Question

Consider a mass M = 25.0kg sitting on an inclined plane whose angle of inclination is 30degrees. A massless core connected to M passes over a friction less pulley and is attached to a mass m= 15.0kg. The mass M slides up the plane at a constant speed. Find the coefficient of kinetic friction between the mass M and the plane

Answer 1

We should determine the coefficient

$\mu.$

Let us consider the forces in this situation. Let the x-axis be directed along the inclined plane towards the motion of M, y-axis be directed perpendicular to x-axis, and y1-axis be directed down along the motion of m.

The projections of forces influencing M onto y-axis are

$N-Mg\cos \alpha=0, N=Mg\cos \alpha.$

The projections of forces influencing M onto x-axis are

$-Mg\sin\alpha - \mu Mg\cos \alpha + T=0,$

where T is the tension of the cord, and sum of forces is 0 due to the constant velocity.

The projections of forces acting on m onto y1-axis are

$T-mg =0.$

From the last equation we get T=mg, from the second equation we get

$Mg\sin\alpha +\mu Mg\cos\alpha =mg,$ so

$\mu = (m-M\sin\alpha) /(M\cos\alpha) =(15. 0-25.0\cdot0.5) /(25.0\cdot\sqrt3/2) \approx 0.115.$