A load weighing 4 kg lying on a horizontal table is acted upon by a horizontally directed force.

A load weighing 4 kg lying on a horizontal table is acted upon by a horizontally directed force. The coefficient of friction between the load and the table surface is 0.3. Determine the modulus of force, under the action of which it will move uniformly on the load?

m = 4 kg.

g = 9.8 m / s ^ 2.

μ = 0.3.

F -?

Let us write Newton’s 2 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 horizontal force, Ffr is the friction force, m * g is the force of gravity, N is the reaction force of the surface.

Since the body moves uniformly, the acceleration is a = 0 m / s ^ 2.

ОХ: 0 = F – Ftr.

OU: 0 = m * g – N.

F = Ftr.

m * g = N.

Ftr = μ * N = μ * m * g.

F = μ * m * g.

F = 0.3 * 4 kg * 9.8 m / s ^ 2 = 11.76 N.

The horizontal force can be directed in any direction.

Answer: a horizontal force F = 11.76 N. must act on the body.