A block is in limiting equilibrium on a rough horizontal surface. If the net contact force is square root of 3 times the normal force, the coefficient of static friction is A square root of 2 B 1 over square root of 2 C 0.5 D 1 over square root of 3
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 stepbystep.
StepbyStep :

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 $f = \mu N$.

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: $$ F_{\text{net}} = \sqrt{N^2 (1 + \mu^2)} = N \sqrt{1 + \mu^2} $$

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: $$ \sqrt{1 + \mu^2} = \sqrt{3} $$

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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