high school physics Torque & Rotational Equilibrium

Torque & Rotational Equilibrium

A door opens more easily when you push far from the hinges because torque grows with lever arm and the force component perpendicular to it. Mechanical equilibrium needs both net force and net torque zero, and on a seesaw that means m₁d₁ = m₂d₂. Heavier loads sit closer to the pivot, lighter ones farther out, to balance. Change the right mass's distance and watch when the seesaw levels out.

The same force does different things depending on where you apply it

🚪 What is torque?

①Push a door at the handle (far from the hinge) and it opens easily

②Torque = distance from the axis × perpendicular component of the force

③For the same force, farther from the axis and more perpendicular means more torque

Balance the seesaw

Distance of right weight from pivot d₂3 m

⚖️ When is it balanced?

①Left torque = 2 kg × 10 × 2 m = 40 N·m (fixed)

②Right torque = 1 kg × 10 × d₂

③Balance is horizontal when the two torques match. Check d₂ on the slider

④Pivot normal force is ΣF=0. This sketch: 20 N + 10 N = 30 N

Torque and the equilibrium conditions

Torque

τ = F · d = F · r sinθ

F=force, d=perpendicular distance from axis to the line of force (=r sinθ)

Conditions for mechanical equilibrium

ΣF = 0 and Στ = 0

both the net force and the net torque must be zero to stay at rest (or move uniformly)

Lever (seesaw) balance

m₁ d₁ = m₂ d₂

the heavier side sits closer, the lighter side farther

From concept to problem

Example 1 — seesaw balance

A 40 kg person sits 2 m to the left of the pivot. How many kg must sit 4 m to the right to balance the seesaw?

Equilibrium: left torque = right torque.

m₁ d₁ = m₂ d₂

Substitute and solve for m₂.

40 × 2 = m₂ × 4 → m₂ = 20

▸ 20 kg

Sitting twice as far from the pivot balances with half the mass.

Example 2 — tension in the rope holding a beam

A uniform beam (length 4 m, weight 60 N) is hinged to a wall at one end; a rope pulls up at the other end to keep it horizontal. Find the tension. (the rope is vertical at the beam end)

Take the hinge as the axis, so the hinge force contributes no torque.

Στ = 0

Weight acts at the center (2 m), tension at the end (4 m).

T × 4 = 60 × 2 → T = 30

▸ 30 N

Choosing the axis where an unknown force acts removes it from the equation and simplifies the work.

Exam Points

exam-style

A light (massless) beam rests on a pivot. A weight W hangs at the left end (2 m from the pivot) and a 30 N weight hangs on the right (d from the pivot), in balance. For d=3 m, choose the correct statement.

①W = 45 N.

②W = 20 N.

③Moving the right weight closer to the pivot keeps the balance.

④The beam's weight determines the balance.

⑤The force the pivot exerts on the beam is zero.

▸ ① W = 45 N.

Equilibrium: W × 2 = 30 × 3 = 90 → W = 45 N → ① correct, ② wrong.

Moving the right weight closer reduces its torque and breaks the balance → ③ wrong.

The beam is massless so its weight is irrelevant (④ wrong); the pivot supports the sum of the weights (75 N) (⑤ wrong).

🎯 Exam Points

①Torque τ = F·d (d = perpendicular distance to axis), larger farther out and more perpendicular

②Equilibrium: ΣF=0 AND Στ=0 (both)