Question

A half meter rod is pivoted at the centre with two weights of $20gf$ and $12gf$ suspended at a perpendicular distance of $6cm$ and $10cm$ from the pivot respectively as shown below.

i) Which of the two force acting on the rigid rod causes clockwise moment?

ii) is the rod in equilibrium?

Answer 1

$\left(ii\right)$COM due to these masses will be at the point of tip of the pivot as,

$X_{cm}=\frac{M_1{X}_1+M_2{X}_1}{M_1+M_2}$

$x_{cm}=\frac{20\times 0+12\times 16}{20+12}$

$\therefore x_{cm}=6cm$, from $20gf$

i.e. at tip of pivot

$\to$ Rod is in equilibrium

$\left(i\right)12gf$, as it forces rod in down-wards direction from right side,

$\therefore$ clockwise direction