Question

A block is in limiting equilibrium on a rough horizontal surface. If the net contact force is $\sqrt 3 $ times the normal force, the coefficient of static friction is:
${\text{A}}{\text{. }}\sqrt 2 $
${\text{B}}{\text{. }}\dfrac{1}{{\sqrt 2 }}$
${\text{C}}{\text{. 0}}{\text{.5}}$
${\text{D}}{\text{. }}\dfrac{1}{{\sqrt 3 }}$

Answer 1

Hint: The contact forces are the normal forces and friction. In this case, the block is in the limiting equilibrium means the static friction is equal to the horizontal force acting on the block and it is not moving, use this concept and solve.

Complete Step-by-Step solution:
We have been given in the question that the given block is in limiting equilibrium on a horizontal surface.
Also, the net contact force is $\sqrt 3 $ times the normal force.
To find- The coefficient of static friction.
Let us assume the normal force be N (acting vertically upward)
And the coefficient of static friction $ = \mu $
So, the static friction is ${f_s} = \mu N$ (acting horizontal)
Therefore, Net contact force is -
$
   = \sqrt {{{(\mu N)}^2} + {N^2}} \\
   = \sqrt {{N^2}({\mu ^2} + 1)} \\
   = N\sqrt {1 + {\mu ^2}} \\
$
Now, given that the static friction = horizontal force acting on the block
$
   \Rightarrow N\sqrt {(1 + {\mu ^2})} = \sqrt 3 N \\
   \Rightarrow \sqrt {(1 + {\mu ^2})} = \sqrt 3 \\
$
Squaring both sides, we get-
$
  1 + {\mu ^2} = 3 \\
   \Rightarrow {\mu ^2} = 2 \\
   \Rightarrow \mu = \sqrt 2 \\
$
Thus, the coefficient of static friction is $\sqrt 2 $ .

Note – Whenever such types of questions appear then first write down the things given in the question. After finding the static friction and normal force and using the condition the net contact force is $\sqrt 3 $ times the normal force, we have found out the coefficient of static friction.