Question

A uniform rod of 20 kg is hanging in horizontal position with the help of two threads. It also supports a 40 kg mass as shown in the figure.
The tensions developed in each thread are:
a) 300N, 200N
b) 400N, 200N
c) 150N, 300N
d) 100N, 300N​

Answer 1

answer : option (b) 400N, 200N

explanation : Let T1 and T2 are tension in string on left side and right side respectively.

at equilibrium,

T1 + T2 = weight of object + weight of rod.

= 40kg × 10m/s² + 20kg × 10m/s²

= 600N .....(1)

Torque = 0

or, torque due to rod + torque due to object + torque due to T1 + torque due to T2 = 0

assuming left end is fixed and right end is rotating.

so, 400N × l/4 (→) + 200N × l/2 (→) + T1 × 0 (←) + T2 × l (←) = 0

or, 100N × l(→) + 100N × l(→) - T2 × l (→) = 0

or, 200 = T2 putting in equation (1),

T1 = 400N

hence, option (b) is correct

Answer 2

Answer:

rotation equilibrium and net torque zero