Question
A block has been placed on an inclined plane with the slope angle $$\theta ,$$ block slides down the plane at constant speed. The coefficient of kinetic friction is equal to
A.
$$\sin \theta $$
B.
$$\cos \theta $$
C.
$$g$$
D.
$$\tan \theta $$
Answer :
$$\tan \theta $$
Solution :
Angle of repose or angle of sliding is defined as the minimum angle of inclination of a plane with the horizontal, such that a body placed on the plane just begins to slide down.
$$AB$$ is an inclined plane such that a body placed on it just begins to slide down
$$\angle BAC = \theta = {\text{angle of repose}}$$
In equilibrium, $$f = mg\,\sin \theta $$
$$\eqalign{ & {\text{and}}\,R = mg\cos \theta \cr & \therefore \frac{f}{R} = \frac{{mg\sin \theta }}{{mg\cos \theta }} = \tan \theta \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\mu = \tan \theta \cr} $$
NOTE
Coefficient of kinetic friction between any two surfaces in contact is equal to the tangent of the angle of inclination between them.
Angle of repose or angle of sliding is defined as the minimum angle of inclination of a plane with the horizontal, such that a body placed on the plane just begins to slide down.
$$AB$$ is an inclined plane such that a body placed on it just begins to slide down
$$\angle BAC = \theta = {\text{angle of repose}}$$
In equilibrium, $$f = mg\,\sin \theta $$
$$\eqalign{ & {\text{and}}\,R = mg\cos \theta \cr & \therefore \frac{f}{R} = \frac{{mg\sin \theta }}{{mg\cos \theta }} = \tan \theta \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\mu = \tan \theta \cr} $$
NOTE
Coefficient of kinetic friction between any two surfaces in contact is equal to the tangent of the angle of inclination between them.
