AP Physics C Mechanics Quiz: Rotational Equilibrium And Newtons First Law

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Rotational Equilibrium And Newtons First LawQuestion 1 of 20

A uniform ladder of mass $m$ and length $L$ leans against a perfectly smooth (frictionless) vertical wall at an angle $\theta$ with the horizontal floor. The floor is rough and provides a static friction force that prevents the ladder from slipping.

Which expression represents the magnitude of the torque produced by the wall's normal force ( $N_w$ ) about the point where the ladder contacts the floor?

N_w L \sin\theta

N_w L \cos\theta

N_w L

Zero, because the wall is frictionless.

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Rotational Equilibrium And Newtons First Law

Practice Rotational Equilibrium And Newtons First Law in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rotational Equilibrium And Newtons First Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which expression represents the magnitude of the torque produced by the wall's normal force ( $N_w$ ) about the point where the ladder contacts the floor?

$N_w L \sin\theta$ (correct answer)
$N_w L \cos\theta$
$N_w L$
Zero, because the wall is frictionless.

Explanation: The torque is given by

\tau = rF\sin\phi

, where

r

is the distance from the pivot to the point of force application,

F

is the force, and

\phi

is the angle between the position vector and the force vector. Here, the pivot is the base of the ladder,

r=L

, and

F = N_w

. The angle between the ladder (position vector) and the horizontal force

N_w

(180^\circ - \theta)

. The lever arm of the force

N_w

about the base is

L\sin\theta

. Thus, the torque is

N_w (L\sin\theta)

Question 2

A non-uniform bar of length $L$ and mass $M$ has its center of mass at a distance $L/4$ from one end, End A. The bar is to be balanced on a single pivot.

Where must the pivot be placed for the bar to be in static equilibrium?

At a distance of $L/4$ from End A. (correct answer)
At a distance of $L/2$ from End A.
At a distance of $3L/4$ from End A.
The pivot position depends on the orientation of the bar.

Explanation: For the bar to be in static equilibrium under its own weight, the pivot must provide a normal force that exactly opposes the gravitational force, and there must be zero net torque. To achieve zero net torque from gravity, the pivot must be placed directly under the center of mass. This way, the lever arm for the gravitational force is zero, resulting in zero torque.

Question 3

A uniform plank of mass $M$ and length $L$ rests on two supports. Support 1 is at the left end ( $x=0$ ) and Support 2 is at the right end ( $x=L$ ). A block of mass $m$ is placed at position $x = L/3$ .

What is the magnitude of the upward force $N_1$ exerted by Support 1 for the plank to be in equilibrium?

$mg/3 + Mg/2$
$2mg/3 + Mg/2$ (correct answer)
$mg/3 + Mg$
$2mg/3 + Mg$

Explanation: To find

N_1

, we sum the torques about Support 2 (

x=L

) and set them to zero. The clockwise torques are from the plank's weight

Mg

x=L/2

and the block's weight

mg

x=L/3

. The counter-clockwise torque is from

N_1

x=0

. The lever arms relative to

x=L

are

L

L/2

, and

2L/3

respectively. So,

N_1(L) - Mg(L/2) - mg(2L/3) = 0

. Solving for

N_1

gives

N_1 = Mg/2 + 2mg/3

Question 4

A helicopter's main rotor is rotating at a constant angular velocity of 300 revolutions per minute during level flight at a constant altitude and velocity.

Which of the following statements best describes the net torque on the rotor?

There is a non-zero constant net torque that maintains the constant angular velocity.
The net torque is zero because the angular velocity is constant. (correct answer)
The net torque is non-zero and increasing to overcome air resistance.
The net torque must be equal to the torque provided by the engine to maintain the rotation.

Explanation: According to Newton's first law for rotation, an object will maintain a constant angular velocity if and only if the net external torque acting on it is zero. The engine provides a forward torque, which is exactly balanced by the resistive torque from air drag, resulting in a net torque of zero and thus a constant angular velocity.

Question 5

A yo-yo rests on a horizontal table. A person pulls the string horizontally to the right with a constant tension $T$ . The string is wrapped around the inner axle of radius $r$ . The outer radius of the yo-yo is $R$ . Static friction between the yo-yo and the table prevents slipping.

For the yo-yo to be in rotational equilibrium while the string is pulled, what must be the magnitude of the static friction force $f_s$ ?

