Question

a uniform metre rule pivoted off centre but maintained in equilibrium by a suspended weight of 2.4 newtons the weight is hung 5 cm from one end of the Meter rule what is the weight of the meter rule when the pivot is at 30cm mark​

Answer 1

Answer:

Torque due to load =4N×40cm=1.6Nm

Torque due to weight of the rule =2N×10cm=0.2Nm

Both are in opposite to each other so net torque =1.6−0.2=1.4Nm