FrictionHard

Question

In the given figure the coefficient of friction between 4kg and 5 kg blocks is 0.2 and between 5 kg block and ground is 0.1 respectively. Choose the correct statements

Options

A.Minimum force needed to cause system to move is 17 N
B.When force is 4N static friction at all surfaces is 4N to keep system at rest
C.Maximum acceleration of 4kg block is 2m/s2
D.Slipping between 4kg and 5 kg blocks start when F is > 17N

Solution

So block ′Q′ is moving due to force while block ′P′ due to friction.
Friction direction on both +Q blocks as shown.

First block ′Q′ will move and P will move with ′Q′ so by FBD taking ′P′ and ′Q′ as system
F − 9 = 0    ⇒    F = 9 N
When applied force is 4 N then FBD

4 kg block is moving due to friction and maximum friction force is 8 N.
So acceleration = 8/4 = 2/m/s2 - amax.
Slipping will start at when Q has +ve acceleration equal to maximum acceleration of P i.e. 2 m/s2.
F − 17 = 5 × 2      ⇒ F = 27 N.

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