The coefficient of static friction bw block B and the horizo

The co-efficient of static friction b/w block B and the horizontal surface and between the rope and surface is 0.40. knowing that m_a = 12 kg determine the smallest mass of block B for which equilibrium is maintained

Solution

Let us consider \'T\' be the tension in the rope.

Now given that m(a) = 12.0 kg

So, frictional force = u*m(a)g = T

To maintain the balance, this must be equal to m(A)*g

Equalizing the two -

u*m(a)g = m(A)*g

=> m(A) = u*m(a) = 0.4*12 = 4.8 kg.

So, the smallest mass of the block A to maintain the equilibrium = 4.8 kg.