To hold the cart on an inclined plane with an inclination angle of 30 degrees, you must apply a force equal to 40 N

To hold the cart on an inclined plane with an inclination angle of 30 degrees, you must apply a force equal to 40 N directed along the inclined plane, and to pull this cart up the inclined plane, you must apply a force equal to 80 N. Determine the coefficient of friction.

Consider the forces acting on a body located on an inclined plane, where α is the angle of its inclination. Along the inclined plane, the following act: friction force Ff = μ ∙ N, where μ is the coefficient of friction and the projection of gravity onto the inclined plane Ftyazh.x = m ∙ g ∙ sinα, where the proportionality coefficient g = 9.8 N / kg. Perpendicular to the inclined plane, the body is affected by: the reaction force of the support N and the projection of the gravity force on the perpendicular to the inclined plane Ftyazh.y = m ∙ g ∙ cosα. These forces cancel each other out, therefore N = m ∙ g ∙ cosα.

To keep the trolley on an inclined plane, in which the angle of inclination is α = 30 °, it is necessary to apply a force F₁ = 40 N, which is helped by Ftr = μ ∙ m ∙ g ∙ cosα and counteracts Fty.x = m ∙ g ∙ sinα, therefore F₁ + Ftr = Ftyl.x or F₁ + μ ∙ m ∙ g ∙ cosα = m ∙ g ∙ sinα. Then F₁ = m ∙ g ∙ sinα – μ ∙ m ∙ g ∙ cosα or F₁ = m ∙ g ∙ 0.5 – μ ∙ m ∙ g ∙ (0.5 ∙ √3).

To pull this cart up the inclined plane, it is necessary to apply a force equal to F₂ = 80 N, which is counteracted by Ftr = μ ∙ N = μ ∙ m ∙ g ∙ cosα and Ftyazh.х = m ∙ g ∙ sinα, then F₂ = m ∙ g ∙ 0.5 + μ ∙ m ∙ g ∙ (0.5 ∙ √3).

We obtain the system of equations: F₁ + F₂ = m ∙ g and F₁ – F₂ = – μ ∙ m ∙ g ∙ √3, then μ = (√3) / 9 ≈ 0.19.

Answer: coefficient of friction ≈ 0.19.