The bar slides down the inclined plane with uniform acceleration. How will the friction force

The bar slides down the inclined plane with uniform acceleration. How will the friction force and acceleration acting on the bar change if another bar is placed on it?

Let m be the mass of the bar, k – the coefficient of friction of the bar against the inclined plane, a – the angle of inclination of the inclined plane, m1 – the mass of the additional load, which was placed on the bar.

Then, in the first case, the friction force will be equal to:

F friction = k * N = k * m * g * cos (a), and in the second case: F friction2 = k * (m + m1) * g * cos (a), since (m + m1)> m, then the friction force will increase.

The acceleration in the first case will be equal to:

m * g * sin (a) – F friction = m * a

m * g * sin (a) – k * m * g * cos (a) = m * a

a = g * sin (a) – k * g * cos (a).

As you can see from the formula, the acceleration does not depend on the mass, and therefore it will be the same in both cases.

Answer: if you put additional weight, the friction force will increase, the acceleration will remain the same.