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Q3 PE

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College Physics (Urone)
Found in: Page 316

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Short Answer

Two children push on opposite sides of a door during play. Both pushes horizontally and perpendicular to the door. One child push 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.

The force exerted by the second child is 23.3 N

See the step by step solution

Step by Step Solution

Step 1: Force

A force is an external factor that can change the rest or motion of a body. It has a size and a general direction.

Step 2: Calculation of the force

The torque by the first child is,

τ1=17.5×0.600 =10.5 N·m

The torque applied by the second child must be the same as the first child to keep the door from moving. The force by the second child is then,

F2=τ1r2 =10.50.450 =23.3 N

Hence, the force is23.3 N.

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