Question

# Two blocks A of 6 kg and B of 4 kg are placed in contact with each other as shown in figure. There is no friction between A and ground and between both the blocks. The coefficient of friction between B and ground is 0.5. A horizontal force F is applied on A. Find the minimum and maximum values of F, which can be applied so that both the blocks can move combinely without any relative motion between them.

A
10 N, 50 N
B
12 N, 50 N
C
12 N, 75 N
D
None of these

Solution

## The correct option is C 12 N, 75 NF − N sin37° = 6a  ⇒F−(3N5)= 6a       (i) N2 = 4g−N cos37° = 40 − 4N5       (ii) N sin37° − f = 4a      (iii)  From Eqs. (i) and (iii):  F − f = 10a ⇒  F − μN2 = 10a              (iv)⇒  F − μ(40−4N5] = 10a      (v) Put the value of μ from Eq. (i) in (v) and also put the value of μ to get a=5F−6042 Now to start motion : a>0   ⇒ F>12 N So the minimum force F to just start the motion is 12 N. Now maximum F will be when N2  just becomes zero. Then from Eq.  (ii): N = 50 N. From Eqs. (i) and (iv), we get F = 75 N (by putting N2 = 0) If we apply F>75 N, then B will start sliding up on A.Physics

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