The free body diagram of the crate can be shown as,
According to free body diagram, the net force acting on the crate is given as,
[tex]F=mg\sin \theta-f\ldots\ldots\text{ (1)}[/tex]The frictional force acting on the crate is,
[tex]f=\mu N[/tex]According to free body diagram, the normal force acting on the crate is,
[tex]N=mg\cos \theta[/tex]Therefore, the frictional force becomes,
[tex]f=\mu mg\cos \theta[/tex]Since, the crate is at rest therefore, according to Newton's second law of motion, the net force acting on the crate is,
[tex]F=0[/tex]Substitute the known values in the equation (1),
[tex]\begin{gathered} 0=mg\sin \theta-\mu mg\cos \theta \\ mg\sin \theta=\mu mg\cos \theta \\ \mu=\frac{\sin \theta}{\cos \theta} \\ =\tan \theta \end{gathered}[/tex]Plug in the known values,
[tex]\begin{gathered} \mu=\tan 32.7^{\circ} \\ \approx0.642 \end{gathered}[/tex]Thus, the coefficient of static friction between the crate and surface of ramp is 0.642.
