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

  • A $\sqrt{2}$
  • B $\frac{1}{\sqrt{2}}$
  • C 0.5
  • D $\frac{1}{\sqrt{3}}$

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Answer

To solve the problem of finding the coefficient of static friction when the net contact force is $\sqrt{3}$ times the normal force, let's break down the problem step-by-step.

Step-by-Step :

  1. Definitions and Notations:

    • Let the normal force be denoted as $N$.
    • Let the coefficient of static friction be symbolized as $\mu$.
    • Static frictional force is given by $f = \mu N$.
  2. Net Contact Force: According to the problem, the net contact force is $\sqrt{3}$ times the normal force. The net contact force on a rough horizontal surface, when a block is in limiting equilibrium, can be calculated using: $$ F_{\text{net}} = \sqrt{N^2 + (\mu N)^2} $$ Simplifying within the square root, we get: $$ F_{\text{net}} = \sqrt{N^2 (1 + \mu^2)} = N \sqrt{1 + \mu^2} $$

  3. Given Condition: It is given that this net contact force equals $\sqrt{3}$ times the normal force: $$ N \sqrt{1 + \mu^2} = \sqrt{3} N $$ Dividing both sides by $N$ (assuming $N \neq 0$), we get: $$ \sqrt{1 + \mu^2} = \sqrt{3} $$

  4. Solving for $\mu$:

    • To eliminate the square root, square both sides of the equation: $$ (\sqrt{1 + \mu^2})^2 = (\sqrt{3})^2 $$
    • Simplifying this, we obtain: $$ 1 + \mu^2 = 3 $$
    • Isolating $\mu^2$, we get: $$ \mu^2 = 3 - 1 $$
    • Therefore: $$ \mu^2 = 2 $$
    • Taking the square root of both sides, we find: $$ \mu = \sqrt{2} $$

Final Answer:

The coefficient of static friction $\mu$ is:

  • Option A) $\sqrt{2}$

So, the correct answer to the question is A) $\sqrt{2}$.


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