Question

a body takes time t to reach the bottom of an inclined plane of angle with horizontal. if plane is rough time taken is 2t. what is the cofficient of friction of rough surface.

Answer 1

Answer on Question #46640, Physics, Mechanics | Kinematics | Dynamics

Question:

a body takes time $t$ to reach the bottom of an inclined plane of angle with horizontal. If plane is rough time taken is 2t. what is the coefficient of friction of rough surface.

Answer:

Suppose length of plane equals $l$ :

$$l = \frac {a t ^ {2}}{2}$$

Therefore:

$$t = \sqrt {\frac {2 l}{a}}$$

In case with friction we have:

Newton's laws of motion:

$$x: \quad m a = m g \sin \theta - F _ {f r}$$ $$y: \quad N = m g \cos \theta$$

Friction force equals $F_{fr} = \mu N = \mu mg\cos \theta$ , $\mu$ - coefficient of friction.

Therefore:

$$a _ {2} = g \sin \theta - \mu g \cos \theta$$

In first case $\mu = 0$:

$$a _ {1} = g \sin \theta$$

Therefore:

$$t _ {1} = t = \sqrt {\frac {2 l}{g \sin \theta}}$$ $$t _ {2} = 2 t = \sqrt {\frac {2 l}{(g \sin \theta - \mu g \cos \theta)}}$$ $$g \sin \theta = 4 \cdot (g \sin \theta - \mu g \cos \theta)$$ $$\mu = 3 \tan \theta$$

Answer: $\mu = 3 \tan \theta$

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