Question

A body of mass $2kg$ requires a force $3F$ to just move it up an inclined plane of angle ${37}^{\circ }$ and a force $F$ to prevent it from sliding down the same plane. Find the coefficient of friction. Take $g=10m/s^2$.

Answer 1

The FBD of the block in both situations is shown below.


From the FBDs, we have
$N=mg\cos {37}^{\circ }$ and
$f=\mu N=\mu mg\cos {37}^{\circ }$ (limiting friction in both cases)
For case (1):
$3F=mg\sin {37}^{\circ }+\mu mg\cos {37}^{\circ }-\left(1\right)$
For case (2):
$mg\sin {37}^{\circ }=F+\mu mg\cos {37}^{\circ }-\left(2\right)$
Substituting $F$ from eq. (2) in (1),
$3(\frac{3mg}{5}-\frac{\mu mg\times 4}{5})=\frac{3}{5}mg+\frac{4}{5}\mu mg$
$\Rightarrow 9-12\mu =3+4\mu \Rightarrow 16\mu =6$
$\Rightarrow \mu =\frac{6}{16}=\frac{3}{8}=0.375$