A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2 Mg force. In this situation, tension in the string is x/5 Mg for x = ______________. Neglect mass of the string and friction of the block (bigger mass) with ice slab.
(Given g = acceleration due to gravity)

Solution

Physics sir · 4 steps

The 2Mg pull beats the hanging Mg, so the small block goes up and the string pulls harder than Mg.

1. Find the direction. The 2Mg pull drags the 4M block to the left. The weight Mg of the hanging block pulls the string the other way. Since 2Mg > Mg, the 4M block moves left and the hanging block M rises.
2. Whole system. Treat both blocks as one system of mass 5M. Tension is internal and the ice is frictionless, so the net driving force is 2Mg - Mg = Mg.

   a = (2Mg - Mg)/(4M + M) = g/5

3. Hanging block. The block M accelerates upward, so the tension must exceed its weight: T - Mg = M × g/5, which gives T = 6/5Mg. As a check on the 4M block: 2Mg - T = 4M × g/5 also gives T = 6/5Mg.
4. So. Comparing T = 6/5Mg with x/5Mg gives x = 6.

The common slip: 4. Assuming the hanging block falls and writing Mg - T = M × g/5 gives T = 4/5Mg, that is 4, not 6.

The idea to keep. Fix the direction of motion first by comparing the driving forces. Then write each block's equation with the acceleration along that direction, and the tension sorts itself out.

Answer from the NTA final answer key. Solution by PrepWiser.