Question

Two identical block A and B each of mass M are connected through a light inextensible string. Coefficient of friction between blocks and surfaces are µ as shown. Initially string is relaxed in its normal length. Force F is applied on block A as shown. Find the force of friction on blocks and tension in the string.

Answer 1

First let us calculate the limiting friction on blocks ‘A’ and ‘B’.

f_sA = µ mg 

f_sB = µ mg

(a) Now when a force of \(\frac 34\) µ mg acts on the block A; it doesn’t cause any motion in A.

Hence; F = f_A = \(\frac 34\) µ mg

And string is left unaltered. Hence tension is zero. And hence f_B = T = zero.

(b) Now when force of \(\frac 32\) µ mg is applied,

Body A will tend to move forward. (F ≥ f_s)

Let us assume that the whole system moves with on acceleration ‘a’.

\(\frac 32\) µ mg – (2 µ mg) = 2 ma

a is negative.

It means that our assumption that both the bodies move is false.

∴ Block B cannot move. Since they both are connected to each other, even A can’t move.

Answer 2

(a) f_A = \(\frac{3}{4} \mu Mg\) , f_B = 0 T = 0

(b) f_A = \(\mu\)Mg, f_B = \(\frac{\mu Mg}{2}\), T = \(\frac{\mu Mg}{2}\)