FrictionHard
Question
The two blocks, m = 10 kg and M = 50 kg are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between M and the ground. A minimum horizontal force F is applied to hold m against M that is equal to :
Options
A.100N
B.50N
C.240N
D.180N
Solution
(C) As m would slip in vertically downward direction, then mg = μN
∴ mg = μN
⇒ N =
= 200 Newton
Same normal force would accelerate M, thus aM =
= 4 m/s2
Taking (m + M) as system
F = (m + M). 4 = 240 N
∴ mg = μN
⇒ N =
= 200 Newton Same normal force would accelerate M, thus aM =
= 4 m/s2 Taking (m + M) as system
F = (m + M). 4 = 240 N
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