$f_s = T$
$f_s = T(r/R)$ (correct answer)
$f_s = T(R/r)$
$f_s = 0$

Explanation: For rotational equilibrium, the net torque about the center of mass must be zero. The tension

T

creates a torque

Tr

(let's say clockwise). The static friction force

f_s

acts at the point of contact with the table (radius

R

) and must create an equal and opposite (counter-clockwise) torque

f_s R

. Therefore,

Tr = f_s R

, which gives

f_s = T(r/R)

Question 6

A door is free to rotate on its hinges. A person pushes on the door with a constant force $F$ perpendicular to the door's surface, causing it to swing open with a constant angular velocity.

Which statement accurately describes the torques acting on the door?

The person applies the only torque, which causes the constant angular velocity.
The net torque is zero, as the person's applied torque is balanced by a frictional torque in the hinges. (correct answer)
The net torque is constant and non-zero, as required for any motion.
The torque from the hinges is greater than the person's applied torque, preventing acceleration.

Explanation: According to Newton's first law for rotation, if an object rotates with a constant angular velocity, its angular acceleration is zero. This implies that the net torque acting on it must be zero. The torque applied by the person must be exactly balanced by an opposing frictional torque from the hinges for the net torque to be zero.

Question 7

A balanced mobile uses a light horizontal bar pivoted at its center, staying at rest. A 2.0 N weight hangs 0.40 m left of the pivot, and a 1.0 N weight hangs 0.10 m right of the pivot. A third weight of 1.5 N is moved along the right side to reestablish equilibrium. Each weight exerts a downward force vector, and each produces torque equal to $rF$ about the pivot. The system is in rotational equilibrium, so the sum of clockwise and counterclockwise torques is zero, consistent with Newton's First Law in rotational form. Considering the forces acting on the object, where should the 1.5 N weight be placed (distance from the pivot on the right) to maintain equilibrium?

$0.47\ \text{m}$ to the right of the pivot (correct answer)
$0.20\ \text{m}$ to the right of the pivot
$0.07\ \text{m}$ to the right of the pivot
$0.47\ \text{m}$ to the left of the pivot

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the net torque about the pivot is zero, meaning clockwise and counterclockwise torques balance perfectly. In this scenario, the 2.0 N weight creates counterclockwise torque (2.0 × 0.40 = 0.80 N·m), while the 1.0 N weight creates clockwise torque (1.0 × 0.10 = 0.10 N·m), leaving a net counterclockwise torque of 0.70 N·m that must be balanced by the 1.5 N weight. Choice A is correct because setting up the balance equation: 0.80 = 0.10 + 1.5 × d, where d is the distance right of the pivot, gives 1.5d = 0.70, so d = 0.70/1.5 = 0.467 m ≈ 0.47 m. Choice C (0.07 m) might result from arithmetic errors, while choice D incorrectly places the weight on the left side. To help students: Create a torque inventory table listing each force, its lever arm, and resulting torque with proper signs, always verify that the sum of all torques equals zero in the final configuration, and use dimensional analysis to ensure the answer has correct units.

Question 8

A seesaw pivots at its center and stays level while two students sit on opposite sides. A $300\,\text{N}$ student sits $1.2\,\text{m}$ to the left of the pivot, and a second student sits to the right with an unknown weight $W$ at a distance $0.80\,\text{m}$ . The forces act vertically downward at their seats, creating torques with lever arms measured from the pivot. The support force at the pivot produces no torque about the pivot. The seesaw is motionless, so Newton's First Law in rotational form requires $\sum \tau_{\text{pivot}}=0$ . Considering the forces acting on the seesaw, calculate the force needed at a specific point to maintain equilibrium.

$W=200\,\text{N}$ , since $300(0.80)=W(1.2)$ .
$W=450\,\text{N}$ , since $300(1.2)=W(0.80)$ . (correct answer)
$W=320\,\text{N}$ , since $300+W=0$ for equilibrium.
$W=288\,\text{N}$ , since $300(1.2)=W(1.0)$ .

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a seesaw with two students sitting on opposite sides of a central pivot, where we need to find the unknown weight to maintain balance. Choice B is correct because applying torque balance about the pivot: clockwise torque = counterclockwise torque, so 300 N × 1.2 m = W × 0.80 m, giving W = 360/0.80 = 450 N. Choice A is incorrect because it reverses the lever arms in the calculation, using 300(0.80) = W(1.2) instead of the correct relationship. To help students: Draw a clear diagram showing the pivot, forces, and lever arms. Emphasize that torque = force × perpendicular distance from pivot, and practice setting up the torque balance equation systematically with proper signs for clockwise vs counterclockwise torques.

