Consider a body of mass mm on an incline with an inclination angle of \thetaθ. To this a force is applied without disturbing the equilibrium. The forces acting on the object are
vec W −→W : Weight of the object acting vertically downward;
vec F_{app} →Fapp : Force applied parallel to the inclined plane,
vecF_f→Ff : Frictional force acting parallel to the plane of the incline,
vec N →N : Normal force acting perpendicular to the plane of the incline,
Consider a coordinate system with its X-axis parallel to the plane downward and Y-axis perpendicular to the plane upward.
Resolve the weight into components that are parallel and perpendicular to the inclined plane
vec W_{||} = -mg\sin\theta; \qquad vec W_{_|_} = -mg\cos\theta
Equilibrium: \qquad vec W + vec F_{app} + vec N + vec F_f = vec 0
(A) Perpendicular Component: \qquad vec N + vec W_{_|_} = vec 0
N - W_{_|_} = 0; \qquad N = mg\cos\theta;
(B) Parallel Component: \qquad vec W_{||} + F_{app} + vec F_f = vec 0
W_{||} + F_{app} - F_f = 0; \qquad F_f = F_{app} + W_{||} ...... (2)
Static Equilibrium: In this case the frictional force is due to static friction.
F_f = \mu_sN = \mu_s.mg\cos\theta; \qquad F_{app} = F_{app}^s = 15 N;
Expanding (2), \mu_s.mg\cos\theta = F_{app}^s + mg\sin\theta
\mu_s = (F_{app}^s/(mg\cos\theta) + \tan\theta) ...... (3)
Dynamic Equilibrium: In this case the frictional force is due to kinetic friction.
F_f = \mu_kN = \mu_k.mg\cos\theta; \qquad F_{app} = F_{app}^k = 5 N;
Expanding (2), \mu_k.mg\cos\theta = F_{app}^k + mg\sin\theta
\mu_k = (F_{app}^k/(mg\cos\theta) + \tan\theta) ...... (4)
Calculate:
\sin\theta = \cos\theta = 1/\sqrt{2}; \qquad \tan\theta = 1; \qquad mg\cos\theta = 62.37 N
mu_s = 1.24; \qquad \mu_k = 1.08
