The bar is pulled along an inclined plane at a constant speed using a cable directed along the inclined plane

The bar is pulled along an inclined plane at a constant speed using a cable directed along the inclined plane. Find the modulus of force tension of the cable, if the mass of the bar is 500 grams, the angle of inclination of the plane is 30 degrees, mu = 0.25

m = 500 g = 0.5 kg.

g = 10 m / s2.

a = 0 m / s2.

∠α = 30 °.

μ = 0.25.

F -?

Let us write Newton’s 2 law in vector form: m * a = F + m * g + N + Ftr, where F is the force with which the load is pulled, m * g is the force of gravity, N is the surface reaction force, Ftr is the friction force.

Since the body is pulled uniformly, then a = 0 m / s2.

ОХ: F – Ftr – m * g * sinα = 0.

OU: 0 = N – m * g * cosα = 0.

F = Ftr + m * g * sinα.

N = m * g * cosα.

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

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

F = 0.5 kg * 10 m / s2 * (0.25 * cos30 ° + sin30 °) = 3.6 N.

Answer: the tension force of the cable is F = 3.6 N.