Question

A uniform metre rule of mass $100g$ is balanced on a fulcrum at mark $40cm$by suspending an unknown mass $m$ at the mark $20cm$.
$\left(i\right)$ Find the value of $m$.
$\left(ii\right)$ To which side the rule will tilt if the mass $m$ is moved to the mark $10cm$?
$\left(iii\right)$ What is the resultant moment now?
$\left(iv\right)$ How can it be download by another mass of $50g$?

Answer 1

Since, uniform metre rule of mass $100g$ is balanced on a fulcrum at mark 40 cm.

so, distance of c.o.m. of rule from balanced point is $10cm.$

Net moment at balancing point should be zero.

So, $m\times 20=100\times 10⟹m=50g$

if m is moved to mark of $10cm$,

then rule will tilt to the side where m is suspended.

Now, for balance this

$50\times 30=100\times 10+50\times (x-40)(x-40)=(1500-1000)/50=10cm$

$⟹X=50cm$.

So, another 50g mass will be suspended at 50cm mark