Rotational Equilibrium and Newton's First Law
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AP Physics C: Mechanics › Rotational Equilibrium and Newton's First Law
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?
$0.07\ \text{m}$ to the right of the pivot
$0.47\ \text{m}$ to the left of the pivot
$0.47\ \text{m}$ to the right of the pivot
$0.20\ \text{m}$ to the right 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.
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.
$W=320,\text{N}$, since $300+W=0$ for equilibrium.
$W=200,\text{N}$, since $300(0.80)=W(1.2)$.
$W=288,\text{N}$, since $300(1.2)=W(1.0)$.
$W=450,\text{N}$, since $300(1.2)=W(0.80)$.
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.
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?
$\sum F=0$ only, regardless of torque
$\sum \tau_{\text{pivot}}\neq 0$ if $\sum F=0$
$\sum \tau_{\text{pivot}}=0$ and $\sum F=0$
$I\alpha=\sum F$ about 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 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.
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?
Move the sign closer to the right end.
Decrease $\theta$ so $T\cos\theta$ increases.
Increase $\theta$ so $T\sin\theta$ increases.
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.
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.
$T_R=75,\text{N}$ downward, to oppose the load’s downward torque.
$T_R=300,\text{N}$ upward, since $\sum F_y=0$ requires $T_R=300$.
$T_R=1200,\text{N}$ upward, from $T_R(1.0)=300(4.0)$.
$T_R=75,\text{N}$ upward, from $T_R(4.0)=300(1.0)$.
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 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.
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.
$F=100,\text{N}$ upward, to increase clockwise torque on the right.
$F=500,\text{N}$ downward, since $250+200=F$ for equilibrium.
$F=100,\text{N}$ downward, since $250(1.0)=200(1.0)+F(0.50)$.
$F=50,\text{N}$ downward, since $250(1.0)=200(0.50)+F(1.0)$.
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 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.
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.
$T=\dfrac{120+200}{\sin 40^\circ}=498,\text{N}$.
$T=\dfrac{(120)(1.5)+(200)(3.0)}{3.0\cos 40^\circ}=524,\text{N}$.
$T=\dfrac{(120)(3.0)+(200)(1.5)}{3.0\sin 40^\circ}=402,\text{N}$.
$T=\dfrac{(120)(1.5)+(200)(3.0)}{3.0\sin 40^\circ}=402,\text{N}$.
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.