Under the action of a horizontal force, a 50 kg body moves uniformly over the surface

Under the action of a horizontal force, a 50 kg body moves uniformly over the surface (the coefficient of friction between the body and the surface is 0.3).

Given:
m = 10 kg, F = 50 N, α = 30∘, μ = 0.1, Ftr−?
The solution of the problem:
Scheme for solving the problem Since the body moves relative to the plane, in this case there is a sliding friction force. It is determined by the formula:

Ftr = μN
It turns out that we need to find the force of the normal reaction of the support N. The body is stationary along the y-axis, which means that the sum of the projections of all forces on this axis according to Newton’s first law is equal to zero.

N + F⋅sinα — mg = 0
N = mg — F⋅sinα
Thus, the general solution to the problem is:

Ftr = μ (mg — F⋅sinα)
Let’s calculate the numerical answer on the calculator.

Ftr = 0.1 (10⋅10-50⋅sin30∘) = 7.5N
Answer: 7.5 N.