Rotational Equilibrium
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
An extended object at rest needs two conditions: ΣF = 0 (no linear acceleration) and Στ = 0 (no angular acceleration). The torque condition holds about any point, so choose the one that makes the algebra easiest.
1. Two conditions
2. A routine that works
1. Draw every force where it acts on the object.
2. Take torques about the point where the most unknowns act: their torques vanish.
3. Use ΣF = 0 for the rest.
Ladder against a frictionless wall: the wall pushes horizontally, the floor pushes up (N) and sideways (friction). Taking torques about the base gives the wall force; then friction equals it.
3. Example, then try it
A 25 kg child sits 1.6 m from a seesaw’s pivot. Where must a 40 kg child sit to balance?
Torques balance: (25)(10)(1.6) = (40)(10)d, so d = 1.0 m on the other side.
A uniform 4.0 m, 20 kg beam rests on supports at both ends. A 60 kg person stands 1.0 m from the left end (the figure below). Find both support forces. (g = 10 m/s²)
Torques about the left end: NR(4.0) = (200)(2.0) + (600)(1.0) = 1000 N·m.
Right support: NR = 250 N.
Forces: NL + NR = 200 + 600, so NL = 550 N.
Check (torques about the right end): NL(4.0) = (200)(2.0) + (600)(3.0) = 2200, so NL = 550 N ✓.
Answer: NL = 550 N, NR = 250 N. The support nearer the person carries more.
4. Common mistakes
Why is it smart to take torques about the point where an unknown force acts?
A force acting at the pivot has r = 0, so its torque is zero. The torque equation then has one fewer unknown.
Ready to practice?
4 questions on Rotational Equilibrium, each with a hint and an explanation for every wrong answer.