Question

A uniform meter rod of 2 N is pivoted at 60 cm mark.The total length of meter rod is 100 cm.If a load of 4 N is suspended at one end 40 cm from pivot point.What is the net torque about the pivot.

Answer 1

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

Answer 2

Answer:

T=R.F

but given length =100cm(1 meter)

Applied force 2N at 60 cm Load at 20 cm (40 cm from pivoted point) Now, T=L.F ;T=1/5.4 =4/5NM

T=3/5.2

=6/5

6/5-4/5

=2/5NM

so, net torque about pivoted point is 2/5NM