AP Physics 1 · Lesson 3

Equilibrium and Newton’s second law for rotation

Know Know it

An object in static equilibrium has a net force of zero and a net torque of zero about every axis. In practice: write ∑F=0\sum F = 0 for each direction, and ∑τ=0\sum\tau = 0 about one well-chosen point. Choose the point where an unknown force acts, so that force drops out of the torque equation.

Treat the weight of a uniform beam as acting at its middle. Then add up the clockwise and counterclockwise torques and set them equal.

When the net torque is not zero, the object’s rotation changes: Newton’s second law for rotation is τnet=Iα\tau_{\text{net}} = I\alpha. More torque gives more angular acceleration; more rotational inertia gives less.

Worked example Easy

A light rod 0.80 m long hangs level from a string tied at its 0.30 m mark. A 0.50 kg toy hangs from the rod’s left end, at the 0 m mark. What mass hung from the right end, at the 0.80 m mark, keeps the rod level?

  1. 0.19 kg
  2. 0.30 kgAnswer
  3. 0.50 kg
  4. 0.83 kg

How to solve it

  1. The left arm is 0.30 m and the right arm is 0.80 - 0.30 = 0.50 m.
  2. Torques balance: (0.50 kg)(g)(0.30)=m(g)(0.50)(0.50 \text{ kg})(g)(0.30) = m(g)(0.50); g cancels.
  3. m=0.15/0.50=0.30m = 0.15/0.50 = 0.30 kg.

Why each choice is right or wrong

  • A. Incorrect. 0.19 kg uses the whole rod length, 0.80 m, as the right-hand lever arm; the arm is 0.50 m from the string.
  • B. Correct. Torques balance about the string: (0.50)(0.30)=m(0.50)(0.50)(0.30) = m(0.50), so m=0.30m = 0.30 kg.
  • C. Incorrect. 0.50 kg matches the toy’s mass, which balances only when the two lever arms are equal.
  • D. Incorrect. 0.83 kg swaps the lever arms, giving the heavier mass to the longer side.

Trap Rotational equilibrium does not mean not rotating

Zero net torque means the angular velocity is constant, which includes spinning at a steady rate. And a nonzero net force with zero net torque, or the reverse, is not equilibrium: a static object needs both sums to be zero.

Explain Explain it

Task verbs: calculate, explain

For a beam on two supports, take torques about one support to find the other support’s force, then use ∑F=0\sum F = 0 to find the first. Check the answer: the support nearer a heavy load should push harder.

For a pulley with mass, treat each object separately: Newton’s second law for each block, and τnet=Iα\tau_{\text{net}} = I\alpha for the pulley, linked by a=rαa = r\alpha when the string does not slip. The tensions on the two sides of the pulley differ, because their difference supplies the pulley’s torque.

Worked example Medium

A uniform footbridge plank is 4.0 m long and weighs 600 N. It rests on supports at its two ends. A 900 N hiker stands 1.0 m from the left end. What upward force does the right support exert?

  1. 225 N
  2. 525 NAnswer
  3. 750 N
  4. 975 N

How to solve it

  1. Take torques about the left support, so its force drops out.
  2. Clockwise: plank weight (600)(2.0)=1,200(600)(2.0) = 1{,}200 N·m and hiker (900)(1.0)=900(900)(1.0) = 900 N·m. Counterclockwise: FR(4.0)F_R(4.0).
  3. FR=2,100/4.0=525F_R = 2{,}100/4.0 = 525 N; then FL=1,500−525=975F_L = 1{,}500 - 525 = 975 N.

Why each choice is right or wrong

  • A. Incorrect. 225 N counts only the hiker’s torque and leaves out the plank’s own 600 N weight at its middle.
  • B. Correct. About the left support: FR(4.0)=(600)(2.0)+(900)(1.0)=2,100F_R(4.0) = (600)(2.0) + (900)(1.0) = 2{,}100 N·m, so FR=525F_R = 525 N.
  • C. Incorrect. 750 N splits the 1,500 N total equally, which is right only if the load is centered.
  • D. Incorrect. 975 N is the left support’s force; the hiker is nearer the left end, so the left support pushes harder.

Worked example Hard

A 1.5 kg block hangs from a light cord wound around a pulley that is a uniform solid disk of mass 6.0 kg (I=12MR2I = \frac{1}{2}MR^2), turning on a frictionless axle. The block is released and the cord does not slip. What is the block’s acceleration?

  1. 2.0 m/s²
  2. 2.5 m/s²
  3. 3.3 m/s²Answer
  4. 9.8 m/s²

How to solve it

  1. Block: mg−T=mamg - T = ma.
  2. Pulley: τ=TR=Iα=12MR2(a/R)\tau = TR = I\alpha = \frac{1}{2}MR^2(a/R), so T=12Ma=3.0aT = \frac{1}{2}Ma = 3.0a.
  3. Add: mg=(m+12M)amg = (m + \frac{1}{2}M)a, so a=(1.5)(9.8)/(1.5+3.0)≈3.3a = (1.5)(9.8)/(1.5 + 3.0) \approx 3.3 m/s².

Why each choice is right or wrong

  • A. Incorrect. 2.0 m/s² treats the pulley’s whole 6.0 kg as if it moved with the block; a disk acts like only half its mass.
  • B. Incorrect. 2.5 m/s² divides the block’s weight by the pulley’s mass alone and leaves out the block’s own mass.
  • C. Correct. For the block, mg−T=mamg - T = ma; for the pulley, TR=12MR2(a/R)TR = \frac{1}{2}MR^2(a/R), so T=12MaT = \frac{1}{2}Ma. Then a=mg/(m+12M)=14.7/4.5≈3.3a = mg/(m + \frac{1}{2}M) = 14.7/4.5 \approx 3.3 m/s².
  • D. Incorrect. 9.8 m/s² is free fall, which ignores the cord’s tension; the pulley must be spun up, so the cord pulls back on the block.

Explain Your turn

A string runs over a pulley that has mass and turns on a frictionless axle. A heavier block hangs on one side and a lighter block on the other, and the heavier block speeds up as it falls. Explain why the string tensions on the two sides of the pulley cannot be equal.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says the pulley has an angular acceleration (its spin speeds up)
  • Says a nonzero net torque is needed because the pulley has rotational inertia
  • Says the tension on the heavier block’s side must be larger

Model answer

As the blocks speed up, the pulley spins faster and faster, so it has an angular acceleration. Because the pulley has mass, its rotational inertia is not zero, and by τnet=Iα\tau_{\text{net}} = I\alpha it needs a net torque to speed up. The only forces with a torque about the axle are the two tensions, which act at the same radius, so they must differ: the tension on the heavier block’s side must be larger.

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