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Q3 PE

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Found in: Page 316

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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 $23.3\mathrm{N}$

See the 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,

$\begin{array}{l}{\tau }_{1}=17.5×0.600\\ =10.5\mathrm{N}·\mathrm{m}\end{array}$

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,

${F}_{2}=\frac{{\tau }_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{10.5}{0.450}\phantom{\rule{0ex}{0ex}}=23.3\mathrm{N}$

Hence, the force is$23.3\mathrm{N}$.

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