In Fig. 6-51, a crate slides down an inclined right-angled trough. The coefficient of kinetic friction between the crate and the trough is${\mathit{\mu }}_{\mathit{k}}$. What is the acceleration of the crate in terms of ${\mathit{\mu }}_{\mathbf{k}}$,$\mathit{\theta }$, and g?

The problem deals with Newton's second law of motion, which states that the acceleration of an object can be expressed in terms of the net force acting upon the object and the mass of the object.First, calculate the resultant normal force, then apply Newton's second law to solve the question.

Formula:

The frictional force is given by,

$f=2{\mu }_k{F}_N$

The free body diagram for the front view is shown below:

As both normal forces make$45°$with the center line, the resultant normal force is:

$$\begin{gathered} F_{Nr}=2F_N\cos \left(45°\right) \\ =\sqrt{2}{F}_N \end{gathered}$$

The free body diagram for the side view is shown below:

Newton's second law of motion for the create (+x direction is down the plane, +y direction is in the direction of $F_{Nr}$):

$x:mg\sin \left(\theta \right)-f=ma$

$y:F_{N}r-mgcos\left(\theta \right)=0$

The frictional force on the crate is given by:

$$\begin{gathered} f=2{\mu }_k{F}_N \\ =2{\mu }_k\left(\frac{F_{Nr}}{\sqrt{2}}\right) \\ =\sqrt{2}{\mu }_k{F}_{Nr} \end{gathered}$$

Substituting frictional force in x component and combining it with y component:

$$\begin{gathered} mg\sin \left(\theta \right)-\sqrt{2}{\mu }_k\left(mg\cos \left(\theta \right)\right)=ma \\ a=g\left(\sin \left(\theta \right)-\sqrt{2}{\mu }_k\cos \left(\theta \right)\right) \end{gathered}$$

Thus, the acceleration of the crate is $a=g\left(\sin \left(\theta \right)-\sqrt{2}{\mu }_k\cos \left(\theta \right)\right)$.