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## Torque Practice Problems with Solutions: AP Physics 1

When you want to rotate a body about an axis or a point, the direction and location of the applied force are also important, in addition to its magnitude.

Let's assume you want to open a door. Is it easier to open the door by applying a force to the doorknob or applying the same force magnitude to a point closer to the hinge?

It is an everyday observation that opening the door by exerting force at a point far away from the hinge is easier.

The distance perpendicular from the line of action of the force to the axis of rotation is called the lever arm or moment arm and is designated by $r_{\bot}$ as shown in the figure below.

In this manner, the torque $\tau$ is defined as the simple product of the lever arm $r_{\bot}$ and the force magnitude $F$, $\tau=r_{\bot}F$ The direction of the torque is found using the right-hand rule.

In the following, we are going to practice some simple problems about torque to deepen our understanding of these concepts.

## Torque Practice Problems

Problem (1): In each of the following diagrams, calculate the torque (magnitude and direction) about point $O$ due to the force $\vec{F}$ of magnitude $10\,\rm N$ applied to a $4-\rm m$ rod. Both the force $\vec{F}$ and the rode lie in the plane of the page.

Solution : As said in the introduction above, the lever arm times the applied force gives us the torque about a point or an axis of rotation. First, we must identify the line of action and then the lever arm $r_{\bot}$.

We reach the line of action of the force by extending the applied force along a straight line in both directions. Now draw a perpendicular line from the point of rotation to that line so that it intersects it at a point. This distance is called the lever arm.

For simplicity in the calculation, the lever arm is always formulated as $r_{\bot}=L\sin\theta$, where $L$ is the distance from the point of application of the force to the axis of rotation and $\theta$ is the acute angle between the force $\vec{F}$ and the line connecting $F$ to the $O$.

We repeat this procedure for each case separately.

(a) In this figure, the line of action of the force is already perpendicular to the axis of rotation. Thus, the lever arm is the full distance between the point of application of the force $F$ and the point $O$, i.e., $r_{\bot}=4\,\rm m$. Therefore, the torque magnitude $\tau$ about point $O$ is calculated as \begin{align*} \tau&=r_{\bot}F \\&=(4)(10) \\&=40\,\rm m.N \end{align*}  (b) To find the torque of this configuration, extend the force $F$ and draw a line perpendicular to it so that it passes through the axis of rotation.

In this case, instead of using geometry to find the lever arm, we use the following formula to understand its application. \begin{align*} r_{\bot}&=L\sin\theta \\ &=4\sin 60^\circ \\ &=2\sqrt{3} \quad \rm m \end{align*} Now, substituting this value into the torque formula, yields \begin{align*} \tau&=r_{\bot}F \\ &=(2\sqrt{3})(10) \\ &=20\sqrt{3}\quad\rm m.N \end{align*} (c) Again, identify the lever arm and compute the magnitude of the torque associated with this force about point $O$. \begin{align*} \tau&=r_{\bot}F \\ &=(L\sin\theta) F \\ &=(4\sin 30^\circ)(10) \\&=20\quad\rm m.N \end{align*}

Problem (2): Two forces ($F_A=12\,\rm N$ and $F_B=8\,\rm N$) are applied to a $5-\rm m$ stick as in the figure below. The rod and the forces are on the plane of the page. Calculate the net torque about point $O$.

Solution : Here, two forces are applied to the rod, causing it to rotate about the point $O$. In such torque problems, we want to find out in which direction the rod (or the object) will eventually rotate.

In all torque practice problems, by convention, counterclockwise rotation is taken to be the positive direction and clockwise the negative direction.

The force $F_A$ rotates the rod with respect to point $O$ counterclockwise, so its corresponding torque is positive with a magnitude of \begin{align*} \tau_A&=r_AF_A\sin\theta \\&=5\times 12\times \sin 90^\circ \\ &=60\quad \rm m.N \end{align*} On the other hand, the force $F_B$ tend to rotate the rod about $O$ clockwise, so we assign a negative to its corresponding torque magnitude, \begin{align*} \tau_B&=r_BF_B\sin\theta \\&=3\times 8\times \sin 37^\circ \\ &=14.4\quad \rm m.N \end{align*} When more than one torque acts on an object, the torques are added and gives the net torque exerted on the object. Therefore, the net torque on this rod exerted by forces $F_A$ and $F_B$ is found to be \begin{align*} \tau_{net}&=\tau_A+\tau_B \\ &=60+(-14.4) \\ &=45.6\quad \rm m.N \end{align*} The net torque is obtained as positive, indicating that the rod will rotate counterclockwise about its axis of rotation $O$.

Problem (3): Calculate the net torque about the axle of the wheel through point $O$ perpendicular to the plane of the page, taking $r=12\,\rm cm$ and $R=24\,\rm cm$.

Solution : First, using the definition of torque, we find its magnitude; then, because torque is a vector quantity in physics, we assign a positive or negative sign to it; and finally, we add torques to obtain the net torque about the desired rotation point.

The force $F_1$ rotates the smaller circle with the lever arm $r_{\bot,1}=0.12\,\rm m$ clockwise, so assign a negative to its torque magnitude. \begin{align*} \tau_1&=r_{\bot,1}F_1 \\&=(0.12)(45) \\&=5.4\quad\rm m.N \end{align*} The force $F_2$ also rotates the bigger circle clockwise, whose torque magnitude would be obtained \begin{align*} \tau_2&=r_{\bot,2}F_2 \\&=(0.24)(15) \\&=3.6 \quad \rm m.N \end{align*} And finally, the force $F_3$ rotates the bigger circle counterclockwise, so by convention assign a positive sign to its torque magnitude: \begin{align*} \tau_3&=r_{\bot,3}F_3 \\&=(0.24)(30) \\&=7.2 \quad \rm m.N \end{align*} Now, add torques with their correct signs to get the net torque about the axle of the wheel: \begin{align*} \tau_{net} &=\tau_1+\tau_2+\tau_3 \\ &=(-5.4)+(-3.6)+(7.2) \\&=-1.8\quad \rm m.N \end{align*} The overall sign of the net torque is obtained as negative, telling us that these forces will rotate the wheel about its axle clockwise.

When a force is applied to the rim of a circle or wheel and makes an angle with the horizontal line, the torque about the center of the wheel (or circle) does not depend on this angle.

By definition, the lever arm is the perpendicular distance from the point of application of force to the axis of rotation. In torque problems involving a wheel (or circle) and forces applying to the rim of it, the lever arm is always the radius of the wheel.

Problem (4): Three forces are applied to a wheel as shown in the figure below. Two forces are tangent to the wheel, while the third forms a $37^\circ$ angle with the tangent to the inner circle. What is the net torque on the wheel due to these three forces about the axle through $O$ perpendicular to the page? Assume that a friction torque of $0.3\,\rm m.N$ opposes the rotation. (take $r=10\,\rm cm$ and $R=20\,\rm cm$)

Solution : Again a wheel and some forces acting on its rim and wanting the net torque about its center.

