A body weighing 10 kg is located on an inclined plane making an angle of 30 ° with the horizon.

A body weighing 10 kg is located on an inclined plane making an angle of 30 ° with the horizon. The coefficient of friction of the body against the plane is μ = 0.6. What force must be applied to the body to move it upward with an acceleration of 2 m / s2?

Given:

m = 10 kilograms – body weight;

y = 30 degrees – the angle of inclination of the plane to the horizon;

k = 0.6 – coefficient of friction;

g = 10 N / kg – acceleration of gravity;

a = 2 m / s ^ 2 – the acceleration with which the body moves up.

It is required to determine what force F (Newton) is required to be applied to the body in order for it to move upward with acceleration a.

Let us expand all the forces into 2 directions: along the inclined plane and perpendicular to it. Then, the forces acting on the body along the inclined plane:

F – F gravity – F friction = ma;

F – m * g * sin (a) – k * N = m * a (1).

Forces acting on the body perpendicular to the inclined plane:

N = m * g * cos (a) (2).

Substitute the value of N from equation (2) into equation (1):

F – m * g * sin (a) – k * m * g * cos (a) = m * a;

F = m * g * sin (a) + k * m * g * cos (a) + m * a;

F = m * (g * sin (a) + k * g * cos (a) + a) = 10 * (10 * 0.5 + 10 * 0.6 * 0.87 + 2) =

= (5 + 5.22 + 2) = 12.22 Newton.

Answer: a force equal to 12.22 Newton must be applied to the body.