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The correct option is:
(a)
It can be seen from the figure above that when a force F acts on block 1, the frictional force acts on the block in the opposite direction. When the block just begins to slide over the one beneath it, the frictional force experienced by it is the static force of frictions.
Therefore, when block 1 just begins to move, F = ƒs
Here, ƒs is the static force of friction between block 1 and block 2.
The static force of friction exerted by block 2 on block 1 is reacted by a force of same magnitude, but in opposite direction. Therefore, block 2 experiences a force ƒs in the forward direction and experiences a static force of friction ƒ’s in the backward direction due to the contact forces between the plastic and its surface.
When block 2 just begins to move with block 1, the force ƒs should be greater than ƒ’s, that is,
Here, N1 is the normal force acting on block 1 and is equal to whereas N2 is the normal force acting on block 2, and is equal to 2mg (m being the mass of block and g the free fall acceleration).
Substitute the same in equation
Thus, the condition that should exist for block 2 to move with block 1 is 2μ2 < μ1.. Therefore (D) is the correct option and the rest are ruled out.
(b)
When the force F acts on the bottom block, the reaction force to the static force of friction experienced by block 2, acts on block 1. Therefore, the net force acting on block 1 is ƒs whereas the net force acting on block 2 is F – ƒs – ƒ’s.
Therefore, the block 1 slides if ƒs > 0
Substitute ƒs = μ1N1,
μ1N1 > 0
μ1 > 0
Therefore, if μ1 > 0, the top block will slide regardless of μ2.
Thus (A) is the correct option, while the rest are ruled out.
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