The forces $F_1$ and $F_2$ rotate the wheel clockwise, which exerts negative torques on the wheel whose magnitudes are found as follows \begin{align*} \tau_1&=r_{\bot,1}F_1 \\&=(0.20)(15) \\&=3\quad \rm m.N \\\\ \tau_2&=r_{\bot,2}F_2 \\&=(0.20)(10) \\&=2\quad \rm m.N \end{align*} The other force $F_3$ that acts at an angle with the rime of the smaller circle apply a positive torque according to the sign conventions for torques (counterclockwise rotation). Its magnitude is \begin{align*} \tau_3&=r_{\bot,3}F_3 \\&=(0.10)(40) \\ &=4\quad\rm m.N \end{align*} Now, sum these torques to find the net torque exerted about the axle of the rotation $O$, being careful not to forget to consider their signs. \begin{align*} \tau_{net}&=\tau_1+\tau_2+\tau_3 \\ &=(-3)+(-2)+(+4) \\ &=-1\quad \rm m.N\end{align*} This is the net torque applied by the external forces that cause the wheel to rotate counterclockwise.

In this question, we are told that the axis of rotation also exerts a friction force, whose corresponding torque has a magnitude of $0.3\,\rm m.N$. This torque, due to a frictional force, opposes the overall rotation of the wheel, which is counterclockwise, so it must be supplied by a positive sign, i.e., $\tau_f=+0.3\,\rm m.N$.

Hence, the total torque with respect to the point $O$ is $\tau_t=-1+(+0.3)=\boxed{-0.7\,\rm m.N}$

Problem (5): A person exerts a force of $50\,\rm N$ on the end of an $86-\rm cm$-wide door to open it. What is the magnitude of the torque if the force is applied (a) perpendicular to the door and (b) at an angle of $53^\circ$ to the plane of the door?

Solution : An overhead view of this configuration is depicted below.

Torque is defined as $\tau=rF\sin\theta$, where $r$ is the distance between the point of application of the force and the point of the axis of rotation, $F$ is the applied force, and $\theta$ is the angle between the applied force and the line connecting the force action point and the rotation point.

(a) In this case, the force is applied to the door perpendicularly. The line joining the force action point (say, the doorknob) and the axis of rotation (the hinge's door), which is actually the same $r$, makes a right angle with the force vector as shown in the figure below, so $\theta=90^\circ$.

Substituting the numerical values into the torque formula gives its magnitude as below: \begin{align*} \tau&=rF\sin\theta \\&=(0.86)(50) \sin 90^\circ \\ &=43\quad\rm m.N \end{align*}  (b) in this part, the angle between $r$ and $F$ is $\theta=53^\circ$ as illustrated in the figure below. Thus, the torque associated with this force is found to be \begin{align*} \tau&=rF\sin\theta \\&=(0.86)(50) \sin 53^\circ \\ &=34.4\quad \rm m.N\end{align*} From this torque question, we can understand the physical concept of torque.

You have seen that the same force applied to the door at two different angles can produce two different torques. But what is this meaning?  You can do this yourself at home and see the result.

Certainly, you will notice that opening a door by applying a force perpendicular to its knob is much easier than applying the same force at some angle.

Therefore, we conclude that the greater the torque produced, the easier the door opens.

Problem (6): In the following figure, all rods have the same length and are pivoted at point $O$. Rank in order, from the smallest to largest, the torques.

Solution : First, calculate the torques corresponding to each applied force. Since the length of the rods was not given, take it as $L$.

(a) A force $F$ is applied to the left end perpendicular to the radial line $r$, such forces create maximum torque whose magnitude is $\tau_a=rF=\boxed{4L}$ (b) In this case, the force $F$ is applied perpendicularly to the middle of the radial line, so the distance between the force action point and the pivot point is $r=\frac L2$ $\tau_b=rF=4(\frac L2 )=\boxed{2L}$ (c) Here, the line of action of the force makes a $45^\circ$ angle with the radial line, $\theta=45^\circ$. On the other hand, the straight distance between the force action point and the pivot point is $r=L$. Thus, the torque associated with this force is computed as \begin{align*} \tau_c&=rF\sin\theta \\&= (L)(4) \sin 45^\circ \\ &=\boxed{2\sqrt{2}L}\end{align*} (d) In this case, the force is pulling straight out from the pivot point $O$ and making a zero angle, $\theta=0$, with the radial line. Hence, the torque of this force is given by $\tau_d=rF\sin\theta=L(4) \sin 0^\circ= \boxed{0}$ Such forces as pulling out from or pushing into the pivot point exert zero torque. Now we are in a position to rank the torques from smallest to largest. $\tau_d <\tau_b < \tau_c <\tau_a$

Problem (7): A person applies a force of $55\,\rm N$ near the end of a $45-\rm cm$-long wrench.  (a) How should the force be applied to produce the maximum torque? (b) What is the maximum torque exerted?

Solution : In the preceding question, we found out that a maximum torque acts on a pivot point when these two conditions are met;

(I) The external force applied to a point where it has the maximum distance from the pivot point (or axis of rotation) and

(II) When the angle between the force action point and the radial line, a straight line that connects the force action point and the pivot point, is $90^\circ$.

Here we are told that the force is applied near the end of the wrench, having a maximum distance from the rotation axis, so the first condition is satisfied.

(a) To satisfy the second condition, the force must be applied at the right angle to the line of the wrench.  (b) With this explanation, the maximum torque is found to be $\tau_{max}=rF=(0.45)(55)=\boxed{24.75\,\rm m.N}$

Problem (8): Find the magnitude and direction of the net torque on a $2-\rm m$-long rod in each of the following cases as shown.  (a) the center of mass of the rod, about point $C$, and (b) through the point $Q$.

Solution : in each case, first, identify the straight distance $r$ between the force action point, where the force acts on the rod, and the pivot point (or the rotation axis).

Next, find the angle $\theta$ between the force $\vec{F}$ and the line connecting the point of application of the force and the pivot point, which is called the radial line, or position vector $\vec{r}$ in your textbooks.

Combining these into the torque formula, $\tau=rF\sin\theta$, to find its magnitude.

(a) Three forces are acting on the rod and causing a torque about the rod's center of mass. Be careful that the point of application of the force $F_3$ does not have distance from the axis of rotation $C$, so the magnitude $r$ of its position vector $\vec{r}$ is zero, i.e., $r=0$.

Consequently, this force cannot rotate the rod, or in other words, the torque due to this force is zero. The other torques are \begin{align*} \tau_1&=rF\sin\theta \\&=(1)(55) \sin 66^\circ \\&=50.24\quad \rm m.N \\\\ \tau_2&=rF\sin\theta \\&=(1)(40) \sin 27^\circ \\ &=18.16\quad \rm m.N\end{align*} The forces $F_2$ and $F_1$ rotate the rod about point $C$ in a counterclockwise direction, so by sign conventions for torques, a positive sign must be assigned to them. Therefore, the net torque about the center of mass of the rod is \begin{align*} \tau_{net}&=\tau_1+\tau_2+\tau_3 \\ &=50.24+18.16+0 \\&=\boxed{+68.4\quad \rm m.N}\end{align*} Consequently, these three forces, applied at different angles to the rod, create a net torque of $68\,\rm m.N$ about the pivot point $C$ and rotate it in a counterclockwise direction (because of the plus sign in front of the net torque).

(b) We want to solve this part by the method of resolving the applied force into its components parallel and perpendicular to the line that connects the axis of the rotation to the point of application of the force, or radial line (this is the same position vector $\vec{r}$).

In this method, the component $F_{\parallel}$ always exerts no torque since it goes through the axis of rotation and thus has a zero lever arm.

Hence, the only component of the force capable of rotating the body about the axis is $F_{\bot}$ which its corresponding torque will be equal to $\tau=rF_{\bot}$ where $r$ is the distance from the axis to the point of application of the force.

