Unit 5.3 | Rotational Equilibrium
- 17 minutes
- 2 formulas
- 20 questions
- 1 videos
Most torque questions on the AP exam are really equilibrium questions. Three steps solve all of them: pick the pivot, write every torque, sum and solve.
Unit 5 Breakdown
In this lesson:
- Newton’s first law for rotation: zero net torque means constant angular velocity
- The two conditions for static equilibrium: [katex] \sum F = 0 [/katex] and [katex] \sum \tau = 0 [/katex]
- A 3-step method that solves every equilibrium problem
- The 4 problem types: seesaws, beams on two supports, hinged beams with a cable, and ladders
Introduction
This lesson covers rotational equilibrium: what happens when the torques on an object cancel, and how to use that to find unknown forces.
In Unit 5.2 you added torques. Now look at the case where they add to zero. Bridges, shelves, cranes, ladders and seesaws all live here, and most “torque” questions on the AP exam are really equilibrium questions.
Newton’s First Law in Rotational Form
An object with zero net torque keeps its angular velocity. If it is not turning, it stays still. If it is turning, it keeps turning at the same rate. A spinning wheel does not need a torque to keep spinning, just as a sliding puck does not need a force to keep sliding.
[katex] \sum \tau = 0 \;\Rightarrow\; \omega = \text{constant} [/katex]
Zero net torque does not mean “not rotating”. It means “not changing its rotation”.
Static Equilibrium: Two Conditions
An object that stays at rest is in static equilibrium. It must not start sliding and it must not start turning, so it needs both:
[katex] \sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum \tau = 0 [/katex]
Zero net force is not enough. Push the left side of a book up and the right side down with equal forces: the forces cancel, but the book spins. The torques do not cancel.
The pivot trick
When an object is in equilibrium, the net torque is zero about every point. So you choose the pivot. Choose it where an unknown force acts: that force has no lever arm, so it drops out of the torque equation.
The 3-Step Method
- Pick the pivot. Put it where an unknown force acts. Any force through the pivot has zero lever arm and drops out.
- Write every torque. For each force, [katex] \tau = rF\sin\theta [/katex] with a sign: counterclockwise +, clockwise −. Put each object’s weight at its center of mass.
- Sum and solve. Set [katex] \sum\tau = 0 [/katex]. If an unknown is left, use [katex] \sum F = 0 [/katex] for it.
Every rotational equilibrium problem, from ladders to seesaws to cranes, is these three steps with different geometry. The four types below are the ones the AP exam uses.
| Problem type | The clue | The setup |
| 1. Seesaws and levers | a board on a fulcrum, “where must she sit” | Pivot at the fulcrum: [katex] m_1 g\,d_1 = m_2 g\,d_2 [/katex] |
| 2. Beams on two supports | two chains, strings or pillars | Pivot at one support, so the other is the only unknown |
| 3. Hinged beams with a cable | a wall hinge and a cable at an angle | Pivot at the hinge, so only [katex] T\sin\theta [/katex] makes torque |
| 4. Ladders | a ladder against a frictionless wall | Pivot at the base; floor friction = wall force |
Problem Type 1: Seesaws and levers
How to spot it: A board or rod balanced on a fulcrum with weights on both sides: “where must she sit”, “how far from the fulcrum”, “what mass balances it”.
Setup: Pivot at the fulcrum. Counterclockwise torques = clockwise torques: [katex] m_1 g\,d_1 = m_2 g\,d_2 [/katex]. If the board has mass and the fulcrum is not at its middle, add the board’s weight at its center.
A [katex] 30\,\text{kg} [/katex] child sits [katex] 2.0\,\text{m} [/katex] from the fulcrum of a seesaw. The fulcrum is at the middle of a uniform board. Where must her [katex] 40\,\text{kg} [/katex] brother sit to balance it?
Pivot at the fulcrum. The board’s weight acts right at the fulcrum, so it makes no torque. [katex] (30)g(2.0) = (40)g\,d [/katex], so [katex] d = 1.5 \text{ m} [/katex] on the other side. Notice that [katex] g [/katex] cancels.
AP trap: Forgetting the board’s own weight when the fulcrum is not at its middle. The weight then acts at the board’s center, some distance from the fulcrum, and adds a torque.
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Problem Type 2: Beams on two supports
How to spot it: A board or beam is held at two points: two chains, two strings, two pillars, or a scale and a pivot. The question asks for a support force or where a load sits.
