Question

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction, when the angle of inclination of the plane is $60°$ is

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

Answer 1

The correct option is C

$\frac{1}{\sqrt{3}}$

Step 1: Given data

The angle of inclination of the plane$\left(\theta \right)$=$60°$

Step 2: Determine the acting frictional force

The formula to compute acting frictional force is as follows:

$F_f=\mu N\dots \dots \dots \dots \dots \dots \left(1\right)$

where $F_f$ is the force of friction, $\mu$ coefficient of friction, $N$ normal force

Step 3: Determine friction force when the motion is upward

Consider the following figure,

Force balance perpendicular to the surface of wedge,

$N=mg\cos 60°$

Substitute the known value of normal reaction in equation (1),

$$\begin{gathered} F_f=\mu mg\times \cos \left(60\right) \\ {F}_f=\frac{\mu mg}{2}\dots \dots \dots \dots \left(2\right) \end{gathered}$$

The motion is up so the friction force will act in the downward direction.

Let $F_u$ be the force required to move a body up a rough inclined plane

As we know that frictional force acts opposite to the direction of motion, balancing in the direction of motion,

Therefore, the net acting force is,

$$\begin{gathered} F_u=mg\sin \left(\theta \right)+F_f \\ {F}_u=mg\sin \left(60\right)+F_f \\ {F}_u=mg\sin \left(60\right)+\frac{\mu mg}{2} \\ {F}_u=mg\left(\frac{\sqrt{3}+\mu }{2}\right)\dots \dots \dots \dots \dots \dots \left(4\right) \end{gathered}$$

Step 4: Determine friction force when the motion is downward

Consider the following figure:

Let $F_d$ be the force required to prevent the body from sliding down the plane.

The motion is downward so the friction force will act upward direction. So, the net downward force is,

$$\begin{gathered} F_d=mg\sin \left(\theta \right)-F_f \\ {F}_d=mg\sin \left(60\right)-\frac{\mu mg}{2} \\ {F}_d=\frac{\sqrt{3}}{2}mg-\frac{\mu mg}{2} \\ {F}_d=\left(\frac{\sqrt{3}-\mu }{2}\right)mg\cdots \cdots \cdots \cdots \left(5\right) \end{gathered}$$

Step 5: Compute the coefficient of friction:

According to the problem, the force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. Therefore, $F_u=2F_d$. So,

From equation (4) and (5)-

$$\begin{gathered} F_u=2F_d \\ mg\left(\frac{\sqrt{3}+\mu }{2}\right)=2mg\left(\frac{\sqrt{3}-\mu }{2}\right) \\ \sqrt{3}+\mu =2\left(\sqrt{3}-\mu \right) \\ \sqrt{3}+\mu =2\sqrt{3}-2\mu \\ 3\mu =\sqrt{3} \\ \mu =\frac{\sqrt{3}}{3} \\ \mu =\frac{1}{\sqrt{3}} \end{gathered}$$

Thus when $\theta =60$, the coefficient of friction will be $\frac{1}{\sqrt{3}}$.

Hence, option C is the correct answer.