In the following figure, the forces are resolved into $F_{\parallel}$ and $F_{\bot}$. Thus, their exerted torques are found to be \begin{align*} \tau_1&=r_1F_{1,\bot} \\&=(0)(55\sin 66^\circ) \\&=0 \\\\ \tau_2&=r_2F_{2,\bot} \\&=(2)(40\sin 27^\circ) \\&=36.32\quad\rm m.N \\\\ \tau_3&=r_3 F_{3,\bot} \\&=(1)(75\sin 53^\circ) \\&=60\quad \rm m.N \end{align*} As you can see, the force $F_1$ is directed at the rotation axis, so $r=0$. The forces $F_2$ and $F_3$ rotate the rod about the point $Q$ in ccw and cw directions, respectively, resulting in a positive and negative torque.   The torque $\tau_2$ is positive since its corresponding force $F_2$ rotates the rod about the point $Q$ counterclockwise (ccw).

The torque $\tau_3$ should be negative since its corresponding force $F_3$ rotates the rod about $Q$ clockwise. Therefore, the net torque about the axis $Q$ is calculated as \begin{align*} \tau_{net}&=\tau_1+\tau_2+\tau_3 \\&=0+(36.32)+(-60) \\ &=\boxed{-23.68\quad\rm m.N} \end{align*} Consequently, the combined forces produce a negative torque that rotates the rod clockwise.

Problem (9): Calculate the net torque (magnitude and direction) applied to the beam in the following figure about (a) the axis through point $O$ perpendicular to the page and (b) the point $C$ perpendicular to the plane of the page.

Solution : As you found out, there are two equivalent ways to calculate torque due to an applied force. One is using the lever arm concept and applying the torque formula, $\tau=r_{\bot}F$, and the other is using the force components, in which only the perpendicular component creates a torque about an axis, $\tau=rF_{\bot}$.

Here, we want to solve this torque AP Physics 1 question by the method of resolving the applied force and applying the formula $\tau=rF_{\bot}$, where $F_{\bot}=F\sin\theta$ and $\theta$ is the angle the force makes with the radial line.

First of all, resolve the forces along $F_{\parallel}$ and perpendicular $F_{\bot}$ to the radial line, the line connecting the point at which the force applies and the pivot point as depicted in the free-body diagram below.

(a) The extension of the radial force component $F_{\parallel}$ passes straight through the pivot point $C$, so it wouldn't create torque.

The extension of the perpendicular force component $F_{\bot}$ always has some finite distance from the pivot point and thus creates torque. The magnitude of torques is found to be \begin{align*} \tau_1 &=rF_{1,\bot} \\&=(3)(20\sin 30^\circ) \\ &=30\quad \rm n.N \\\\ \tau_2 &=rF_{2,\bot} \\&=(0)(30\sin 53^\circ) \\ &=0 \\\\ \tau_3 &=rF_{3,\bot} \\&=(3)(44\sin 45^\circ) \\ &=92.4\quad \rm n.N \end{align*} Notice that for torque due to the force $F_2$, the angle between $F_2$ and the vertical line is given, not the radial line, which is favored. This force applies straight to the axis of rotation and exerts no torque.

Those were the magnitudes of the torques; now determine their correct signs, which indicate the direction of rotations, since torque is a vector quantity in physics, having both a magnitude and a direction.

The torque $tau_1$ acts to rotate the rod clockwise, so a negative is assigned to it. On the other hand, the torque $tau_3$ rotates the rod counterclockwise, so it must be accompanied by a positive sign according to the convention of signs for torques.

The sum of these torques gives the net torque exerted on the pivot point $C$: \begin{align*} \tau_{net} &=\tau_1+\tau_2+\tau_3 \\ &=(-30)+0+(92.4) \\&=62.4\quad \rm m.N \end{align*} Ultimately, the rod will rotate counterclockwise due to applying these forces since its net torque is positive.

(b) Now, we want to find the net torque due to the same forces but about point $O$. In this case, the force $F_3$ exerts no torque as it passes straight through the axis of the rotation $O$, $\tau_3=0$. The magnitude of the torques of the other forces about point $O$ is calculated as below \begin{align*} \tau_1&=r_1F_{1,\bot} \\&=L(F_1 \sin 30^\circ) \\&=(6)(20\times 0.5) \\&=60\quad \rm m.N \\\\ \tau_2&=r_2F_{2,\bot} \\&=(L/2)(F_2 \sin 53^\circ) \\&=(3)(30\times 0.8) \\&=72\quad \rm m.N \end{align*} Therefore, the net torque about point $O$ by considering the correct sign for each torque (positive torque for counterclockwise and negative for clockwise direction) is \begin{align*} \tau_{net}&=\tau_1+\tau_2+\tau_3 \\ &=(-60)+(+72)+0 \\&=+12\quad\rm m.N\end{align*} Thus, this combination of forces rotates the rod in a counterclockwise direction about point $O$, resulting in a net positive torque.

Overall, from this important problem, we learned that torques must always be calculated with reference to a specific point. In part (a), the torque of $F_2$ was zero about point $C$ but not about point $O$. The same reasoning is also true for the force $F_3$ about these two pivot points.

Therefore, the true statement for describing torques due to some applied forces is "the torque of force $F$ about (or with respect to) point $X$".

Problem (10): Two blocks of mass $m$ are attached to a massless rod that pivots as shown in the figure below. Considering the rod is held initially in the horizontal position and released, what is the net torque (magnitude and direction) on the pivot when it is just released?

Solution : two equal masses are standing on a level rod pivoted at a point. First, calculate the magnitude of torques associated with each mass exerted on the rod, then assign a positive or negative sign to each torque to indicate their direction.

Each mass applies a weight force of $w=mg$ to the rod perpendicularly. According to the sign conventions for torques, the left mass rotates the rod counterclockwise about the pivot point with a positive torque and the right mass clockwise with a negative torque.

The magnitude of each torque is calculated by the general torque equation as below \begin{align*} \tau_1&=rF\sin\theta \\&=\mathcal l_1 (mg) \sin 90^\circ \\&=\mathcal l_1 mg \\\\ tau_2&=rF\sin\theta \\&=\mathcal l_2 (mg) \sin 90^\circ \\&=\mathcal l_2 mg \end{align*} The net torque about the pivot point is the sum of the torques due to the applied forces: \begin{align*} \tau_{net}&=\tau_1+\tau_2 \\&=+\mathcal l_1 mg + (-\mathcal l_2 mg) \\ &=mg( \mathcal l_1-\mathcal l_2) \end{align*} In the last step, $mg$ is factored out. Since the lever arm for $m_2$ is greater than $m_1$ or $\mathcal l_2 >\mathcal l_1$, the net torque about the pivot point will be negative. In other words, this combination of masses on the rod just after releasing leads to a clockwise rotation with respect to the support.

Problem (11): A mechanic is loosening a nut using a $25-\rm cm$-long wrench by applying a force of $20\,\rm N$ at an angle of $30^\circ$ to the end of the handle.  (a) What torque does the mechanic apply to the center of the nut? (b) In which direction should he exert this force to obtain maximum torque, and with what magnitude?

Solution : The angle between the force applied to the wrench and the radial line is given by $30^\circ$. Here, the distance between the point at which the force acts and the nut (axis of rotation) is $r=0.25\,\rm m$.

(a) Use the general equation for torque, $\tau=rF\sin\theta$, to find its magnitude as follows \begin{align*} \tau&=rF\sin\theta \\ &=(0.25)(20\times \sin 30^\circ) \\&=2.5\quad \rm m.N \end{align*}  (b) Once the applied force is resolved into its radial $F_{\parallel}$ and perpendicular $F_{\bot}$ components, the $F_{\bot}$ points in the counterclockwise direction, so it exerts a positive torque by our sign convention.

