Question

A uniform metal rod of negligible mass 1m long carries a load of 2kg at one end and 3kg at the other end .find the point about which the rod remains horizonta

Answer 1

Let the distance between the load of 2kg and the fulcrum be x
Thus the distance between load of 3kg and the fulcrum is (1-x)
According to principle of moments,
In equilibrium,
Sum of Clockwise moment = Sum of Anticlockwise moments
3*(1-x)=2*x
3-3x=2x
3=5x
x=3/5
x=0.6m
Therefore to make the scale horizontal the scale must be balanced at a distance of 0.6m from the 2kg end

Answer 2

Let us take the length between l of 2kg and f be x
Then, distance between l of 3kg and f = 1-x
A.T.Q.
3(1-x) = 2×x
3-3x = 2x
3 = 2x + 3x
3 = 5x
3÷5 = x
0.6 = x
x = 0.6
Thus the point about which the rod remains horizonta is 0.6m