Down an inclined plane that forms an angle of 40 degrees with the horizon, the coefficient of friction is 0.3.

Down an inclined plane that forms an angle of 40 degrees with the horizon, the coefficient of friction is 0.3. It is necessary to determine the acceleration of the body.

We are given the angle of inclination of the plane α = 40º, the coefficient of friction μ = 0.3.

You need to find the acceleration a of the bar with which it slides down.

Three forces act on our bar: gravity, friction and support reaction force. The vector sum of these forces determines the acceleration of our bar. According to Newton’s second law, the vector sum: Ftr + Ftzh + N = m * a.

Let’s find the projections of all vectors onto the x-axis parallel to the horizontal surface of the earth:

friction force projection: -μ * m * g * cosα,

gravity projection: zero,

projection of the reaction force vector of the support: m * g * sіnα,

acceleration vector projection: a * cosα.

Substituting all the projections into the equation of Newton’s second law, we get:

-μ * m * g * cosα + m * g * sіnα = m * a * cosα.

Hence:

a = (g * sіnα – μ * g * cosα) / cosα = g * (tga – μ) = 10 m / s ^ 2 * (0.84 – 0.3) = 5.4 m / s ^ 2