Author : Dr. Ali Nemati Published : Mar 20, 2023

• 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
• 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

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.

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.

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

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

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.

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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• Authors: William Moebs, Samuel J. Ling, Jeff Sanny
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• Book title: University Physics Volume 1
• Publication date: Sep 19, 2016
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• Section URL: https://openstax.org/books/university-physics-volume-1/pages/12-2-examples-of-static-equilibrium

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## Physics library

Course: physics library   >   unit 7.

• Introduction to torque
• Moments (part 2)
• Finding torque for angled forces
• Rotational version of Newton's second law
• More on moment of inertia
• Rotational inertia
• Rotational kinetic energy
• Rolling without slipping problems
• Angular momentum
• Constant angular momentum when no net torque
• Angular momentum of an extended object
• Ball hits rod angular momentum example
• Cross product and torque

## What is torque?

How is torque calculated, how is torque measured, what role does torque play in rotational kinematics, what is rotational equilibrium, how does torque relate to power and energy, how can we increase or decrease torque, data sources, want to join the conversation.

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• it is not moving in any linear direction AND
• it is not rotating..

If both of these conditions are met, then scientists and engineers say, "The body is in mechanical equilibrium. " There are two other forms of eqilibrium called static and dynamic equilibrium. Static equilibrium means the body is moving at a constant velocity. Mechanical equilibrium is a subset of static equilibrium. Dynamic equilibrium is where a body is accelerating and moving at constant velocity at the same time. An example of this is a ball that is spinning while not moving linearly. Every spinning object experiences a centripetal acceleration. In the example the object is spinning but not experiencing any translational motion.

• The red car is in mechanical and static equilibrium.
• The blue car is only in static equilibrium
• The spinning "eight ball" is in dynamic equalibrium

This unit will only deal with mechanical equilibrium.

When you solved free body problems, in a previous unit , you were applying one of the conditions of mechanical and dynamic equilibrium by summing up the forces. This is called the first condition of equilibrium. It looked like this...

This YouTube video is here, http://youtu.be/PhOCrlQqI3A

When do you use the summing up the torques technique ?

• Use summing up the forces and torques when the question asks how far away from a point a force is applied.This includes questions that ask...
• how far a person can walk on a board or beam.
• Or where can a weight be placed on beam
• Or where the lifting force needs to be placed. All these questions involve a force(s) and a distance. -and that's a torque.
• Solution on Paper
• Video Solution

This YouTube video is found at, http://youtu.be/e_T0s-gpjnE

This video can be is on YouTube at, http://youtu.be/b7EA_F38hn4

## Chapter: 11th Physics : UNIT 5 : Motion of System of Particles and Rigid Bodies

Solved Example Problems for Torque

## Example 5.7

If the force applied is perpendicular to the handle of the spanner as shown in the diagram, find the (i) torque exerted by the force about the center of the nut, (ii) direction of torque and (iii) type of rotation caused by the torque about the nut.

Arm length of the spanner, r  =  15 cm = = 15×10 −2 m

Force, F = 2.5 N

Angle between r and F, θ = 90 o

(i) Torque, τ θ = rF sin

(ii) As  per  the  right  hand  rule,  the direction of torque is out of the page.

(iii) The type of rotation caused by the torque is anticlockwise.

## Example 5.9

A crane has an arm length of 20 m inclined at 30 o  with the vertical. It carries a container of mass of 2 ton suspended from the top end of the arm. Find the torque produced by the gravitational force on the container about the point where the arm is fixed to the crane. [Given: 1 ton  =  1000 kg; neglect the weight of the arm. g  =  10 ms -2 ]

The force F at the point of suspension is due to the weight of the hanging mass.

We can solve this problem by three different methods.

The angle (θ) between the arm length (r) and the force (F) is, θ  =  150 o

The torque ( τ ) about the fixed point of the arm is,

Method – II

Let us take the force and perpendicular distance from the point where the arm is fixed to the crane.

Method – III

Let us take the distance from the fixed point and perpendicular force.

All the three methods, give the same answer

Solved Example Problems for Torque about an Axis

## Example 5.10

Three mutually perpendicular beams AB, OC, GH are fixed to form a structure which is fixed to the ground firmly as shown in the Figure. One string is tied to the point C and its free end D is pulled with a force F. Find the magnitude and direction of the torque produced by the force,

i. about the points D, C, O and B

ii. about the axes CD, OC, AB and GH.

The torque of a force about an axis is independent of the choice of the origin as long as it is chosen on that axis itself. Th is can be shown as below.

Let O be the origin on the axis AB, which is the rotational axis of a rigid body. F is the force acting at the point P. Now, choose another point O’ anywhere on the axis as shown in Figure 5.10.

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## 5.3: The Second Condition for Equilibrium

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Learning Objectives

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

• State the second condition that is necessary to achieve equilibrium.
• Explain torque and the factors on which it depends.
• Describe the role of torque in rotational mechanics.

Definition: Torque

The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges.

Several familiar factors determine how effective you are in opening the door (Figure $$\PageIndex{1}$$). First of all, the larger the force, the more effective it is in opening the door—obviously, the harder you push, the more rapidly the door opens. Also, the point at which you push is crucial. If you apply your force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. The most effective direction is perpendicular to the door—we push in this direction almost instinctively.

The magnitude, direction, and point of application of the force are incorporated into the definition of the physical quantity called torque. Torque is the rotational equivalent of a force. It is a measure of the effectiveness of a force in changing or accelerating a rotation (changing the angular velocity over a period of time). For forces applied strictly at right angles, we can express the magnitude of the torque in equation form as

$\tau = r_{\perp}F$

where $$\tau$$ (the Greek letter tau) is the symbol for torque,  $$r_{\perp}$$ is the perpendicular lever arm, and $$F$$ is the magnitude of the force.

The perpendicular lever arm $$r_{\perp}$$ is the shortest distance from the pivot point to the line along which $$F$$ acts. Note that the line segment that defines the distance $$r_{\perp}$$ is perpendicular to $$F$$, as its name implies.

The SI unit of torque is newtons times meters, usually written as N$$\cdot$$m. For example, if you push perpendicular to the door with a force of 40 N at a distance of 0.800 m from the hinges, you exert a torque of 32 N$$\cdot$$m (0.800 m $$\times$$ 40 N) relative to the hinges. If you reduce the force to 20 N, the torque is reduced to 16 N$$\cdot$$m, and so on.

The torque is always calculated with reference to some chosen pivot point. For the same applied force, a different choice for the location of the pivot will give you a different value for the torque. Any point in any object can be chosen to calculate the torque about that point. The object may not actually pivot about the chosen “pivot point.”

Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point.

Now, the second condition necessary to achieve equilibrium is that the net external torque on a system must be zero . An external torque is one that is created by an external force. You can choose the point around which the torque is calculated. The point can be the physical pivot point of a system or any other point in space—but it must be the same point for all torques. If the second condition (net external torque on a system is zero) is satisfied for one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest.  The second condition necessary to achieve equilibrium is stated in equation form as

$net \, \tau = 0$

where net means total. Torques which are in opposite directions are assigned opposite signs. A common convention is to call counterclockwise (ccw) torques positive and clockwise (cw) torques negative.

When two children balance a seesaw as shown in Figure $$\PageIndex{3}$$, they satisfy the two conditions for equilibrium. Most people have perfect intuition about seesaws, knowing that the lighter child must sit farther from the pivot and that a heavier child can keep a lighter one off the ground indefinitely.

