The body slides at a constant speed down an inclined plane with an angle of inclination

The body slides at a constant speed down an inclined plane with an angle of inclination of 450 to the horizon. Determine the coefficient of friction.

To determine the coefficient of sliding friction of a body with mass m along an inclined plane, consider the forces acting on it: these are the gravity force F, the reaction force of the support N and the friction force Ftr. If we introduce a coordinate system with the center located at the center of mass of the body, with the abscissa axis directed along the inclined plane in the direction of the body movement, and the ordinate axis directed upward, perpendicular to the inclined plane, then the forces will have the following coordinates: sinα; – m ∙ g ∙ cosα), N (0; N) and Ffr (- μ ∙ N; 0), where α is the angle of inclination of the plane, coefficient g = 9.8 N / kg, μ is the coefficient of sliding friction of the body on an inclined plane. According to Newton’s 2 law, the resultant of all forces (m ∙ a; 0) is equal to the sum of these forces, that is, the abscissa of the resultant will be m ∙ a = m ∙ g ∙ sinα – μ ∙ N and the ordinate of the resultant will be 0 = – m ∙ g ∙ cosα + N, or N = m ∙ g ∙ cosα and m ∙ а = m ∙ g ∙ sinα – μ ∙ m ∙ g ∙ cosα. From the resulting system of equations, we express the coefficient of friction: μ = (g ∙ sinα – a) / (g ∙ cosα).
From the condition of the problem it is known that the body slides at a constant speed v down an inclined plane with an angle of inclination α = 45 ° to the horizon. From the fact that the speed is constant, we get that the acceleration of the body a = 0 m / s².
Substituting the values ​​of physical quantities in the calculation formula, we obtain μ = (g ∙ sinα – 0) / (g ∙ cosα) = tgα = tg45 ° = 1.
Answer: 1 is the value of the coefficient of friction.