Setup: Put the pivot at the support you are not asked about. The other support force is then the only unknown in [katex] \sum\tau = 0 [/katex]. Use [katex] \sum F_y = 0 [/katex] for the second support.
A uniform [katex] 6.0\,\text{m} [/katex] plank weighing [katex] 200\,\text{N} [/katex] rests on supports at its two ends. A [katex] 600\,\text{N} [/katex] painter stands [katex] 2.0\,\text{m} [/katex] from the left end. Find both support forces.
Pivot at the left support. Counterclockwise: [katex] F_R(6.0) [/katex]. Clockwise: [katex] (200)(3.0) + (600)(2.0) = 1800 \text{ N·m} [/katex]. So [katex] F_R = 300 \text{ N} [/katex]. Then [katex] \sum F_y = 0 [/katex] gives [katex] F_L = 800 – 300 = 500 \text{ N} [/katex]. The painter is closer to the left end, so the left support carries more.
AP trap: Putting the pivot in the middle of the beam out of habit. Then both support forces stay in the torque equation and you have two unknowns.
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Problem Type 3: Hinged beams held by a cable
How to spot it: A beam or shelf is attached to a wall by a hinge (or pin) and held up by a cable, wire or strut at an angle. A sign or crate may hang from it.
Setup: Pivot at the hinge: the hinge force has an unknown size and direction, and it drops out. The cable’s torque is [katex] TL\sin\theta [/katex], with [katex] \theta [/katex] between the beam and the cable. Then [katex] \sum F = 0 [/katex] gives the hinge force.
A uniform [katex] 2.0\,\text{m} [/katex] shelf of mass [katex] 10\,\text{kg} [/katex] is hinged to a wall and held level by a cable from its far end to the wall, at [katex] 30^\circ [/katex] above the shelf. Find the tension and the force from the hinge.
Pivot at the hinge: [katex] T(2.0)\sin 30^\circ = (10)(9.8)(1.0) [/katex], so [katex] T = 98 \text{ N} [/katex]. Vertical forces: [katex] H_y + T\sin 30^\circ = 98 [/katex], so [katex] H_y = 49 \text{ N} [/katex] up. Horizontal forces: the cable pulls the shelf toward the wall, so the hinge pushes it away with [katex] H_x = T\cos 30^\circ \approx 85 \text{ N} [/katex].
AP trap: Assuming the hinge force is zero, or that it points along the beam. Here the hinge holds up half the shelf’s weight and also pushes out from the wall.
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Problem Type 4: Ladders
How to spot it: A ladder leans against a “smooth” or “frictionless” wall on a rough floor: “minimum coefficient of friction”, “how high can the painter climb”, “force from the wall”.
Setup: Four forces: the weight at the middle, the floor’s normal force [katex] N [/katex] and friction [katex] f [/katex] at the base, and the wall’s push [katex] N_w [/katex] at the top. Pivot at the base. Then [katex] \sum F_x = 0 [/katex] gives [katex] f = N_w [/katex] and [katex] \sum F_y = 0 [/katex] gives [katex] N = Mg [/katex].
A uniform [katex] 6.5\,\text{m} [/katex] ladder of mass [katex] 15\,\text{kg} [/katex] leans against a frictionless wall. Its base is [katex] 2.5\,\text{m} [/katex] from the wall and its top is [katex] 6.0\,\text{m} [/katex] up. Find the force from the wall and the friction from the floor.
Use lever arms. The weight [katex] (15)(9.8) = 147 \text{ N} [/katex] acts straight down at the middle, a horizontal distance of [katex] 1.25\,\text{m} [/katex] from the base. The wall force is horizontal, so its lever arm is the height, [katex] 6.0\,\text{m} [/katex]. [katex] N_w(6.0) = (147)(1.25) [/katex], so [katex] N_w \approx 31 \text{ N} [/katex]. Then [katex] f = N_w \approx 31 \text{ N} [/katex], pointing toward the wall, and [katex] N = 147 \text{ N} [/katex].
AP trap: Drawing floor friction pointing away from the wall. The base of the ladder wants to slide away from the wall, so friction pushes it back toward the wall.
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Lesson 5.4 Preview
In the final lesson the torques no longer cancel. We meet rotational inertia and use Newton’s second law for rotation, [katex] \sum\tau = I\alpha [/katex], on pulleys and swinging rods.
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This article was written by Jason