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?

Question

Physics

Martin Eaton

24 September, 07:18

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?

+4

Answers (2)

Know the Answer?

Sandra Benjamin · Answer 1

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 = ?

Answer:

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.

Pablo Acevedo · Answer 2

40N force.

Explanation:

Let the force of the first child be F1 and that of the second child be F2

F1 = 30N and is perpendicular to the door

F2 = unknown

Moment arm for F1 = 0.4m

And that for F2 is 0.6m

The Force F2 has a component perpendicular to the door too. This component is equato F2Cos60° since F2 is directed to the door at 60°.

Taking moments about the hinge of the door, (Perpendicular force times the moment arm) the force F1 has a clockwise moment while the perpendicular component of F2 has an anticlockwise moment about the hinge.

So for stability, all clockwise moment about the hinge must balance anti clockwise moment

So

F1*0.4 = F2Cos60° * 0.6

30 * 0.4 = F2 * 0.5*0.6

12 = 0.3F2

F2 = 12/0.3 = 40N

F2 = 40N