Question

Bob is pulling a $30kg$filing cabinet with a force of $200N$, but the filing cabinet refuses to move. The coefficient of static friction between the filing cabinet and the floor is $0.80$. What is the magnitude of the friction force on the filing cabinet?

Short answer

The magnitude of the friction force on the filling cabinet is$f_s=235.2N$

Step-by-step solution

Content Introduction

The friction force is proportional to the normal force N exerted by the wall against the block. According to Newton's third law, normal force is equal to and opposite in direction to the force exerted on the wall by the block, which is dependent on F and $\theta$. The friction force will be directed in the opposite direction as the motion.

Content Explanation

We are given:

$m=30kg$is mass,$F_N=200N$is force,${\mu }_s=0.80$is the coefficient of friction

The magnitude of the friction force on the filling cabinet is

$$\begin{gathered} f_s={\mu }_s{F}_N \\ \sum _{}{f}_y=0 \end{gathered}$$

Since all forces are in equilibrium, we got

$$\begin{gathered} F_N-mg=0 \\ {F}_N=30\times 9.81 \\ {F}_N=294N \end{gathered}$$

Solve for the magnitude of friction force on the filling cabinet,

$$\begin{gathered} f_s=0.8\times 294 \\ {f}_s=235.2N \end{gathered}$$