A load weighing 97 kg is moved evenly along a horizontal surface using a rope

A load weighing 97 kg is moved evenly along a horizontal surface using a rope, forming an angle of 30 ° with the horizon. Determine the tension on the rope if the coefficient of friction is 0.2.

m = 97 kg.

g = 9.8 m / s ^ 2.

∠α = 30 °.

μ = 0.2.

F -?

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

ОХ: 0 = F * cosα – Ftr.

OU: 0 = F * sinα – m * g + N.

N = m * g – F * sinα.

Ftr = μ * N = μ * (m * g – F * sinα).

Ftr = F * cosα.

μ * (m * g – F * sinα) = F * cosα.

μ * m * g – μ * F * sinα = F * cosα.

μ * m * g = μ * F * sinα + F * cosα.

F = μ * m * g / (μ * sinα + cosα).

F = 0.2 * 97 kg * 9.8 m / s ^ 2 / (0.2 * sin30 ° + cos30 °) = 196.8 N.

Answer: the tension force of the rope is F = 196.8 N.