Question 9

A seesaw is in rotational equilibrium about a frictionless pivot, remaining horizontal. A 500 N person sits 1.0 m to the left of the pivot, and a 250 N person sits 2.0 m to the right. Both forces act downward, and their lever arms are measured from the pivot along the plank. The plank's weight acts at the pivot and produces negligible torque. The torques balance so that clockwise torque equals counterclockwise torque, consistent with Newton's First Law in rotational form ( $\sum \tau=0$ ). Based on the system described, which condition must be met for the system to remain in rotational equilibrium?

$\sum \tau_{\text{pivot}}=0$ and $\sum F=0$ (correct answer)
$\sum F=0$ only, regardless of torque
$I\alpha=\sum F$ about the pivot
$\sum \tau_{\text{pivot}}\neq 0$ if $\sum F=0$

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium requires that both the net torque and net force on an object equal zero, ensuring no linear or angular acceleration. In this scenario, the seesaw experiences torques from two people: 500 N × 1.0 m = 500 N·m counterclockwise and 250 N × 2.0 m = 500 N·m clockwise, which balance perfectly. Choice A is correct because for complete static equilibrium, both conditions must be satisfied: Στ = 0 (no angular acceleration) and ΣF = 0 (no linear acceleration). Choice B is incorrect because torque balance is essential for rotational equilibrium, choice C incorrectly applies the rotational dynamics equation to a static situation, and choice D contradicts the equilibrium condition. To help students: Emphasize that static equilibrium requires two separate conditions to be met simultaneously, practice identifying all forces and their points of application, and reinforce that torque depends on both force magnitude and lever arm distance.

Question 10

A 1.8 m uniform beam is hinged at the left end and supports a 180 N sign at the right end. The beam's weight is 90 N acting at its center. A cable attaches to the beam 1.8 m from the hinge and pulls at an angle $\theta$ above the beam, providing an upward component that creates counterclockwise torque. The beam remains horizontal, so clockwise and counterclockwise torques must balance. With zero angular acceleration, Newton's First Law in rotational form requires $\sum \tau_{\text{hinge}}=0$ . Based on the system described, which change increases the counterclockwise torque from the cable without changing the cable tension magnitude?

Increase $\theta$ so $T\sin\theta$ increases. (correct answer)
Decrease $\theta$ so $T\cos\theta$ increases.
Move the sign closer to the right end.
Increase the beam's mass to raise $I$ .

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium requires the net torque about any point to be zero, with torque depending on both the force magnitude and its perpendicular distance from the pivot. In this scenario, the cable creates counterclockwise torque through its vertical component T sin(θ), which must balance the clockwise torques from the beam and sign weights. Choice A is correct because increasing θ increases sin(θ), thereby increasing the vertical component of the tension and its torque without changing the tension magnitude T. Choice B is incorrect because cos(θ) relates to the horizontal component which doesn't contribute to torque about a horizontal axis, choice C would increase clockwise torque making balance harder, and choice D incorrectly references moment of inertia which doesn't affect static torque. To help students: Use vector decomposition to identify which force components create torque, remember that torque depends on the perpendicular distance between the force line and pivot, and practice varying one parameter while holding others constant to see the effect.

Question 11

A horizontal beam of length $4.0\,\text{m}$ is held in static equilibrium by two vertical support cables at its ends. A $300\,\text{N}$ load hangs $1.0\,\text{m}$ from the left end, and the beam's own weight is negligible. The upward tensions $T_L$ and $T_R$ act at the left and right ends, respectively, and the load's weight acts downward at its attachment point. Taking torques about the left end, the lever arm for $T_R$ is $4.0\,\text{m}$ and for the load is $1.0\,\text{m}$ . The beam does not rotate, so Newton's First Law in rotational form implies $\sum \tau_{\text{left}}=0$ . Considering the forces acting on the object, calculate the force needed at a specific point to maintain equilibrium.

$T_R=75\,\text{N}$ upward, from $T_R(4.0)=300(1.0)$ . (correct answer)
$T_R=1200\,\text{N}$ upward, from $T_R(1.0)=300(4.0)$ .
$T_R=300\,\text{N}$ upward, since $\sum F_y=0$ requires $T_R=300$ .
$T_R=75\,\text{N}$ downward, to oppose the load's downward torque.

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a horizontal beam with a load closer to the left support, requiring us to find the right support tension using torque balance. Choice A is correct because taking torques about the left end: clockwise torque from load = counterclockwise torque from right tension, so 300 N × 1.0 m = TR × 4.0 m, giving TR = 300/4.0 = 75 N upward. Choice C is incorrect because it assumes equal sharing of the load, ignoring that the load is closer to the left support, which must carry more of the weight. To help students: Demonstrate that the support closer to the load carries more weight. Practice choosing convenient pivot points (like one support) to eliminate unknown forces from the torque equation.

