A body with a mass of 10 kg moves uniformly along a horizontal plane under the action of a force equal to 50N

A body with a mass of 10 kg moves uniformly along a horizontal plane under the action of a force equal to 50N, directed at an angle of 30 degrees to the horizon. If the coefficient of sliding friction between the body and the plane is 0.1, then the friction force acting on the body is?

m = 10 kg. F = 50 N. g = 9.8 m / s ^ 2. ∠α = 30 “. Μ = 0.1. Ftr -? The friction force Ffr is determined by the formula: Ftr = μ * N, where μ is the coefficient of friction between surfaces, N is the reaction force of the support along which the body slides. directions: N + F * sinα = m * g. The reaction force of the support will have the form: N = m * g – F * sinα. Ftr = μ * (m * g – F * sinα). Ftr = 0.1 * ( 10 kg * 9.8 m / s ^ 2 – 50 N * sin30 “) = 7.3 N.
Answer: the body is affected by the friction force Ffr = 7.3 N.