In the figure, m A = 2 kg and m B = 4 kg. For what minimum value of F, A starts slipping over B: (g = 10 m/s 2 ) –
Text Solution
Verified by ExpertsThe correct answer is:
B
Maximum frictional force between A and B could be
f 1 = µ 1 m A g = (0.2) (2) (10) N
f 1 = 4 N
Hence, maximum common acceleration till both the blocks move with same acceleration is
a = 
= 
= 2 m/s 2 Now, taking (A + B) as the system:
(From weight = upthrust)
(f 2 ) max = µ 2 (m A + m B )g = 24 N
F – 24 = (m A + m B ) a = 6 × 2 = 12
F = 36 N
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