5.5 Rotational equilibrium (Newton's first law for rotation)

Four little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. The playground seesaw

Read the story as text

You (30 kg) and your big brother (45 kg) want to play on a seesaw. When you both sit at the very ends, his side slams to the ground and you are stuck in the air.

He shuffles in toward the middle. At one special spot the seesaw floats level and stays there, even though he is heavier than you.

Where is that spot, and how much is the pivot pushing up on the seesaw then?

2. Further out turns it more

Quick check: a 30 kg kid sits 1.0 m from the pivot. Where must a 15 kg kid sit, on the other side, to balance?

3. The pivot holds everyone up

Quick check: a balanced seesaw holds a 30 kg kid and a 20 kg kid. The plank is 10 kg. How hard does the pivot push up?

4. Check yourself

Think of your answer first, then tap to see it.

a) A 20 kg child sits 2.0 m from the middle of a seesaw. Where must a 40 kg child sit to balance?

Show answer

1.0 m from the middle, on the other side. Twice the mass, half the distance: 20 × 2.0 = 40 × 1.0.

b) A seesaw floats level and stays still. What is the net torque on it? The net force?

Show answer

Both are zero. Not starting to turn means Στ = 0. Not starting to move means ΣF = 0.

c) The seesaw is balanced. Then the small kid slides further out. What happens?

Show answer

Her end goes down. Her weight now acts further from the pivot, so her turning effect (torque) is bigger than his.

Already know this?

Three quick questions. Get all three right first time and you can skip ahead.

A 25 kg child sits 2.0 m from the middle of a seesaw. Where must a 50 kg adult sit to balance?

An object stays at rest, not moving and not turning. Which must be true?

In a balance problem, a smart choice of axis (pivot point for torques) is:

An object that is not speeding up its spin, and not speeding up in a straight line, has zero net torque and zero net force.

Two conditions

In pictures

  • A seesaw that stays level is not turning: the turning effects on the two sides cancel.
  • A turning effect is weight × distance from the pivot. A small kid far out can balance a big kid close in.
  • The pivot pushes up as hard as everything pulls down, so the seesaw does not fall either.
  • No turn and no fall: that is equilibrium.

Below is the same idea in words, and then with numbers.

An object is in static equilibrium when it stays at rest: no linear acceleration and no angular acceleration. That needs two things at once:

ΣF = 0   and   Στ = 0

ΣF = 0 stops it from moving off. Στ = 0 stops it from starting to turn. This is Newton's first law in rotational form: with zero net torque, ω stays constant (here, zero). The same rules hold for an object turning at constant ω, which is also "rotational equilibrium".

30 kg45 kg N (pivot) 294 N441 N 1.5 m1.0 m

How to solve any balance problem:

  1. Draw the object and every force, at the point where it acts. A uniform beam's weight acts at its middle.
  2. Pick an axis. Any point works when Στ = 0, so pick the point where an unknown force acts: its torque is zero and it drops out.
  3. Write Στ = 0 with signs (counterclockwise +). Solve.
  4. Write ΣF = 0 (up = down, left = right) to find the rest.

With numbers: the seesaw

The seesaw's own weight acts at the pivot (it is uniform and pivoted in the middle), so it makes no torque. Counterclockwise is positive.

Show all steps as text
  1. Your torque: +(30 kg)(9.8 m/s²)(1.5 m) = +441 N·m.Your weight on the left turns it counterclockwise.
  2. Brother at distance x on the right: −(45)(9.8)x.His weight turns it clockwise.
  3. Στ = 0: 441 − 441x = 0, so x = 1.0 m.g cancels: m₁x₁ = m₂x₂, so 30 × 1.5 = 45 × 1.0.
  4. Pivot force: ΣF = 0, N = (30 + 45 + 10 kg seesaw)(9.8) = 833 N up.The pivot holds up everything, including the 10 kg plank.

Balance the seesaw

A 4.0 m, 10 kg plank on a central pivot 0.50 m high. Set each child's mass and distance from the pivot. The scene shows each torque and the net torque. Press Play to release the seesaw: if the net torque is not zero, it tips until one end hits the ground. Start values are balanced; change one and predict.

