IIT JEE , AIEEE Forum - Physics , Mechanics

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Since the direction and magnitude of force is constant, we have

Fsin a - m₁ g ? T sin a = m₁ a _y ---------- (1)

Where m₁ is the mass of that part of the chain, which already got lifted and T is the tension in that part of the chain, which is the intersection of the hanging part, and the part still on the ground. ?m₂? is that part of the chain which is still on the ground.

Fcos a - T cos a = m₁ a _x -------------- (2)

T cos a = m₂ a_x -------------- (3)

From 2 and 3 we get

A_x = Fcos a / m ---------------- (4)

Putting 4 in 3 we get

T = m₂ F / m ----------------------- (5)

Putting 5 in 1 we get

A_y = (F sin a / m) ? g -------------------- (6)

From 4 and 6 we get the velocities in x and y directions as

V_x = F cos a t / m --------------- (3)

V_y = (Fsin a - mg) t / m --------------- (4)

Also since a _y is constant by the time the chain is totally lifted the C.M. of the entire chain will be at a height of h = L sin a / 2. And from the equation of

h = a _y t² / 2 , we get the value of ?t?. Therefore from 3 and 4 we get the values of

V_x and V_y . Then the resultant velocity is given by the square root of the sum of the squares of V_x and V_y

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