Question

The diagram below shows a lever of uniform mass, supported at the middle point. Four coins of equal masses are placed at mark 4 on the left hand side. Where should be 5 coins of same mass, as that of previous coins be located to balance the lever?

Answer 1

Let the mass of each coin is 'm'.
Then the weight of each coin $=mg$.
The distance of 5 coins on right sides is x, from the mid-point.
Let hand side moment by 4 coins at mark 4 is.
$LHM=4mg\times \mathrm{4.....}\left(1\right)$
Right hand side moment by 5 coins at mark x is,
$RHM=x\times 5mg$
According to the law of moments,
$LHM=RHM.$
$4mg\times 4=x\times 5mg$
$\Rightarrow x=\frac{16}{5}=3.2$
5 coins should be located at 3.2, on right side.