Try: make the right child 60 kg. Where must they sit? (30 × 1.5 = 60 × x, so x = 0.75 m.) Then set it and press Play: the graph stays flat.

Worked examples

Where to sit basic

A 20 kg child sits 2.0 m from the centre of a seesaw. Where should a 40 kg child sit to balance?

Show solution
Show all steps as text
  1. m₁x₁ = m₂x₂: (20)(2.0) = (40)x, so x = 1.0 m on the other side.Twice the mass, half the distance.

Plank on two supports medium

A uniform 4.0 m plank of mass 12 kg rests on supports at its two ends. A 60 kg person stands 1.0 m from the left end. Find both support forces.

Show solution
Show all steps as text
  1. Weights: plank (12)(9.8) = 117.6 N at 2.0 m; person (60)(9.8) = 588 N at 1.0 m.Uniform plank: weight at the middle.
  2. Axis at the left support (removes FL): FR(4.0) − 117.6(2.0) − 588(1.0) = 0.FR turns counterclockwise; the weights clockwise.
  3. FR = (235.2 + 588) / 4.0 = 206 N.Solve for the one unknown.
  4. ΣF = 0: FL + 206 = 117.6 + 588, so FL = 500 N.Check: the left support is closer to the person, so it carries more. ✓

Off-centre metre stick medium

A 0.20 kg uniform metre stick balances on a pivot at the 40 cm mark when a 0.10 kg mass hangs from it. Where is the mass?

Show solution
Show all steps as text
  1. The stick's weight acts at 50 cm, which is 0.10 m right of the pivot: torque −(0.20)(9.8)(0.10) = −0.196 N·m.When the pivot is not at the middle, the beam's own weight makes a torque.
  2. The mass must be on the left at distance d: +(0.10)(9.8)d.It has to turn the other way.
  3. 0.98d = 0.196, so d = 0.20 m: at the 20 cm mark.40 cm − 20 cm.

Hinged beam and cable: all the forces AP

A uniform 2.0 m beam weighing 50 N sticks out horizontally from a wall hinge. A cable from its far end goes up to the wall at 30° above the horizontal. Find the tension and the hinge force components.

Show solution
Show all steps as text
  1. Torques about the hinge: T(2.0) sin 30° − 50(1.0) = 0, so T = 50 N.Hinge force has no torque about the hinge.
  2. Horizontal: the cable pulls toward the wall with T cos 30° = 43.3 N, so the hinge pushes away from the wall with H = 43 N.ΣFx = 0.
  3. Vertical: V + T sin 30° − 50 = 0, so V = 50 − 25 = 25 N up.ΣFy = 0.
  4. Check with torques about the beam's far end: V(2.0) − 50(1.0) = 0 gives V = 25 N. ✓Any axis must give the same answer in equilibrium.

Practice

  1. A 25 kg child sits 2.0 m from the centre of a balanced seesaw. A 50 kg adult on the other side must sit:

    Show answer

    (25)(2.0) = (50)x → x = 1.0 m.

  2. An object is in static equilibrium. Which must be true?

    Show answer

    Both: no linear acceleration needs ΣF = 0, no angular acceleration needs Στ = 0. Forces can act; they just balance.

  3. A person walks along a plank resting on two supports, moving toward the right support. What happens to the support forces?

    Show answer

    ΣF = 0 fixes the total (it equals the total weight). Torques about the left support: as the person's lever arm grows, FR must grow, so FL shrinks.

  4. A light metre stick pivots at the 50 cm mark. A 2.0 N weight hangs at the 10 cm mark. Where must a 4.0 N weight hang to balance it?

    Show answer

    (2.0)(0.40) = (4.0)d → d = 0.20 m on the other side: 50 + 20 = 70 cm. (30 cm is on the same side as the 2.0 N weight, which would not balance.)

  5. Can an object have zero net torque but a non-zero net force?

    Show answer

    Push a box right through its centre: the line of action passes through the centre of mass, so there is no torque about it, but ΣF ≠ 0 and the box accelerates. The two conditions are independent.

