26. Weights of 4 kgf and 6 kgf are suspended from the ends of a uniform iron rod of length 50 cm and weight 4 kgf.
At what point will the rod balance on a supporting peg?
[Ans. at 28 6 cm from the 4 kalend!
Answers
Given:
Weight = W1 = 4kg
Weight = W2 = 6kg
Length of the rod = 50cm
To Find:
Point at which the rod balance on a supporting peg
Solution:
Let the distance where supporting peg is placed = L.
Thus,
Total momentum on 4kg -
= 4 × L + (L x 4/50) × L/2, ( momentum due weight of L at a distance of L/2 as iron rod is uniform ) -- 1
Similarly,
Total momentum on 6kg -
= 6 × (50-L) + [(50-L) × 4/50] × (50-L)/2, ---2
As the rod is in an equilibrium state, thus both side momentum must be equal.
Therefore, equation both equations -
4 × L + (L × 4/50) × L/2 = 6 × (50-L) + [(50-L) × 4/50] x (50-L)/2
= 4L + 4L2/100 = 300 - 6L + 4(50-L)2/100
= 4L + 4L2/100 = 300 - 6L + 4(2500 +L2 -100L)/100
= 4L = 300 -6L +100 -4L
= 14L = 400
L = 400/14
= 28.57
Answer: The rod balance on a supporting peg at 28 6cm