take a step-by-step approach to solve the numerical.

Conceptualize Think about any ladders you have climbed. Remember that a large friction force between the bottom of the ladder and the surface is required to keep the leaning ladder in a balanced condition. If the friction force is zero, then the ladder won’t be able to stay up. You may simulate a ladder with a ruler leaning against a vertical surface. You will find that the ruler slips at some angles and stay up at others. Categorize We do not wish the ladder to slip, so we model it as a rigid object in equilibrium. Analyze The free-body diagram (FBD) showing all the external forces acting on the ladder is illustrated in Figure 2 .

how to solve ladder problem physics

The force exerted by the ground on the ladder is the vector sum of a normal force n and the force of static friction f s .

The force P exerted by the wall on the ladder is horizontal because the wall is frictionless.

Apply the first condition for equilibrium to the ladder:

Σ F x = f s – P = 0 ……… (1)

Σ F y = n – mg = 0 ….. (2)

Solve Equation (1) for P : P = f s ……… (3)

Solve Equation (2) for n : n = mg ……..(4)

When the ladder is on the verge of slipping, the force of static friction must have its maximum value, which is given by f s ,max = μ s n.

Combine this equation with Equations (3) and (4):

P = f s = μ s n = μ s mg …………. (5)

Apply the second condition for equilibrium to the ladder, taking torques about an axis through O :

Σ τ = P l sin θ min – mg (l/2) cos θ min = 0

Solve for tan θ min :

sin θ min / cos θ min = tan θ min = mg/ (2P) = mg/(2 μ s mg) = 1/( 2 μ s ) = 1/(2 x 0.4) = 1.25

hence, θ min = tan -1 (1.25) = 51 °

Observation Notice that the angle depends only on the coefficient of friction, not on the mass or length of the ladder.

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Physics Network

How a ladder leaning at a wall is in equilibrium?

At top of the ladder there is a normal force due to the wall. The gravitational force is acting at the center of the ladder. There are torques due to these forces. If the angle is large enough, the net torque and the net force are zero, and the ladder is in equilibrium.

How do you solve ladder problems in physics?

How do you solve a ladder friction problem?

Which equations of equilibrium are used in ladder problem?

This problem is statically determinate; i.e. the three reaction forces (R1,R2,R3) are uniquely solvable from the three equations of static equilibrium in which the net horizontal force, net vertical force, and rotational moment about any point at rest all vanish.

How do you find the resistance of an infinite ladder?

How far along the ladder can a 60.0 kg person climb before the ladder starts to slide?

So, the length of ladder the man climb =2. 7m.

What is ladder friction?

The ladder is acted upon by the following set of forces:1 Weight W acting downwards at its mid point. 2 Normal reaction Rh and friction force Fh = µRh at the end B leaning against the wall. Since the ladder has a tendency to slip downwards the friction force will be acting upwards.

How do mechanics solve friction problems?

What is the equation for static friction?

The formula to calculate the static friction is given as: Static Friction = Normal Force x Static Friction coefficient. Static friction = 60 N.

What are the 3 equations of equilibrium?

In order for a system to be in equilibrium, it must satisfy all three equations of equilibrium, Sum Fx = 0, Sum Fy = 0 and Sum M = 0. Begin with the sum of the forces equations. The simplest way to solve these force systems would be to break the diagonal forces into their component pars.

What are the three conditions of equilibrium?

A solid body submitted to three forces whose lines of action are not parallel is in equilibrium if the three following conditions apply : The lines of action are coplanar (in the same plane) The lines of action are convergent (they cross at the same point) The vector sum of these forces is equal to the zero vector.

How do you find the angle of a ladder?

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

The ground being frictionless would make it impossible to climb the ladder at all. The ladder wouldn’t even rest up against the wall. It would slide down and lie flat immediately.

How do you find the force exerted by a ladder?

What is the equivalent resistance between A and B in the given infinite ladder?

The equivalent resistance between A and B is RAB and between A and C is RAC.

How do you find the equivalent resistance in A and B?

