A 20 kg load is pulled along an inclined plane with an upward slope of 30 degrees with an acceleration

A 20 kg load is pulled along an inclined plane with an upward slope of 30 degrees with an acceleration of 0.6 m / s ^ 2. The coefficient of friction of the load on the plane is 0.46. The force applied in the direction of motion is?

m = 20 kg.

a = 0.6 m / s ^ 2.

g = 10 m / s ^ 2.

∠α = 30 “.

μ = 0.46.

F -?

We write 2 Newton’s law for the projections of forces on an inclined plane and perpendicular to it.

m * a = F – Ftr – m * g * sinα;

N = m * g * cosα.

The friction force Ffr is determined by the formula: Ffr = μ * N, where μ is the friction coefficient, N is the support reaction force.

m * a = F – μ * m * g * cosα – m * g * sinα.

F = m * a + μ * m * g * cosα + m * g * sinα.

F = 20 kg * 0.6 m / s ^ 2 + 0.46 * 20 kg * 10 m / s ^ 2 * 0.87 + 20 kg * 10 m / s ^ 2 * 0.5 = 192 N.

Answer: the load is pulled with a force of F = 192 N.