A box with 15 kg milk cartons stands on an inclined plane. Determine the coefficient of friction

A box with 15 kg milk cartons stands on an inclined plane. Determine the coefficient of friction of the box on this plane if its maximum inclination angle at which the box does not slide is 60. What is the friction force and the reaction force of the support?

Given:

m = 15 kilograms is the mass of a box with milk cartons standing on an inclined plane;

y = 60 degrees – the limiting angle of inclination of the plane at which the box does not slide;

g = 9.8 m / s ^ 2 – acceleration of gravity.

It is required to determine the coefficient of friction k, the friction force F friction (Newton) and the reaction force of the support N (Newton).

Since the box remains at rest, it means that all the forces acting on it are equal. Then:

F friction = F gravity;

k * N = m * g * sin (y);

k * m * g * cos (y) = m * g * sin (y);

k * cos (y) = sin (y);

k = sin (y) / cos (y) = tg (y) = tg (60) = 1.7.

N = m * g * cos (y) = 15 * 9.8 * cos (60) = 147 * 0.5 = 73.5 Newton.

F friction = k * N = 1.7 * 73.5 = 125 Newton.

Answer: the coefficient of friction is 1.7, the friction force is 125 Newton, the reaction force of the support is 73.5 Newton.