  • ∵6Ω and 3Ω are in parallel.
  • Also, 4Ω and 12Ω are in parallel.
  • Therefore, equivalent Resistance between A and B is 5Ω

What is the value of resistance between A and B?

The value of effective resistance between A and B is (R= 2kΩ)

How far up a ladder can you go?

A person’s maximum safe reaching height is approximately 4′ higher than the height of the ladder. For example, a typical person can safely reach an 8′ ceiling on a 4′ ladder*. Extension ladders should be 7 to 10 feet longer than the highest support or contact point, which may be the wall or roof line.

How do you find the coefficient of friction for a ladder?

What is the magnitude of the contact force on the ladder at the base if the knight has climbed two thirds of the way up the ladder?

What are the magnitude and direction of the contact force on the ladder at the base if the knight has climbed two-thirds of the way up the ladder? Answer: 510 N.

What is P in friction?

The direction of the frictional force is opposite to the pending relative motion of the surfaces. If the force (P) increases, friction force (F) will also increase, until F Fs (limiting static frictional force).

What is triangular law force?

Triangle Law of Forces It states, “If two forces acting simultaneously on a particle, be represented in magnitude and direction by the two sides of a triangle, taken in order; their resultant may be represented in magnitude and direction by the third side of the triangle, taken in opposite order.”

What is screw friction?

The concept of an applied force in the direction of impending motion works for either (1) a force applied in the impending motion direction of a screw, or (2) a force applied to the impending motion direction of a nut.

How do you solve Block Block friction problems?

What is the formula of angle of friction?

Formula Used:The expression for angle of friction is given as: tanα=μ Here α is the angle of friction and μ is the coefficient of limiting friction. The expression for angle of repose is given as: tanθ=μ Here β is the angle of repose and μ is the coefficient of limiting friction.

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  1. How to Solve a Ladder Problem (Equilibrium, Forces, & Torque)

    how to solve ladder problem physics

  2. Torque Example #3: Leaning Ladder Problem

    how to solve ladder problem physics

  3. Leaning Ladder Equilibrium Problem: Find Minimum Angle

    how to solve ladder problem physics

  4. Solving a Physics Problem: Ladder Facing a Wall

    how to solve ladder problem physics

  5. Statics Practice Problem 8-16: Friction exampe of a ladder leaning against a wall

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  6. Physics, Torque (12 of 13) Static Equilibrium, Ladder Problem

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COMMENTS

  1. A leaning ladder

    Task number: 654 A ladder is leant against the wall. The coefficient of the static friction μsw between the ladder and the wall is 0.3 and the coefficient of the static friction μsf between the ladder and the floor is 0.4. The centre of mass of the ladder is in the middle of it.

  2. Worked example 10.3: Leaning ladder

    Answer: The angle subtended by the ladder with the ground satisfies Let be the normal reaction at the wall, let be the normal reaction at the ground, and let be the frictional force exerted by the ground on the ladder, as shown in the diagram. Consider the torque acting on the ladder about the point where it meets the ground.

  3. Static Equilibrium

    15K 1.2M views 7 years ago New Physics Video Playlist This physics video tutorial explains the concept of static equilibrium - translational & rotational equilibrium where everything is at rest...

  4. Leaning Ladder Equilibrium Problem: Find Minimum Angle

    Physics Ninja looks at the leaning ladder problem. Newton's laws are used to find the minimum angle where the ladder remain in equilibrium. The forces and...

  5. Torque Ladder Problems

    0:00 / 8:45 Torque Ladder Problems James Dann, Ph.D. 11.3K subscribers 66K views 12 years ago The classic ladder problem is solved using torque and forces. By James Dann for ck12.org...

  6. 12.2 Examples of Static Equilibrium

    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.

  7. 9.3: Virtual Work

    A method of solving such problems, therefore, is to write down an expression for the potential energy and put the derivatives equal to zero. ... Let's start by doing a simple ladder problem by the method of virtual work. The usual uniform ladder of high school physics, of length \( 2l\) and weight \( mg\), is leaning in limiting static ...

  8. Climbing a Ladder

    Select the Relation Solve Understand In order for you to climb the ladder safely, it must remain in place as you climb. In other words, you want the motion of the ladder to remain at rest both translationally and rotationally as the location of the forces on the ladder changes.