Question 12

A seesaw is level about a central pivot while two downward forces act at different distances. A $250\,\text{N}$ force acts $1.0\,\text{m}$ to the left of the pivot, and a $200\,\text{N}$ force acts $1.0\,\text{m}$ to the right. A third downward force $F$ is added $0.50\,\text{m}$ to the right of the pivot, changing the net torque. The pivot contact force acts at the pivot and produces no torque about it. The board remains at rest only if the clockwise and counterclockwise torques balance, consistent with Newton's First Law in rotational form. Based on the system described, calculate the force needed at a specific point to maintain equilibrium.

$F=100\,\text{N}$ downward, since $250(1.0)=200(1.0)+F(0.50)$ . (correct answer)
$F=50\,\text{N}$ downward, since $250(1.0)=200(0.50)+F(1.0)$ .
$F=100\,\text{N}$ upward, to increase clockwise torque on the right.
$F=500\,\text{N}$ downward, since $250+200=F$ for equilibrium.

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a seesaw with two initial forces creating an imbalance, requiring a third force to restore equilibrium. Choice A is correct because taking torques about the pivot: counterclockwise torque from left = clockwise torques from right, so 250 N × 1.0 m = 200 N × 1.0 m + F × 0.50 m, giving F = (250-200)/0.50 = 100 N downward. Choice C is incorrect because an upward force on the right would create counterclockwise torque, worsening the imbalance rather than correcting it. To help students: Emphasize the importance of force direction in determining torque direction. Practice identifying whether a force creates clockwise or counterclockwise torque based on its position relative to the pivot and its direction.

Question 13

A uniform $3.0\,\text{m}$ beam supports a hanging sign and is pinned to a wall at the left end, remaining horizontal. The beam's weight is $120\,\text{N}$ acting at its center, and the sign's weight is $200\,\text{N}$ acting at the right end. A cable attaches to the right end and makes a $40^\circ$ angle above the beam, exerting tension $T$ along its force vector. Taking torques about the wall pin, the pin's reaction forces create no torque, while the weights create clockwise torque balanced by the cable's counterclockwise torque. The system is static, so Newton's First Law in rotational form requires $\sum \tau_{\text{pin}}=0$ . Considering the forces acting on the object, calculate the force needed at a specific point to maintain equilibrium.

$T=\dfrac{(120)(1.5)+(200)(3.0)}{3.0\sin 40^\circ}=402\,\text{N}$ . (correct answer)
$T=\dfrac{(120)(1.5)+(200)(3.0)}{3.0\cos 40^\circ}=524\,\text{N}$ .
$T=\dfrac{(120)(3.0)+(200)(1.5)}{3.0\sin 40^\circ}=402\,\text{N}$ .
$T=\dfrac{120+200}{\sin 40^\circ}=498\,\text{N}$ .

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a beam pinned to a wall with its weight and a sign's weight creating clockwise torques that must be balanced by the cable's tension. Choice A is correct because the vertical component of tension (T sin 40°) creates the counterclockwise torque: T sin 40° × 3.0 m = 120 N × 1.5 m + 200 N × 3.0 m, giving T = 780/(3.0 sin 40°) = 402 N. Choice B is incorrect because it uses cos 40° instead of sin 40° for the vertical component of tension that creates the torque. To help students: Draw force diagrams showing tension components and emphasize that only the perpendicular component of a force contributes to torque. Practice decomposing angled forces and identifying which component creates torque about the chosen pivot.

Question 14

A system consists of a uniform disk that can rotate about its center. An external agent applies a torque $\tau(t) = (4.0 \text{ N} \cdot \text{m/s})t - 8.0 \text{ N} \cdot \text{m}$ . The disk is initially rotating with a constant angular velocity.

At what time $t$ is the disk momentarily in rotational equilibrium?

$t = 0.5$ s
$t = 1.0$ s
$t = 2.0$ s (correct answer)
$t = 4.0$ s

Explanation: An object is in rotational equilibrium when the net torque acting on it is zero, which means its angular acceleration is zero. Setting the applied torque to zero:

\tau(t) = 0

. So

(4.0)t - 8.0 = 0

, which gives

4t = 8

, so

t = 2.0

s. At this moment, the net torque is zero and the disk is momentarily in rotational equilibrium.

Question 15

A uniform meter stick of mass $M$ is supported by a pivot at the 50 cm mark. A mass $m_1 = 2M$ is placed at the 20 cm mark.

