Question

Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.

Answer 1

Answer:

The second child must exert a force of magnitude 23.3N to keep the door from moving.

Explanation:

We have to find the moment that the first child exerts and then match it to that exercised by the second child.

F1= 17.5N

d1= 0.6m

F2= ?

d2= 0.45m

M= F * d

M1= 17.5N * 0.6m

M1= 10.5 N.m

M1=M2

M2= F2 * 0.45m

10.5 N.m= F2 * 0.45m

10.5 N.m/0.45m = F2

F2=23.3 N

Answer 2

The force that the second child must exert to keep the door from moving is 23.33 N.

What is a balanced force?

A balanced force occurs when an object subjected to different forces are at equilibrium.

Torque applied to the door

F1r1 = F2r2

(17.5 x 0.6) = F2(0.45)

F2 = 23.33 N

Thus, the force that the second child must exert to keep the door from moving is 23.33 N.

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