Question

two masses M1 equals to 5 kg M2 equals to 10 kg connected by an inextensible string over a frictionless Pulley a moving as shown in the figure the coefficient of friction of the horizontal surface is 0.15 the minimum weight and that should be put on the top of them to do stop the motion is

Answer 1

For the motion to stop, the frictional force should be equal to the normal reaction force exerted by the masses.

That means the coefficient should be M₁ / M₂

5 / 10 = 0.5

ω ( coefficient of friction) = Frictional force / Normal Force.

Given the coefficient = 0.15 and the total mass = 15 kg we can get the frictional force.

Weight of the two masses = 15 × 10 = 150N


0.15 = F / 150

F = 0.15 × 150 = 22.5 N

From the coefficient 0.5 we can get the required frictional force :

150 × 0.5 = 7.5 N

The force required is thus :

75 N - 22.5 N = 52.5 N