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Grade 11-12 (age 16-18)

high school physics Torque & Rotational Equilibrium

A door opens easily when you push on the handle, far from the hinges, because torque, the turning effect of a force, is the distance from the axis times the perpendicular component of the force. For mechanical equilibrium, where an object neither moves nor turns, both the net force and the net torque must be 0, and for a seesaw m₁d₁ = m₂d₂. That is why the heavier person must sit close to the pivot and the lighter one far away for the seesaw to level out. In the picture, changing the distance d₂ to the right-hand weight levels the seesaw at the spot where the two torques are equal.

The Same Force Has a Different Effect Depending on Where It Acts

🚪 What Torque Is
①A door opens easily when you push the handle, far from the hinges
②Torque is the distance from the axis to where the force acts × the perpendicular component of the force
③For the same force, the farther from the axis and the more perpendicular it is applied, the larger the torque

Leveling the Seesaw

3 m
⚖️ When Is It Level?
①The left torque stays at 20 N × 2 m = 40 N·m
②The right torque is 10 N × d₂, so it grows in proportion to d₂
③At d₂ = 4 m, where the two torques are equal, the seesaw is level; closer than that it tilts left, and farther it tilts right
④When it is level, the force with which the pivot holds up the seesaw follows from ΣF = 0: 20 N + 10 N = 30 N

Torque and the Conditions for Equilibrium

Torque
τ = F · d = F · r sinθ
F is the force and d is the perpendicular distance from the axis to the line of action of the force (= r sinθ)
Conditions for mechanical equilibrium
ΣF = 0 and Στ = 0
Both the net force and the net torque must be 0 for an object to stay at rest (or keep moving and turning at a steady rate)
Balance of a Lever (Seesaw)
m1 d1 = m2 d2
The heavier side goes close to the pivot, the lighter side far from it

From Concept to Problem

Example 1: Balancing a Seesaw
On the left, 2 m from the pivot, sits a 40 kg person. What mass must sit 4 m to the right of the pivot for the seesaw to balance?
1
The condition for balance is left torque = right torque.
m1 d1 = m2 d2
2
Put in the values and solve for m₂.
40 × 2 = m2 × 4 → m2 = 20
▸ 20 kg
Sitting twice as far from the pivot lets half the weight keep the balance.
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)
1
Taking the hinge as the axis, the force from the hinge produces no torque.
Στ = 0
2
The beam's weight acts at its middle (2 m), and the tension acts at its end (4 m).
T × 4 = 60 × 2 → T = 30
▸ 30 N
Choosing the axis at the point where an unknown force acts removes that force from the equation and makes the calculation easier.

Exam Points

CSAT-style problem (Physics II), adapted
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.
1
From the balance condition W × 2 = 30 × 3 = 90, W = 45 N. So ① is correct and ② is wrong.
2
Moving the right-hand object closer to the pivot reduces the right torque and breaks the balance, so ③ is wrong.
3
The beam's mass is ignored, so its weight plays no part (④ is wrong), and the pivot supports the sum of the two weights, 75 N (⑤ is wrong).
🎯 Exam Points
①Torque τ = F·d (d is the perpendicular distance to the axis) is larger farther from the axis and when the force is perpendicular
②Mechanical equilibrium needs both ΣF = 0 and Στ = 0
③Balance of a lever: m₁d₁ = m₂d₂
④Choosing the axis where an unknown force acts simplifies the calculation
⑤The force at the pivot comes from the vertical force balance (ΣF = 0)
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