Answer:
x = 45 cm
Explanation:
Given that,
The length of a rod, L = 50 cm
Mass, m₁ = 0.2 kg
It is at 40cm from the left end of the rod.
We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.
The centre of mass of the rod is at 25 cm.
Taking moments of both masses such that,
[tex]15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm[/tex]
The distance from the left end is 40+5 = 45 cm.
Hence, at a distance of 45 cm from the left end it will balance the rod.
