Question

The diagram shows a uniform metre rule weighing $100gf$, pivoted at its centre $O$. Two weights $150gf$ and $250gf$ hang from the points $A$ and $B$ of the metre rule such that $OA=40cm$ and $OB=20cm$. Calculate:
(i) the total anticlockwise moment about $O$,
(ii) the total clockwise moment about $O$;
(iii) the total clockwise and anticlockwise moments; and
(iv) the distance from $O$ where a $100gf$ weight should be placed to balance the metre rule.

Answer 1

$\left(I\right)$ Total anticlockwise moment about $0$

$=150gf\times 40cm=6000gfcm$

$\left(II\right)$ Total clockwise moment about $0$

$=250gf\times 20cm=5000gfcm$

$\left(III\right)$ The difference of anti-clockwise and clockwise moment

$=6000-5000=1000gfcm$

$\left(IV\right)$ From the principle of moments,

anti-clockwise moment$=$clockwise moment

To balance it, $100gf$ weight should be kept on right hand side so as to produce a clockwise moment about the $0$.

Let its distance from the point $0$ be $dcm$

Then, $150gf\times 40cm=250gf\times 20cm+100gf\times d$

$6000gfcm=5000gfcm+100gf\times d$

$\Rightarrow 1000gfcm=100gf\times d$

$\therefore d=\frac{1000gfcm}{100gf}=10cm$ on the right side of $0$