A 2 kg block is placed on a 3 kg block. Ground is smooth and coefficient of static friction between blocks is 0.5. A force F is applied on the lower block at 37° below the horizontal. Take g = 10 m/s², sin37° = 3/5 and cos37° = 4/5. Maximum F for which the upper block does not slip is:
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Question

A 2 kg block is placed on a 3 kg block. Ground is smooth and coefficient of static friction between blocks is 0.5. A force F is applied on the lower block at 37° below the horizontal. Take g = 10 m/s², sin37° = 3/5 and cos37° = 4/5. Maximum F for which the upper block does not slip is:

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G1 Physics · Accepted Answer

For no slip, maximum common acceleration is μg = 5 m/s² because the upper block is accelerated by friction. Horizontal external force on the two-block system is F cos37° = 0.8F. Thus a = 0.8F/5. At limiting condition 0.8F/5 = 5, so F = 31.25 N. The downward component of F acts on the lower block and does not change the normal between the two blocks.