Question

A mass of 4 Kg rest on the horizontal plane. The plane is gradually inclined until at an angle of $\theta ={15}^{\circ }$ with the horizontal,the mass just being to slide what is the coffeficient of static friction between the block & the surface.

Answer 1

We know that,

Mass $m=4kg$

angle $\theta ={15}^0$

Now, the normal force is

$N=mg$

$N=4\times 10$

$N=40N$

Now, frictional force always opposes direction of motion

So,

$F_f=mg\cos \theta$

$\mu N=4\times 10\times \cos {15}^0$

$\mu \times 40=40\times 0.96$

$\mu =0.96$

Hence, the friction is $0.96$