Example $$\PageIndex{1}$$: She Saw Torques On A Seesaw

The two children shown in Figure $$\PageIndex{3}$$ are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple—more involved examples will follow.) The first child has a mass of 26.0 kg and sits 1.60 m from the pivot.

• If the second child has a mass of 32.0 kg, how far is she from the pivot?
• What is $$F_p$$, the supporting force exerted by the pivot?

Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then identify all external forces acting on the system.

Solution (a)

The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the torque produced by each. Torque is defined to be

$\tau = r_{\perp}F\nonumber$

The torques exerted by the three forces are first,

$\tau_1 = r_1w_1\nonumber$

$\tau_2 = -r_2w_2\nonumber$

\begin{align*} \tau_p &= r_pF_p \\[5pt] &= 0 \cdot F_p \\[5pt] &= 0. \end{align*}

Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. Since $$F_p$$ acts directly on the pivot point, the distance $$r_p$$ is zero. A force acting on the pivot cannot cause a rotation, just as pushing directly on the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. Therefore

$\tau_2 = -\tau_1,\nonumber$

$r_2w_2 = r_1w_1.\nonumber$

Weight is mass times the acceleration due to gravity. Entering $$mg$$ for $$w$$, we get

$r_2m_2g = r_1w_1g.\nonumber$

Solve this for the unknown $$r_2$$:

$r_2 = r_1\dfrac{m_1}{m_2}.\nonumber$

The quantities on the right side of the equation are known; thus, $$r_2$$ is

\begin{align*} r_2 &= (1.60 \, m)\dfrac{26.0 \, kg}{32.0 \, kg} \\[5pt] &= 1.30 \, m \end{align*}

As expected, the heavier child must sit closer to the pivot (1.30 m versus 1.60 m) to balance the seesaw.

Solution (b)

This part asks for a force $$F_p$$. The easiest way to find it is to use the first condition for equilibrium, which is

$net \, F = 0.\nonumber$

The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as

$net \, F_y = 0 \nonumber$

where we again call the vertical axis the y -axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that

$F_p - w_1 - w_2 = 0.\nonumber$

This equation yields what might have been guessed at the beginning:

$F_p = w_1 + w_2. \nonumber$

So, the pivot supplies a supporting force equal to the total weight of the system:

$F_p = m_1g + m_2g. \nonumber$

Entering known values gives

\begin{align*} F_p &= (26.0 \, kg)(9.80 \, m/s^2) + (32.0 \, kg)(9.80 \, m/s^2) \\[5pt] &= 568 \, N. \end{align*}

The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other than the location of the seesaw’s actual pivot!

Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since $$F_p$$ is exerted on the pivot point, its lever arm is zero. Hence, the torque exerted by the supporting force $$F_p$$ is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem.

Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case . Always enter the correct forces—do not jump ahead to enter some ratio of masses.

Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation—the distances $$r_{\perp}$$ and $$r_2$$ are the distances to points directly below the center of gravity of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point.

Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion. We will examine this in the next chapter.

Take-Home Experiment

• Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies?
• The second condition assures those torques are also balanced. Torque is the rotational equivalent of a force in producing a rotation and is defined to be $\tau = r_{\perp}F \nonumber$ where $$\tau$$ is torque, $$r_{\perp}$$ is the perpendicular distance from the pivot point to the line containing the point where the force is applied, and $$F$$ is magnitude of the force.
• The perpendicular lever arm $$r_{\perp}$$ is the shortest distance from the pivot point to the line along which $$F$$ acts. The SI unit for torque is the newton-meter N$$\cdot$$m. The second condition necessary to achieve equilibrium is that the net external torque on a system must be zero: $net \, \tau = 0 \nonumber$ By convention, counterclockwise torques are positive, and clockwise torques are negative.

Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a  Creative Commons Attribution License (by 4.0) .

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## Chapter 15.3: Solving Equilibrium Problems

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## Learning Objective

• To solve quantitative problems involving chemical equilibriums.

There are two fundamental kinds of equilibrium problems: (1) those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and (2) those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.

## Calculating an Equilibrium Constant from Equilibrium Concentrations

We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of CaCO 3 (s) to CaO(s) and CO 2 (g) is K = [CO 2 ]. At 800°C, the concentration of CO 2 in equilibrium with solid CaCO 3 and CaO is 2.5 × 10 −3 M. Thus K at 800°C is 2.5 × 10 −3 . (Remember that equilibrium constants are unitless.)

A more complex example of this type of problem is the conversion of n -butane, an additive used to increase the volatility of gasoline, to isobutane (2-methylpropane). This reaction can be written as follows:

$$n-butane \left ( g \right ) \rightleftharpoons isobutane \left ( g \right ) \tag{15.3.1}$$

and the equilibrium constant K = [isobutane]/[ n -butane]. At equilibrium, a mixture of n -butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n -butane. Substituting these concentrations into the equilibrium constant expression,

$$K= \dfrac{isobutane}{n-butane}=\dfrac{0.041\;\cancel{M}}{0.016 \; \cancel{M}} = 2.6 \tag{15.3.2}$$

Thus the equilibrium constant for the reaction as written is 2.6.

## Example 15.3.1

The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:

$$2SO_{2}\left ( g \right ) + O_{2}\left ( g \right ) \rightleftharpoons 2SO_{3}\left ( g \right )$$

A mixture of SO 2 and O 2 was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10 −2 M SO 3 , 3.5 × 10 −3 M O 2 , and 3.0 × 10 −3 M SO 2 . Calculate K and K p at this temperature.

Given: balanced equilibrium equation and composition of equilibrium mixture

Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain K .

Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,

$$K=\dfrac{\left [ SO_{3} \right ]^{2}}{\left [ SO_{2} \right ]^{2}\left [ O_{2} \right ]}=\dfrac{\left ( 5.0\times 10^{-2} \right )^{2}}{\left ( 3.0\times 10^{-3} \right )^{2}\left ( 3.5\times 10^{-3} \right )}=7.9\times 10^{4}$$

To solve for K p , we use Equation 15.2.17 , where Δ n = 2 − 3 = −1:

$$K_{p}= K\left ( RT \right )^{\Delta n}$$ $$=7.9\times 10^{4}\left [ \left (0.082606\; L\cdot atm/mol\cdot \cancel{K} \right ) \left ( 800 \; \cancel{K} \right )\right ]$$ $$=1.2\times 10^{3}$$

Hydrogen gas and iodine react to form hydrogen iodide via the reaction

$$H_{2}\left ( g \right ) + I_{2}\left ( g \right ) \rightleftharpoons 2HI\left ( g \right )$$

A mixture of H 2 and I 2 was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10 −2 M HI, 6.47 × 10 −3 M H 2 , and 5.94 × 10 −4 M I 2 . Calculate K and K p for this reaction.

Answer: K = 48.8; K p = 48.8

Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example 9 shows one way to do this.

## Example 15.3.2

A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl 2 . Calculate K at this temperature. The equation for the decomposition of NOCl to NO and Cl 2 is as follows:

$$2NOCl \left ( g \right ) \rightleftharpoons 2NO\left ( g \right ) + Cl_{2}\left ( g \right )$$

Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium

A Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).

B Calculate all possible initial concentrations from the data given and insert them in the table.

C Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.

D Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.

A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:

$$K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}}$$

To obtain the concentrations of NOCl, NO, and Cl 2 at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.

B Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl] i = 1.00 mol/2.00 L = 0.500 M. The initial concentrations of NO and Cl 2 are 0 M because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl 2 in a 2.00 L container, so [Cl 2 ] f = 0.056 mol/2.00 L = 0.028 M. We insert these values into the following table:

C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl 2 , the substance for which initial and final concentrations are known:

Δ[Cl 2 ] = [0.028 M (final) − 0.00 M (initial)] = +0.028 M

According to the coefficients in the balanced chemical equation, 2 mol of NO are produced for every 1 mol of Cl 2 , so the change in the NO concentration is as follows:

$$\Delta \left [ NO \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=0.056\; M$$

Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl 2 produced, so the change in the NOCl concentration is as follows:

$$\Delta \left [ NOCl \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{-2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=-0.056\; M$$

We insert these values into our table:

D We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl:

[NO] f = 0.000 M + 0.056 M = 0.056 M [NOCl] f = 0.500 M + (−0.056 M) = 0.444 M

We can now complete the table:

We can now calculate the equilibrium constant for the reaction:

$$K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}}=\dfrac{\left ( 0.056 \right )^{2}\left ( 0.028 \right )}{0.444}^{2}=4.5\times 10^{-4}$$

The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH 3 ) by reacting 0.1248 M H 2 and 0.0416 M N 2 at about 500°C. At equilibrium, the mixture contained 0.00272 M NH 3 . What is K for the reaction N 2 + 3 H 2 ⇌ 2NH 3 at this temperature? What is K p ?

Answer: K = 0.105; K p = 2.61 × 10 −5

The original laboratory apparatus designed by Fritz Haber and Robert Le Rossignol in 1908 for synthesizing ammonia from its elements. A metal catalyst bed, where ammonia was produced, is in the large cylinder at the left. The Haber-Bosch process used for the industrial production of ammonia uses essentially the same process and components but on a much larger scale. Unfortunately, Haber’s process enabled Germany to prolong World War I when German supplies of nitrogen compounds, which were used for explosives, had been exhausted in 1914.

## Calculating Equilibrium Concentrations from the Equilibrium Constant

To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n -butane to isobutane (Equation 15.26), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n -butane, we can determine the concentration of n -butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example 9.

The initial concentrations of the reactant and product are both known: [ n -butane] i = 1.00 M and [isobutane] i = 0 M. We need to calculate the equilibrium concentrations of both n -butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as + x , then the change in the concentration of n -butane is Δ[ n -butane] = − x . This is because the balanced chemical equation for the reaction tells us that 1 mol of n -butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.

Substituting the expressions for the final concentrations of n -butane and isobutane from the table into the equilibrium equation,

$$K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{x}{1.00-x}=2.6$$

Rearranging and solving for x ,

$$x = 2.6\left ( 1.00-x \right )=2.6-2.6x$$ $$x + 2.6x =2.6$$ $$x = 0.72$$

We obtain the final concentrations by substituting this x value into the expressions for the final concentrations of n -butane and isobutane listed in the table:

[ n -butane] f = (1.00 − x ) M = (1.00 − 0.72) M = 0.28 M [isobutane] f = (0.00 + x ) M = (0.00 + 0.72) M = 0.72 M

We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same K that we used in the calculation:

$$K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{0.72 \; \cancel{M}}{0.28 \; \cancel{M}}=2.6$$

This is the same K we were given, so we can be confident of our results.

Example 10 illustrates a common type of equilibrium problem that you are likely to encounter.

## Example 15.3.3

The water–gas shift reaction is important in several chemical processes, such as the production of H 2 for fuel cells. This reaction can be written as follows:

$$H_{2}\left ( g \right ) + CO_{2}\left ( g \right ) \rightleftharpoons H_{2}O\left ( g \right ) + CO\left ( g \right )$$

K = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H 2 and 0.0150 M CO 2 is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?

Given: balanced equilibrium equation, K , and initial concentrations

A Construct a table showing what is known and what needs to be calculated. Define x as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of x . From the values in the table, calculate the final concentrations.

B Write the equilibrium equation for the reaction. Substitute appropriate values from the table to obtain x .

C Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain K .

A The initial concentrations of the reactants are [H 2 ] i = [CO 2 ] i = 0.0150 M. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H 2 O as x , then Δ[H 2 O] = + x . We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of x . For example, 1 mol of CO is produced for every 1 mol of H 2 O, so the change in the CO concentration can be expressed as Δ[CO] = + x . Similarly, for every 1 mol of H 2 O produced, 1 mol each of H 2 and CO 2 are consumed, so the change in the concentration of the reactants is Δ[H 2 ] = Δ[CO 2 ] = − x . We enter the values in the following table and calculate the final concentrations.

B We can now use the equilibrium equation and the given K to solve for x :

$$K=\dfrac{\left [ H_{2}O] \right ] \left [ CO \right ]}{\left [ H_{2} \right ]\left [ CO_{2} \right ]}=\dfrac{\left (x \right )\left ( x \right ) }{\left ( 0.0150-x \right )\left ( 0.0150-x \right )}=\dfrac{x^{2}}{\left ( 0.0150-x \right )^{2}}=0.160 \notag$$

We could solve this equation with the quadratic formula, but it is far easier to solve for x by recognizing that the left side of the equation is a perfect square; that is,

$\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106 \notag$

(The quadratic formula is presented in Essential Skills 7 in Section 15.7 .) Taking the square root of the middle and right terms,

$\dfrac{x^2}{(0.0150−x)^2} =(0.106)^{1/2}=0.326 \notag$

$x =(0.326)(0.0150)−0.326x \notag$

$1.326x=0.00489 \notag$

$x =0.00369=3.69 \times 10^{−3} \notag$

C The final concentrations of all species in the reaction mixture are as follows:

• $$[H_2]_f=[H_2]_i+Δ[H_2]=(0.0150−0.00369) \;M=0.0113\; M$$
• $$[CO_2]_f =[CO_2]_i+Δ[CO_2]=(0.0150−0.00369)\; M=0.0113\; M$$
• $$[H_2O]_f=[H_2O]_i+Δ[H_2O]=(0+0.00369) \;M=0.00369\; M$$
• $$[CO]_f=[CO]_i+Δ[CO]=(0+0.00369)\; M=0.00369 \;M$$

We can check our work by inserting the calculated values back into the equilibrium constant expression:

$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107 \notag$

To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.

Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:

$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)} \notag$

K = 54 at 425°C. If 0.172 M H 2 and I 2 are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?

Answer: [HI] f = 0.270 M; [H 2 ] f = [I 2 ] f = 0.037 M

In Example 10, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example 11.

## Example 15.3.4

In the water–gas shift reaction shown in Example 10, a sample containing 0.632 M CO 2 and 0.570 M H 2 is allowed to equilibrate at 700 K. At this temperature, K = 0.106. What is the composition of the reaction mixture at equilibrium?

Given: balanced equilibrium equation, concentrations of reactants, and K

Asked for: composition of reaction mixture at equilibrium

A Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations ( x ) and the final concentrations.

B Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for x .

C Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain K .

A [CO 2 ] i = 0.632 M and [H 2 ] i = 0.570 M. Again, x is defined as the change in the concentration of H 2 O: Δ[H 2 O] = + x . Because 1 mol of CO is produced for every 1 mol of H 2 O, the change in the concentration of CO is the same as the change in the concentration of H 2 O, so Δ[CO] = + x . Similarly, because 1 mol each of H 2 and CO 2 are consumed for every 1 mol of H 2 O produced, Δ[H 2 ] = Δ[CO 2 ] = − x . The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.

