Two children push on opposite sides of a door. The first child pushes with a force of 30 N perpendicular to the door at a position 0.4 m from the hinges of the door. With what force should the second child push in order to maintain static equilibrium if the second child pushes at an angle 60° relative to the door at a position 0.6 m from the hinges?

Given Information:

Force exerted by first child = F₁ = 30 N

Lever arm of first child = r₁ = 0.4 m

Angle = θ₁ = 90°

Lever arm of second child= r₂ = 0.6 m

Angle = θ₂ = 60°

Required Information:

Force exerted by second child = F₂ = ?

Force exerted by second child = F₂ = 23.12 N

Explanation:

In order to maintain static equilibrium, the net torque on the door must be equal to zero.

∑τ = 0

τ₁ = τ₂

We know that torque is a measure of rotational force of an object and is given by

τ = rFsin(θ)

Where F is the applied force, r is the lever arm that is the perpendicular distance between the rotation of axis and applied force and θ is the angle between applied force and lever arm.

The torque exerted by first child is

τ₁ = r₁F₁sin(θ₁)

τ₁ = 0.4*30*sin(90°)

τ₁ = 0.4*30*1

τ₁ = 12 N.m

The torque exerted by second child is

τ₂ = r₂F₂sin(θ₂)

τ₂ = 0.6*F₂*sin(60°)

τ₂ = 0.6*F₂*0.866

τ₂ = 0.519F₂

Finally applying the equilibrium condition

τ₁ = τ₂

12 = 0.519F₂

F₂ = 12/0.519

F₂ = 23.12 N

Therefore, the second child is required to apply a force of 23.12 N to maintain static equilibrium.

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