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Asked by jhajuhi19 | 29 Aug, 2019, 10:14: AM
Expert Answer
Let the system of three blocks of mass m1 , m2 and m3 move with acceleration a due to applied force F.
Then acceleration is given by, a = F/(m1 + m2 + m3 ) ......................(1)
Let us find the normal force across the contact surface between m2 and m3 from the free body diagram of mass m3 as shown from figure.
N = m3 a = m3 × [ F / (m1 + m2 + m3 ) ] ...................(2)
Now as shown in free body diagram of mass m2 , block of mass m2 will not slip if μN ≥ m2 g
hence, μ m3 × [ F / (m1 + m2 + m3 ) ] ≥ m2 g or F ≥ (m1 + m2 + m3 ) [ m2 g / ( μ m3 ) ]
Answered by Thiyagarajan K | 29 Aug, 2019, 02:05: PM
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