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A Monkey of mass m climbs up to a rope hung over a fixed pulley with acceleration g/4. The opposite end of the rope is tied to a block of mass M lying on a rough horizontal plane. The coefficient of friction between the block and horizontal plane is μ. Find the tension in the rope.
As monkey is accelerating upwards by g/4 it must be applying its mass times its acceleration force downwards.
THUS
Force ON ROPE
by monkey :- m*(g/4) downwards + its mass times acceleration due to gravity i.e m*g
Fnet=m(g+g/4)= 5mg/4 which equals the tension.
till Fnet≤µMg
I tried this question and was able to solve!
Acceleration of Monkey/Rope = g/4
Let the acceleration of ''M'' = a
So, the acceleration of the rope = a
(a)monkey = (a)monkey/rope + (a)rope
= (a-g/4)
For mass M, T - μmg = Ma ...(1)
For Monkey, mg - T = m(a - g/4) ...(2)
From equation (1) and (2),
mg - μMg = a(M + m) - mg/4
Then,
a = (5m/4 - μM)g / (M + m) = (5m - 4μM)g / 4(M + m)
Therefore, T = [M(5m - 4μM)g / 4(M + m)] - μMg
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