  6. In a beam problem, why do we usually take torques about the point where an unknown force acts?

    Show answer

    In equilibrium Στ = 0 about every point, so we are free to choose. Choosing where an unknown acts removes it (r = 0) and leaves fewer unknowns.

  7. Short answer. A painter (70 kg) stands 1.0 m from the left end of a uniform 3.0 m scaffold plank (20 kg) supported at both ends. Find the force from each support.

    Show answer

    Torques about the left end: FR(3.0) = (20)(9.8)(1.5) + (70)(9.8)(1.0) = 294 + 686 = 980 N·m, so FR = 327 N.

    ΣF = 0: FL = (90)(9.8) − 327 = 882 − 327 = 555 N.

  8. Short answer. A uniform 4.0 m, 20 kg plank rests on supports at x = 0 (left end) and x = 3.0 m, so 1.0 m hangs over. How far past the right support can a 60 kg person walk before the plank tips?

    Show answer

    At the point of tipping the left support force becomes zero and the plank pivots on the right support.

    Torques about x = 3.0 m: plank weight (20 g) acts at 2.0 m, 1.0 m to the left: 20g(1.0). Person at d to the right: 60g d.

    20(1.0) = 60d, so d = 0.33 m past the support (x ≈ 3.33 m). Any farther and the net torque tips the plank.

AP question types for this topic

Four short free response questions, one of each AP Physics 1 type: Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. Try each one on paper before you open the worked answer.

1. Mathematical Routines (5 points)

Sofia (mass m1) sits a distance d1 from the centre of a uniform seesaw pivoted at its middle. Her brother Luca (mass m2) wants to sit on the other side so it balances level.

  1. Derive an expression for Luca's distance d2 from the pivot.
  2. m1 = 30 kg, d1 = 1.2 m, m2 = 40 kg. Calculate d2.
  3. The seesaw plank has mass 10 kg. Calculate the force the pivot exerts on the plank when it balances.
Show worked answer and scoring

(a) Take torques about the pivot (the plank's weight and the pivot force have zero lever arm). Net torque zero: m1g d1 = m2g d2, so d2 = m1d1/m2.

(b) d2 = 30 × 1.2/40 = 0.90 m.

(c) Net force zero: N = (10 + 30 + 40) × 9.8 = 784 N ≈ 780 N, upward.

Scoring (5 points): (a) 2 points: 1 point for net torque zero about the pivot, 1 point for the expression. (b) 1 point for 0.90 m. (c) 2 points: 1 point for net force zero including the plank, 1 point for 784 N.

2. Translation Between Representations (6 points)

A uniform 20 kg plank, 4.0 m long, rests on supports A and B at its two ends. Chen, mass 60 kg, walks slowly from A to B. x is his distance from A.

FBx
A
FBx
B
FBx
C
FBx
D

In words: Four graphs of the force from support B against x. A is a straight line rising from the origin. B is a straight line rising from a value above zero. C is a straight line falling. D is a flat line.

  1. Which graph (A to D) shows the force FB from support B against x? Justify.
  2. Describe the graph of FA against x, and what FA + FB equals at every x.
  3. Calculate FA and FB when Chen is 1.0 m from A.
Show worked answer and scoring

(a) B. Torques about A: FBL = Mg(L/2) + mgx, so FB = Mg/2 + (mg/L)x. That is a straight line with positive slope that starts at Mg/2 = 98 N, not at zero, because B holds half the plank even when Chen is at A.

(b) FA is a straight line falling from 686 N at x = 0 to 98 N at x = L. FA + FB = (20 + 60) × 9.8 = 784 N at every x (net force zero), so the two lines are mirror images.

(c) FB = 98 + 588 × 1.0/4.0 = 245 N. FA = 784 − 245 = 539 N. Chen is nearer A, so A carries more.

Scoring (6 points): (a) 2 points: 1 point for B, 1 point for the torque reason with the nonzero start. (b) 2 points: 1 point for FA falling linearly, 1 point for the constant 784 N sum. (c) 2 points: 1 point for each force.

