Chapter 4, Problem 54P

A 25.0-kg box is released on a 27° incline and accelerates down the incline at 0.30 m/s². Find the friction force impeding its motion. What is the coefficient of kinetic friction?

To determine

The coefficient of kinetic friction between the box and the surface of the incline, and the friction force acting on the box.

Answer to Problem 54P

Solution:

The coefficient of kinetic friction between the box and the surface of the incline is found to be 0.48.

The friction force acting on the sliding box is found to be $\text{1}\text{.1}\times {\text{10}}^2\text{ N}$.

Explanation of Solution

A box of mass m is released down the incline at an angle $\theta =27°$ to the horizontal. The free body diagram is as shown below.

The weight mg of the box acts vertically downwards. The normal force $F_N$ acts perpendicular to the inclined plane. The weight mg is resolved into two components, $mg\sin \theta \text{ and }mg\cos \theta$ parallel and perpendicular to the inclined plane. The force of kinetic friction $F_{fr}$ acts opposite to the direction of motion of the box. But there is a net force F acting on the box in the downward direction parallel to the inclined plane and the box accelerates downwards with an acceleration a.

The system is in vertical equilibrium. Therefore,

    $F_N=mg\cos \theta$……(1)

The force of friction is related to the normal force as,

    $F_{fr}={\mu }_k{F}_N$……(2)

The coefficient of kinetic friction is ${\mu }_k$

The frictional force $F_{fr}$ is given by,

    $F_{fr}={\mu }_{k}mg\cos \theta$……(3)

For the downward accelerated motion,

    $F=mg\sin \theta -F_{fr}$

Using equation (3)in the above equation,

    $\begin{array}{c}F=mg\sin \theta -F_{fr}\\ =mg\sin \theta -{\mu }_{k}mg\cos \theta \\ =mg\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$……(4)

Using Newton’s second law, $F=ma$, write the expression for ${\mu }_k$.

    $\begin{array}{c}ma=mg\left(\sin \theta -{\mu }_k\cos \theta \right)\\ a=g\left(\sin \theta -{\mu }_k\cos \theta \right)\\ {\mu }_k=\tan \theta -\frac{a}{g\cos \theta }\end{array}$……(5)

Given:

The mass of the box $m=25.0\text{ kg}$

The angle of the incline $\theta =27°$

The acceleration of the box down the incline $\theta =27°$

Formula used:

    $F_{fr}={\mu }_{k}mg\cos \theta$

    ${\mu }_k=\tan \theta -\frac{a}{g\cos \theta }$

Calculation:

Substitute the given values of $\theta$, a and 9.8 m/s² in the expression for ${\mu }_k$.

$\begin{array}{l}{\mu }_k=\tan \theta -\frac{a}{g\cos \theta }\\ =\tan 27°-\frac{\left(0.30{\text{ m/s}}^2\right)}{\left(9.8{\text{ m/s}}^2\right)\left(\cos 27°\right)}\\ =0.5095-0.0343\end{array}$

                  =0.4752

The coefficient of kinetic friction correct to 2 figures is 0.48.

Use this value of ${\mu }_k$

in the expression for friction force $F_{fr}={\mu }_{k}mg\cos \theta$ and calculate the value of the friction force.

$\begin{array}{l}{F}_{fr}={\mu }_{k}mg\cos \theta \\ =\left(0.48\right)\left(25.0\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)\left(\cos 27°\right)\\ =1.048\times {10}^2\text{N}\end{array}$

The friction force is $\text{1}\text{.1}\times {\text{10}}^2\text{ N}$.