At which mark must a second mass $m_2 = 3M$ be placed to keep the meter stick in static equilibrium?

the 60 cm mark
the 70 cm mark (correct answer)
the 80 cm mark
the 90 cm mark

Explanation: For static equilibrium, the net torque about the pivot must be zero. Let the pivot be the origin (50 cm). The torque from

m_1

is counter-clockwise:

\tau_1 = m_1 g r_1 = (2M)g(50-20\text{ cm})

. The torque from

m_2

is clockwise:

\tau_2 = -m_2 g r_2 = -(3M)g(x-50\text{ cm})

. Setting the sum of torques to zero:

(2M)g(30) - (3M)g(x-50) = 0

. This simplifies to

60 = 3(x-50)

, so

20 = x-50

, and

x = 70

cm.

Question 16

A rigid body is subjected to a set of external forces. The vector sum of these forces is found to be zero.

Which of the following motions is impossible for the body?

The body is at rest and remains at rest.
The body is moving with a constant linear velocity.
The body is rotating with a constant angular velocity.
The body is moving with a constant non-zero linear acceleration. (correct answer)

Explanation: If the vector sum of external forces is zero (

\sum F = 0

), the body is in translational equilibrium. This means its linear acceleration must be zero (

\sum F = ma

). Therefore, it is impossible for the body to have a constant non-zero linear acceleration. It can be at rest, move with constant velocity, or rotate, as long as the net torque condition is also met for that state of rotational motion.

Question 17

A uniform rod of length $L$ and weight $W$ is hinged to a wall at one end. A horizontal force $F$ is applied to the other end to hold the rod at an angle $\theta$ above the horizontal.

For the rod to be in rotational equilibrium, what must be the relationship between the applied force $F$ and the weight $W$ ?

$F L \cos\theta = W \frac{L}{2} \sin\theta$
$F L \sin\theta = W \frac{L}{2} \cos\theta$ (correct answer)
$F L = W \frac{L}{2}$
$F L \cos\theta = W L \cos\theta$

Explanation: To be in rotational equilibrium, the net torque about the hinge must be zero. The weight

W

acts at the center of the rod (

L/2

) and its lever arm is

(L/2)\cos\theta

, creating a clockwise torque. The force

F

acts at the end (

L

) and its lever arm is

L\sin\theta

, creating a counter-clockwise torque. Setting the magnitudes of these torques equal:

F(L\sin\theta) = W(\frac{L}{2}\cos\theta)

Question 18

A body is in static equilibrium.

Which of the following statements must be true?

No forces are acting on the body.
The body must be unaccelerated, but it may have a net torque acting on it.
The net force and net torque on the body are both zero. (correct answer)
The body may be accelerating linearly, but its angular acceleration must be zero.

Explanation: Static equilibrium is defined by two conditions. First, the object must have zero linear acceleration, which means the vector sum of all external forces is zero (translational equilibrium). Second, the object must have zero angular acceleration, which means the vector sum of all external torques about any point is zero (rotational equilibrium).

Question 19

A student holds a bicycle wheel by its axle and gets it spinning rapidly. The student then sits on a stool that is free to rotate and holds the wheel with its axle vertical.

The student-stool-wheel system is now rotating with a constant angular velocity. What can be concluded about the net torque on the system?

A net torque from the spinning wheel keeps the system rotating.
The net torque on the system is zero. (correct answer)
A net torque must be supplied by the student to maintain the rotation.
The net torque is non-zero and points vertically.

Explanation: The system is rotating with a constant angular velocity. According to the rotational version of Newton's first law, if the angular velocity of a system is constant, its angular acceleration is zero, and therefore the net external torque acting on the system must be zero. Frictional torques in the stool's axle are assumed to be negligible in this idealized scenario.

Question 20

A uniform beam of mass $M$ and length $L$ is pivoted at its center. Two forces are applied: a force $F_1 = 20\text{ N}$ downward at a distance $L/3$ from the pivot, and a force $F_2$ upward at a distance $L/4$ from the pivot on the opposite side. If the beam remains in rotational equilibrium, what is the magnitude of $F_2$ ?

$15\text{ N}$
$26.7\text{ N}$ (correct answer)
$30\text{ N}$
$80\text{ N}$

Explanation: For rotational equilibrium, the net torque about the pivot must be zero. Taking counterclockwise as positive:

\tau_{net} = F_2 \cdot \frac{L}{4} - F_1 \cdot \frac{L}{3} = 0

. Substituting values:

F_2 \cdot \frac{L}{4} = 20 \cdot \frac{L}{3}

. Solving:

F_2 = \frac{20 \cdot 4}{3} = \frac{80}{3} = 26.7\text{ N}

. Choice A uses

F_2 = F_1 \cdot \frac{3}{4}

incorrectly. Choice C assumes equal lever arms. Choice D multiplies instead of dividing by the ratio.