3. Experimental Design and Analysis (6 points)

Mira wants the mass M of a small rock. She balances a metre stick on a pivot at its centre and hangs the rock 0.10 m to the left of the pivot. She hangs a known mass m on the right and slides it to the distance x where the stick balances level.

m (kg)0.0500.1000.1500.2000.250
x (m)0.3000.1510.0990.0750.061

In words: Masses of 0.050, 0.100, 0.150, 0.200 and 0.250 kilograms balanced at 0.300, 0.151, 0.099, 0.075 and 0.061 metres.

  1. Why does she balance the stick at its centre first, and how should she judge that it is level?
  2. What should she graph to get a straight line? What does the slope equal?
  3. Use the data to find the slope and the rock's mass.
Show worked answer and scoring

(a) With the pivot at the stick's centre of mass, the stick's own weight has zero lever arm, so only the two hanging masses make torque. Judge level with a spirit level, or by lining the stick up with a horizontal mark, and nudge the mass both ways to find the balance point.

(b) Balance: mgx = Mg(0.10), so x = (0.10 M)(1/m). Graph x against 1/m: a line through the origin with slope = 0.10 M.

(c) 1/m = 20, 10, 6.7, 5.0, 4.0 kg⁻¹. Slope ≈ (0.300 − 0.061)/(20 − 4.0) ≈ 0.0149 kg·m, so M = 0.0149/0.10 ≈ 0.15 kg.

Scoring (6 points): (a) 2 points: 1 point for the stick's weight having no torque, 1 point for a level check. (b) 2 points: 1 point for x against 1/m, 1 point for slope = 0.10 M. (c) 2 points: 1 point for the slope, 1 point for M ≈ 0.15 kg.

4. Qualitative/Quantitative Translation (6 points)

Kofi holds a 5.0 kg bag of rice in his hand with his forearm horizontal. His biceps attaches to the forearm 0.040 m from the elbow, and the bag hangs 0.35 m from the elbow. He claims: “My biceps must pull with far more force than the bag weighs.” (Ignore the forearm's own weight; treat the biceps pull as vertical.)

  1. Explain in words why Kofi is right.
  2. Derive an expression for the biceps force in terms of m, g, the bag distance d and the biceps distance b, and evaluate it.
  3. Compare the biceps force with the bag's weight, and connect the ratio to your explanation in (a).
Show worked answer and scoring

(a) The forearm is a lever pivoting at the elbow. The bag's weight acts far from the pivot but the biceps acts very close to it. To make an equal and opposite torque with a tiny lever arm, the biceps needs a large force.

(b) Torques about the elbow: Fbiceps b = mg d, so Fbiceps = mgd/b = 5.0 × 9.8 × 0.35/0.040 ≈ 430 N.

(c) The bag weighs 5.0 × 9.8 = 49 N; 429/49 = 8.75 = d/b. The force ratio is exactly the lever arm ratio from (a): 0.35 m is 8.75 times 0.040 m, so the biceps pulls 8.75 times the weight.

Scoring (6 points): (a) 2 points: 1 point for torques about the elbow, 1 point for the short biceps lever arm. (b) 2 points: 1 point for F = mgd/b, 1 point for about 430 N. (c) 2 points: 1 point for the ratio 8.75, 1 point for linking it to d/b.

Common mistakes

The mistake: "balanced means equal masses on each side".

Why it is wrong: balance needs equal torques, m₁x₁ = m₂x₂, so a heavier person sits closer.

How to spot it: the distances are different.

The mistake: leaving out the beam's own weight.

Why it is wrong: unless the pivot is exactly under the beam's centre, its weight makes a torque (acting at the middle).

How to spot it: the problem gives the beam's mass, or calls it "uniform".

The mistake: using only Στ = 0 and forgetting ΣF = 0 (or the reverse).

Why it is wrong: you need both to find all unknown forces, e.g. the pivot or hinge force.

How to spot it: you still have an unknown force left after the torque equation.

The mistake: giving every torque a plus sign.

Why it is wrong: torques that turn opposite ways must cancel to give zero.

How to spot it: your Στ = 0 equation has all + terms, so it cannot equal zero.