Question

A house is built on the top of a hill with \[45^\circ \]slope. Due to the sliding of material and sand from top to the bottom of the hill, the slope angle has been reduced. If the coefficient of static friction between sand particles is 0.75, what is the final angle obtained by hill? (${\tan ^{ - 1}}0.75 = 37^\circ $ ).

A) $8^\circ $
B) $45^\circ $
C) $37^\circ $
D) $30^\circ $

Answer 1

Hint: Here we have to use the concept of resolution of forces. Here, we need to find the equilibrium of the sand. So to do that we need to find equations for all the forces so that the net forces are equal to zero.

Complete step by step solution:
Resolve forces and find the angle:

Resolve the vertical and the horizontal forces,
$N = mg\cos \theta $; …(m = mass, g = acceleration due to gravity)
Now, resolve the horizontal forces equate them together.
$\mu N = mg\sin \theta $; …(Frictional force = $\mu N$)
Put the value of N in the above equation:
$\mu \times mg\cos \theta = mg\sin \theta $;
Solve for the coefficient of friction and cancel out the terms with same values.
$\mu = \dfrac{{mg\sin \theta }}{{mg\cos \theta }}$;
$\mu = \tan \theta $;
Solve for the angle;
${\tan ^{ - 1}}\mu = \theta $;
Put the value of coefficient of friction which is 0.75.
${\tan ^{ - 1}}0.75 = \theta $;
The angle is:
$\theta = 37^\circ $;

Hence, Option (C) is correct. The final angle attained by the hill is $37^\circ $.

Note: Here the sand is sliding that means the forces on the sand are not in equilibrium. When the sand stops sliding that means the net force on the sand is equal to zero. Here, draw a diagram specifying each of the forces acting on the sand. Find the unknown angle.