Question

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.

Short answer

The force exerted by the second child is$23.3\mathrm{N}$

Step-by-step solution

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.

Calculation of the force

The torque by the first child is,

$\begin{array}{l}{\tau }_1=17.5\times 0.600\\ =10.5\mathrm{N}·\mathrm{m}\end{array}$

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,

$$\begin{gathered} F_2=\frac{{\tau }_1}{r_2} \\ =\frac{10.5}{0.450} \\ =23.3\mathrm{N} \end{gathered}$$

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