B We can now use the equilibrium equation and the known K value to solve for x :

$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106 \notag$

In contrast to Example 10, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:

$x^2 = 0.106(0.360 − 1.20x + x^2) \notag$

Collecting terms on one side of the equation,

$0.894x^2 + 0.127x − 0.0382 = 0 \notag$

This equation can be solved using the quadratic formula:

$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \notag$

$x =0.148 \text{ and } −0.290 \notag$

Only the answer with the positive value has any physical significance, so Δ[H 2 O] = Δ[CO] = +0.148 M, and Δ[H 2 ] = Δ[CO 2 ] = −0.148 M.

• $$[H_2]_f[ = [H_2]_i+Δ[H_2]=0.570 \;M −0.148\; M=0.422 M$$
• $$[CO_2]_f =[CO_2]_i+Δ[CO_2]=0.632 \;M−0.148 \;M=0.484 M$$
• $$[H_2O]_f =[H_2O]_i+Δ[H_2O]=0\; M+0.148\; M =0.148\; M$$
• $$[CO]_f=[CO]_i+Δ[CO]=0 M+0.148\;M=0.148 M$$

We can check our work by substituting these values into the equilibrium constant expression:

$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107 \notag$

Because K is essentially the same as the value given in the problem, our calculations are confirmed.

The exercise in Example 8 showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which K = 54 at 425°C. If a sample containing 0.200 M H 2 and 0.0450 M I 2 is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?

Answer: [HI] f = 0.0882 M; [H 2 ] f = 0.156 M; [I 2 ] f = 9.2 × 10 −4 M

In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small ( K ≤ 10 −3 ) or very large ( K ≥ 10 3 ), which means that the change in the concentration (defined as x ) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example 12.

## Example 15.3.5

Atmospheric nitrogen and oxygen react to form nitric oxide:

$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)} \notag$

K p = 2.0 × 10 −31 at 25°C. What is the partial pressure of NO in equilibrium with N 2 and O 2 in the atmosphere (at 1 atm, P{N 2 } = 0.78 atm and P{O 2 } = 0.21 atm

Given: balanced equilibrium equation and values of K p , P{O 2 } and P{N 2 }

Asked for: partial pressure of NO

A Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.

B Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration ( x ).

C Calculate the partial pressure of NO. Check your answer by substituting values into the equilibrium equation and solving for K .

A Because we are given K p and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of O 2 is 0.21 atm and that of N 2 is 0.78 atm. If we define the change in the partial pressure of NO as 2 x , then the change in the partial pressure of O 2 and of N 2 is − x because 1 mol each of N 2 and of O 2 is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.

B Substituting these values into the equation for the equilibrium constant,

$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31} \notag$

In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the x value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − x ) = 0.78 and (0.21 − x ) = 0.21. Substituting these expressions into our original equation,

$\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31} \notag$

$\dfrac{4x^2}{0.16} =2.0 \times10^{−31} \notag$

$x^2=\dfrac{0.33 \times 10^{−31}}{4} \notag$

$x^=9.1 \times 10^{−17} \notag$

C Substituting this value of x into our expressions for the final partial pressures of the substances,

• $$P_{NO}=2x \; atm=1.8 \times 10^{−16} \;atm$$
• $$P_{N_2}=(0.78−x) \;atm=0.78 \;atm$$
• $$P_{O_2}=(0.21−x) \;atm=0.21\; atm$$

From these calculations, we see that our initial assumption regarding x was correct: given two significant figures, 2.0 × 10 −16 is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if x is less than about 5% of the total, or 10 −3 > K > 10 3 , then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic NO, an ingredient of smog, does not form from atmospheric concentrations of N 2 and O 2 to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:

$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 times 10^{−31} \notag$

The final K p agrees with the value given at the beginning of this example.

Under certain conditions, oxygen will react to form ozone, as shown in the following equation:

$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \notag$

K p = 2.5 × 10 −59 at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere P(O 2 ) =0.21 atm ?

Answer: 4.8 × 10 −31 atm

Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large ( K ≥ 10 3 ). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example 13.

## Example 15.3.6

The chemical equation for the reaction of hydrogen with ethylene (C 2 H 4 ) to give ethane (C 2 H 6 ) is as follows:

K = 9.6 × 10 18 at 25°C. If a mixture of 0.200 M H 2 and 0.155 M C 2 H 4 is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?

Given: balanced chemical equation, K , and initial concentrations of reactants

A Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.

B Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for x (the change in concentration).

C Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.

A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define − x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is + x . The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.

B Substituting values into the equilibrium constant expression,

$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18} \notag$

Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus x is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + x ) = 0.045 and (0.155 − x ) = 0.155] as follows:

$K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18} \notag$

$x=3.6 \times 10^{−19} \notag$

C The small x value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:

• $$[C_2H_6]_f = (0.155 − x)\; M = 0.155 \; M$$
• $$[C_2H_4]_f = x\; M = 3.6 \times 10^{−19} M$$
• $$[H_2]_f = (0.045 + x) \;M = 0.045 \;M$$

We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:

$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18} \notag$

This K value agrees with our initial value at the beginning of the example.

Hydrogen reacts with chlorine gas to form hydrogen chloride:

$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \notag$

K p = 4.0 × 10 31 at 47°C. If a mixture of 0.257 M H 2 and 0.392 M Cl 2 is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?

• $$[H_2]_f = 4.8 \times 10^{−32}\; M$$
• $$[Cl_2]_f = 0.135\; M$$
• $$[HCl]_f = 0.514\; M$$

When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.

## Key Takeaway

• Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture.

## Conceptual Problems

Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.

Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when K is (a) very large and (b) very small? Illustrate this technique using the system A + 2B ⇌ C for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?

## Numerical Problems

Please be sure you are familiar with the topics discussed in Essential Skills 7 ( Section 15.7 ) before proceeding to the Numerical Problems.

In the equilibrium reaction A + B ⇌C, what happens to K if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction A ⇌B + C?

The following table shows the reported values of the equilibrium P{O 2 } at three temperatures for the reaction Ag 2 O(s) ⇌ 2 Ag(s) + 1/2 O 2 (g) for which Δ H ° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?

Given the equilibrium system N 2 O 4 (g) ⇌ 2 NO 2 (g), what happens to K p if the initial pressure of N 2 O 4 is doubled? If K p is 1.7 × 10 −1 at 2300°C, and the system initially contains 100% N 2 O 4 at a pressure of 2.6 × 10 2 atm, what is the equilibrium pressure of each component?

At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium according to the following equation: H 2 (g) + I 2 (g) ⇌2HI(g) At equilibrium, [H 2 ] = 0.047 M and [HI] = 0.345 M. What are K and K p for this reaction?

Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: CO(g) + 2H 2 (g) ⇌ CH 3 OH(g) and K p = 1.3 × 10 −4 . If 56.0 g of CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?

Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction A (s ⇌ 2 B(g) + C(g), what is K p ?

The decomposition of ammonium carbamate to NH 3 and CO 2 at 40°C is written as NH 4 CO 2 NH 2 (s) ⇌ 2NH 3 (g) + CO 2 If the partial pressure of NH 3 at equilibrium is 0.242 atm, what is the equilibrium partial pressure of CO 2 ? What is the total gas pressure of the system? What is K p ?

At 375 K, K p for the reaction SO 2 Cl 2 (g) ⇌ SO 2 (g) + Cl 2 g) is 2.4, with pressures expressed in atmospheres. At 303 K, K p is 2.9 × 10 −2 .

