Question

Figure shows a system of two blocks of 10 kg and 5 kg mass, connected by ideal strings and pulleys.
Here ground is smooth and friction coefficient between the two blocks is u = 0.5. A horizontal force F is
applied on lower block as shown. The minimum value of Frequired to start sliding between the blocks
is: (Take g = 10 m/s2)
5 kg
10 kg
(A) 12.5 N
(B) 25 N
(C) 50 N
(D) 100 N​

Answer 1

Answer:

The answer is 25N.

Explanation:

Here at the instant of sliding limiting friction f = µN will act on the two blocks for limiting equilibrium,

Therefore we use : F = T + µ N … (1)

and   T = µ N … (2)

and   2T + N = 50 … (3)

Solving equation (2) and (3) we get

N = 50/2µ + 1 = 25 N

Now from equation (1) & (2)

F = 2 µN = 2 × 0.5 × 25 = 25 N.

Thus the minimum value of F required to start sliding between the blocks is 25N.