Question

A Block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar 1 m long. what force is required to balance it at its Center of gravity by applying the force at the other end of a bar ?​

Answer 1

Explanation:

mass= 10 kg

distance=20cm which will be 0.2 m

f=?

f= mass / distance

f=10kg/0.2m

=00.2

Answer 2

Answer:

40N

Explanation:

>m=20kg

L=1m

w1=L 1 =0.2m and w2=L 2=0.5m

Clockwise moments=Anticlockwise moments

F1×L1=F2×L2

mg×0.2=F2×0.5

F2=20÷0.5

F2=40N