Chapter 2, Problem 47A

Two children push on a heavy crate that rests on a basement floor. One pushes horizontally with a force of 150 N and the other pushes in the same direction with a force of 180 N. The crate remains stationary. Show that the force of friction between the crate and the floor is 330 N.

To determine

To show: Friction force between crate and floor is 330 N.

Explanation of Solution

Given:

Force of one children, $F_1=150\text{N}$

Force of other children, $F_2=180\text{N}$

Direction of both forces are same and crate is stationary.

Formula Used:

For a system in static mechanical equilibrium, the net force is zero.

Mathematically, $\sum F=0$

Calculation:

Since the crate is in static mechanical equilibrium, so the sum of all the forces on it will be zero.

As the force of friction will always be in opposite direction of motion or the tendency of motion.

So, using $\sum F=0$

  $\begin{array}{l}{F}_1+F_2-f=0\\ \Rightarrow f=F_1+F_2\\ =150+180\\ =330\text{N}\end{array}$

Conclusion:

Therefore, it is proved that friction force between crate and floor is 330 N