In the figure shown, the coefficient of static friction between block B and the wall is 2/3 and the coefficient of kinetic friction between B and the wall is 1/3. Other contacts are smooth. Find the minimum force ‘F’ required to lift B, up. Now if the force applied on A is slightly increased than the calculated value of minimum force, then find the acceleration of B. Mass of A is 2m and the mass of B is m. Take tan θ = ¾
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) F min = 
mg, (ii) b = 
Sol. (i) The F.B.D. of A and B are
For A to be in equilibrium A; F = N sin θ ............(1)
For B to just lift off B N cos θ = mg + µ s N ′ ...............(2)
For horizontal equilibrium of B; N ′ = N sin θ ............(3)
From (2) and (3)
N (cos θ – µ s sin θ ) = mg or N = 
mg or N = 
mg ...............(4)
From equation (1) F = N × 
⇒ ∴ F = 
mg
(ii) The acceleration of the block A be a and B be b
F – N sin θ = 2ma ...............(1)
N cos θ – mg – µ k N ′ = mb ...............(2)
N ′ = N sin θ ................(3)
From constraint =
a sin θ = b cos θ ................(4)
Solving (1), (2), (3) and (4) we get ⇒ b = 
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