Question

A body takes time t to reach the bottom of an inclined plane of angle θ will the horizontal. if the plane made rough, time takes now is 2t. The coefficient of friction of the rough surface is

Answer 1

Step 1: Given data

1. The angle of the inclined plan is $\theta$.
2. The initial velocity of the body is zero.
3. Time is taken by the body to reach from top to bottom on a smooth inclined plane is t sec.
4. Time is taken by the body to reach from top to bottom on a rough inclined plane is 2t sec.

Step 2: Exerted force on a body placed on an inclined plane

1. For a block on an inclined plane, there are basically three types of force exerted, frictional force, normal force, and gravitational force.
2. The frictional force is a force, that is directed opposite to the body's motion and that prevents the movement of the body at the point of contact.
3. The frictional force is defined by the form, $F=\mu N$, where $\mu$is the coefficient of friction and n is the normal force.

Step 3: Diagram

Step 4: Finding the coefficient of friction for the smooth inclined plane

the distance (hypotenuse) between the top point and bottom point is, $s_{smooth}=\frac{1}{2}g\sin \theta \times t^2$ ……………..(1)

(applying the formulae of kinematics)where g is the acceleration due to gravity and t is the taken time.

Step 5: Finding the coefficient of friction for the Rough inclined plane

the normal force (N) on the body is,

$N=mg\cos \theta$ ……………(2)

Now, the frictional force of the body on the rough plane is,

$$\begin{gathered} F=\upsilon N=\mu mg\cos \theta \\ orF=\mu mg\cos \theta ...................\left(3\right) \end{gathered}$$

And the acceleration of the body is,

$$\begin{gathered} a=g\sin \theta -\mu g\cos \theta \\ ora=g(\sin \theta -\mu \cos \theta )................\left(4\right) \end{gathered}$$

Again,

$$\begin{gathered} s_{rough}=\frac{1}{2}\left(\sin \theta -\mu \cos \theta \right)\times g\times {\left(2t\right)}^2. \\ or{s}_{rough}=\left(\sin \theta -\mu \cos \theta \right)\times g\times 2t^2...............\left(5\right) \end{gathered}$$

Now, comparing equations (1) and (5) we get,

$$\begin{gathered} \frac{1}{2}g\sin \theta \times t^2=\left(\sin \theta -\mu \cos \theta \right)\times g\times 2t^2. \\ or\left(\sin \theta -\mu \cos \theta \right)=\frac{\sin \theta }{4} \\ or\frac{\left(\sin \theta -\mu \cos \theta \right)}{\sin \theta }=\frac{1}{4} \\ or1-\mu cot\theta =\frac{1}{4} \\ or\mu cot\theta =\frac{3}{4} \\ or\mu =\frac{3}{4cot\theta }=\frac{3}{4}\tan \theta \\ or\mu =\frac{3}{4}\tan \theta \end{gathered}$$

So, the coefficient of friction is $\mu =\frac{3}{4}\tan \theta$.