The mass of the part of the string below A in figure is m. Find the tension of the string at the lower end and at A.

Explanation

To get the tension at the lower end we need the force exerted by the string on the block.
Take block as the system. The forces on it are
a. pull of the string T, upward
b. pull of the earth Mg, downward,
The free body diagram for the block is shown in ure. s the acceleration of theblocfk is zero, these forces should add to zero. Hence the tension at the lower end is T=Mg.

To get the tension T at A we need the force exerted by the upper part of the string on the lower part of the string. For this we many write the equation of motion for the lower part of the string. So take the string below A as the system. The forces acting on this part of the string
a. T', upward by the upper pasrt of the string
b. mg, downward by the block.
c. T, downward by the block
Note that c we have written T for by the block on the string. We have already used the symbol T Newton's third law here. Teh force exerted by the block on the string is equal in magnitude to the force exerted by the string on the block.
The free body diagram for this part is hsown in ure part of the string) is an equilibrium, Newton's first law gives
`T'T+mg`
` But T=Mg hence, T'=(M+m)g`