• What is K for the reaction at each temperature?
• If a sample at 375 K has 0.100 M Cl 2 and 0.200 M SO 2 at equilibrium, what is the concentration of SO 2 Cl 2 ?
• If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?

For the gas-phase reaction aA ⇌ bB, show that K p = K ( RT ) Δ n assuming ideal gas behavior.

For the gas-phase reaction I 2 ⇌2I, show that the total pressure is related to the equilibrium pressure by the following equation:

$P_T=\sqrt{K_pP_{I_2}} + P_{I_2} \notag$

Experimental data on the system Br2(l) ⇌ Br 2 (aq) are given in the following table. Graph [Br 2 ] versus moles of Br 2 (l) present; then write the equilibrium constant expression and determine K .

Data accumulated for the reaction n- butane(g) ⇌ isobutane(g) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n -butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n -butane and isobutane?

Solid ammonium carbamate (NH 4 CO 2 NH 2 ) dissociates completely to ammonia and carbon dioxide when it vaporizes:

$NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)} \notag$

At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is K p ? If the concentration of CO 2 is doubled and then equilibrates to its initial equilibrium partial pressure + x atm, what change in the NH 3 concentration is necessary for the system to restore equilibrium?

The equilibrium constant for the reaction COCl 2 (g) ⇌ CO(g) + Cl 2 (g) is K p = 2.2 × 10 −10 at 100°C. If the initial concentration of COCl 2 is 3.05 × 10 −3 M, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?

Aqueous dilution of IO 4 − results in the following reaction:

$IO^−_{4(aq)}+2H_2O_{(l)} \rightleftharpoons H_4IO^−_{6(aq)} \notag$

and K = 3.5 × 10 −2 . If you begin with 50 mL of a 0.896 M solution of IO 4 − that is diluted to 250 mL with water, how many moles of H 4 IO 6 − are formed at equilibrium?

Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation:

$I_{2(g)}+Br_{2(g)} \rightleftharpoons 2IBr_{(g)} \notag$

K p = 1.2 × 10 2 . If you begin the reaction with 7.4 g of I 2 vapor and 6.3 g of Br 2 vapor in a 1.00 L container, what is the concentration of IBr(g) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?

For the reaction

$C_{(s)} + 12N_{2(g)}+\frac{5}{2}H_{2(g)} \rightleftharpoons CH3NH2(g) \notag$

K = 1.8 × 10 −6 . If you begin the reaction with 1.0 mol of N 2 , 2.0 mol of H 2 , and sufficient C(s) in a 2.00 L container, what are the concentrations of N 2 and CH 3 NH 2 at equilibrium? What happens to K if the concentration of H 2 is doubled?

## Contributors

Modified by Joshua B. Halpern

#### IMAGES

1. Torque Seesaw Balance Example (Static Equilibrium

2. Torque

3. Torque Example #3: Leaning Ladder Problem

4. Static Equilibrium Problem with Angled Torque

5. Torque & Rotational Equilibrium

6. Static Equilibrium Sample Problem 2

1. 9: Statics and Torque (Exercises)

Repeat the seesaw problem in Example with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium. Solution

2. Torque Practice Problems with Solutions: AP Physics 1

Problem (1): In each of the following diagrams, calculate the torque (magnitude and direction) about point O O due to the force \vec {F} F of magnitude 10\,\rm N 10N applied to a 4-\rm m 4− m rod. Both the force \vec {F} F and the rode lie in the plane of the page.

3. PDF Lecture 8 Torque

Method 1: If you're given r and θ, use formula for torque (magnitude) τ = r F sinθ (Note: sinθ = sinφ, ∴ it doesn't matter which angle you use) Calculating torque (2) Method 2: If you're given d the "perpendicular distance" from axis to the "line of action", then use formula τ = d F

4. 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.

5. PDF Physics 101:Lecture 15 Torque, F=ma for rotation, and Equilibrium

r COM θ θ Fparallel F CW rotation is negative Two ways to compute torque: 1. Put r and F vectors tail-to-tail and compute = rFsinq. F r 2. Decompose F into components parallel and perpendicular to r, and take: t = rF

6. Torque and equilibrium review (article)

Key terms [If work and torque have the same units, does that mean torque is energy?] Equations How to visualize the torque equation A wrench produces a torque on a nut if a force is applied to it correctly (see Figure 1). The equation for torque is: τ = r F sin θ Figure 1. Variables of the torque equation shown for a wrench and nut.

7. Torque Equilibrium

Torque Equilibrium Examples. Most equilibrium problems require the application of force as well as torque for their solution, but the examples below illustrate equilibrium of torque. Index. Torque concepts. HyperPhysics ***** Mechanics ***** Rotation. R Nave.

8. 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...

9. Torque & Equilibrium Practice Problems

1 PRACTICE PROBLEM A female worker in the workplace has a safe limit to lift objects weighing 160 N (16 kg) at places of work. The worker can lift weights greater than 160 N using a trolley. You may consider a 1.5 m long trolley, weighing 100 N, and the center of gravity located at 0.6 m from the axle.

10. Torque (article)

The concept of rotational equilibrium is particularly useful in problems involving multiple torques acting on a rotatable object. In this case it is the net torque which is important. If the net torque on a rotatable object is zero then it will be in rotational equilibrium and not able to acquire angular acceleration.

11. Torque and the Second Condition of Equilibrium

General strategy when solving equilibrium problems. Draw an extended free body diagram, "EFBD." Sum up the forces in the "x" and "y" directions. ... Wherever the point is located, all force that point towards or away from the point will not exert a torque about this point. Mathematically this means that the force in toque equaiton will dissappear.

12. Solving Torque Equilibrium Problems: How to Find Tension and Satisfy

1. Explain how it is possible for a large force to produce a small torque, and how it is possible for a small force to produce a large torque? 2. Find the tension force in the cord supporting the meter stick for each part of the experiment.

13. Torque & Equilibrium Video Tutorial & Practice

Energy in Connected Objects (Systems) Adding Mass to a Moving System. Collisions & Motion (Momentum & Energy) Collisions with Springs. Intro to Center of Mass. More Connect Wheels (Bicycles) Learn Torque & Equilibrium with free step-by-step video explanations and practice problems by experienced tutors.

14. Torque & Equilibrium Video Tutorial & Practice

A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw of FIGURE EX12.32. H... 7. Friction, Inclines, Systems. 8. Centripetal Forces & Gravitation. Learn Torque & Equilibrium with free step-by-step video explanations and practice problems by experienced tutors.

15. Rotational Equilibrium

Today, we look at how to solve rotational static equilibrium problems in physics. These are problems such as a balance beam or a lever arm or a draw bridge. ...

16. Solved Example Problems for Torque

Solution The force F at the point of suspension is due to the weight of the hanging mass. We can solve this problem by three different methods. Method - I The angle (θ) between the arm length (r) and the force (F) is, θ = 150o The torque (τ) about the fixed point of the arm is, Method - II

17. 5.3: The Second Condition for Equilibrium

If the second condition (net external torque on a system is zero) is satisfied for one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest. The second condition necessary to achieve equilibrium is stated in equation form as. netτ = 0 (5.3.2) (5.3.2) n e t τ = 0.

18. Physics: Mastering Static Equilibrium with Tension, Torque, and More

Tension and torque are essential concepts in understanding the stability of objects in physics. Understanding torque and tension in static equilibrium can help solve complex physics problems involving multiple variables and equations. The ability to calculate tension forces and components of forces can lead to a deeper understanding of physics ...

19. Chapter 15.3: Solving Equilibrium Problems

The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define −x as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is +x ...