  9. A Simple Demonstration for the Static Ladder Problem

    In teaching the ladder problem we begin with a simple diagram of the leaning ladder and ask the students to identify the forces on the ladder and where they act. Students will readily identify three forces: the ladder's weight (mg), the normal force of the ground on the ladder, N, and the force of the wall on the ladder, F wall, as shown in Fig. 1.

  10. Equilibrium in 2D

    Multiple Choice. A ladder of mass 20 kg (uniformly distributed) and length 6 m rests against a vertical wall while making an angle of Θ = 60° with the horizontal, as shown. A 50 kg girl climbs 2 m up the ladder. Calculate the magnitude of the total contact force at the bottom of the ladder (Remember:You will need to first calculate the ...

  11. Physics, Torque (12 of 13) Static Equilibrium, Ladder Problem

    Static Equilibrium, The Ladder Problem; Shows how to use static equilibrium to determine the force of friction between the bottom of the ladder and the groun...

  12. Rotation of a slipping ladder

    18 If the ladder is slipping on the floor as well as the wall, then the point of rotation is where the two normal forces intersect. This comes from the fact that reaction forces must pass through the instant center of motion, or they would do work. In the diagram below forces are red and velocities blue.

  13. PDF Torque Analyses of a Sliding Ladder

    1 Problem The problem of a ladder that slides without friction while touching a floor and wall is often used to illustrate Lagrange's method for deducing the equation of motion of a mechanical system. Suppose the ladder has mass m, length 2l, and makes angle θ to the vertical.

  14. The Ladder Problem

    Here is the solution—in video form. Find the coefficient of friction to prevent a ladder from sliding Share Watch on But wait! There's more. Let's do another problem. Suppose you have the same ladder and the same human. However, in this case the coefficient of static friction between the ladder and the ground is 0.55.

  15. Physics of a Sliding Ladder

    Using results from related rate problems, some calculus books suggest that a ladder leaning against a wall and sliding under the influence of gravity will reach speeds that approach infinity. This Demonstration is built from the actual equations that govern the motion of the ladder as determined by the theory of rigid body mechanics. It shows that a sliding ladder never reaches very high speeds. T

  16. Equilibrium in 2D

    Problem. A ladder of mass 20 kg (uniformly distributed) and length 6 m rests against a vertical wall while making an angle of Θ = 60° with the horizontal, as shown. A 50 kg girl climbs 2 m up the ladder. Calculate the magnitude of the total contact force at the bottom of the ladder (Remember:You will need to first calculate the magnitude of N ...

  17. How to solve leaning ladder equilibrium numerical?

    By Anupam M In this post, we will solve a numerical problem based on the equilibrium of a leaning ladder. To do this you need to first understand the conditions of equilibrium. If you require some revision then you can quickly go through our physics tutorial on the equilibrium conditions first, before attempting this numerical.

  18. Physics 15 Torque Example 7 (7 of 7) The Ladder Problem ...

    Visit http://ilectureonline.com for more math and science lectures!In this seventh of the seven part series we will find out a) Will the ladder slip when the...

  19. How do you solve ladder problems in physics? [Answered!]

    How do you solve ladder problems in physics? September 29, 2022 by George Jackson Spread the love Table of Contents show How do you solve a ladder friction problem? What are the forces acting on a ladder against wall? A ladder is leaning against the wall.

  20. PHYS 210 The Ladder Problem

    This video shows how to solve one of the most common types of static equilibrium problems in Physics, the "Ladder" problem.

  21. How a ladder leaning at a wall is in equilibrium? [FAQ!]

    The ladder is acted upon by the following set of forces:1 Weight W acting downwards at its mid point. 2 Normal reaction Rh and friction force Fh = µRh at the end B leaning against the wall. Since the ladder has a tendency to slip downwards the friction force will be acting upwards. How do mechanics solve friction problems?

  22. How to solve a ladder rotational equilibrium physics problem

    How much friction is needed so that the ladder does NOT slip?This is the ladder equilibrium problem commonly asked in a physics class. Students will need to ...