Answer to Question #102544 in Mechanics | Relativity for Megha

Question

A box of mass 75 kg is placed on an inclined plane. When the angle of the plane is
increased to 30º, the box begins to slide downwards. Calculate the coefficient of static
friction between the plane and the box. Draw the free body diagram. Take g=10.0 ms^-2.

Answer 1

As per the given question,

mass of the box=75kg

Angle of inclination$=30^\circ$

g $=10m/sec^2$

The block diagram of the question is given below,

$mg\sin30=f_s$

$\Rightarrow mg\sin30=\mu N--------(i)$

Now,

$N=mg\cos 30=75\times10\times \dfrac{\sqrt{3}}{2}------(ii)$

From equation (i) and (ii)

$\Rightarrow 75\times 10\times \dfrac{1}{2}=75\times 10\times \dfrac{\sqrt{3}}{2}\mu$

$\mu=\dfrac{1}{\sqrt{3}}$