Consider the situation shown in the figure. The horizontal surface below the bigger block is smooth. The coefficient of friction between the blocks is u. Find the minimum and maximum force that can be applied in order to keep the smaller block at rest w.r.t. the bigger block.
Answers
Answer:
Conditions
If no force is applied, the block A will slip on C towards right and block B will move downward.
Suppose small force required in preventing slipping is F, taking A+B+C as the system,
The only external horizontal force on the system is F.
Hence the acceleration of the system is
a = F/2m+M ……………………..(1)
Explanation:
Let ‘block A’ as the system. The force on A are
1) Tension (T) by the string towards right
2) Friction (f) by the block C towards left
3) Weight (mg) downward and
4) Normal force (N) upward.
For vertical equilibrium N = mg.
Minimum force to prevent slipping and similarly friction is limiting-
f = mN
f=mmg
As the block moves towards right with an acceleration a,
T –f = ma
Or T - mmg = ma ----------------- (2)
Now let ‘block B’ as the system then force on B are-
1) Tension (T) upward.
2) Weight (mg) downward
3) Normal force (N1) towards right
4) Friction (f1) upward.
On right side movement of block with an acceleration a,
N1 = ma.
Work of limited friction
f1= mN1 = mma
For vertical equilibrium
T + f1 = mg
T + mma = mg ----------------- (3)
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Eliminating T from (1) and (3)
= (1-u)*g/(1+ u)
On putting large force on block A, it slips on block C towards left and similarly, block B slips on block C in upwards direction. So, friction on A is towards right and on B is downwards.
Thus the minimum and maximum force should be between
(1-u)(M+2m)*g/(1+u) and
(1+ u)(M+2m)*g/(1- u)
to keep the smaller block at rest w.r.t. the bigger block.