On an inclined plane 13m long and 5m high there is a load weighing 26kg. The friction coefficient is 0.5.

On an inclined plane 13m long and 5m high there is a load weighing 26kg. The friction coefficient is 0.5. What force must be applied to the load along the plane in order to pull the load in? To pull off the load?

S = 13 m.

h = 5 m.

m = 26 kg.

g = 10 m / s2.

μ = 0.5.

Fв -?

Fн -?

We will assume that the body is raised and lowered uniformly, that is, its acceleration is a = 0 m / s2.

For the movement of a load on an inclined plane, Let us write Newton’s 2 law in vector form: Ft + m * g + N + Ftr = 0, where F is the force applied to the load, m * g is the force of gravity, N is the reaction force of the surface of the inclined plane , Ftr – friction force.

1) The load is pulled up.

ОХ: Fв – Ftr – m * g * sinα = 0.

ОУ: – m * g * cosα + N = 0.

Fv = Ftr + m * g * sinα.

N = m * g * cosα.

The friction force Ffr is expressed by the formula: Ffr = μ * N = μ * m * g * cosα.

Fв = μ * m * g * cosα + m * g * sinα = m * g (μ * cosα + sinα).

sinα = h / S.

sinα = 5 m / 13 m = 0.385.

cosα = √ (1 – sin2α).

cosα = √ (1 – (0.385) ^ 2) = 0.93.

Fw = 26 kg * 10 m / s2 * (0.5 * 0.93 + 0.385) = 221 N.

2) The load is pulled down.

ОХ: Fн – Ftr + m * g * sinα = 0.

ОУ: – m * g * cosα + N = 0.

Fн = Ftr – m * g * sinα.

N = m * g * cosα.

Fн = μ * m * g * cosα – m * g * sinα = m * g * (μ * cosα – sinα).

Fн = 26 kg * 10 m / s2 * (0.5 * 0.93 – 0.385) = 21 N.

Answer: Fw = 221 N, Fn = 21 N.