Question

An elephant weighs $7500kg$ and has a height of $2.7m$.
If you could stack $73.5kg$ capybaras each $50cm$ in height on top of the elephant, at how many capybaras added would the center of mass of the system be vertically outside the elephant, that is to say $y_{CM}>2.7m$?
For the purposes of this problem, treat the elephant and capybaras as a uniform masses (each animal has its center of mass at its vertical center).

• A. $n=6$
• B. $n=24$
• C. $n=52$
• D. $n=103$
• E. $n=206$

Answer 1

The correct option is B $n=24$

Given : $M=7500$ kg $m=73.5$ kg $y=1.35$ m $y_1=2.95$ m $y_2=3.45$ m $y_3=3.95$ m

Let $n$ numbers of capybaras are added.

Using $y_{CM}=\frac{My+(m{y}_1+m{y}_2+...m{y}_n)}{M+nm}$ where $y_{CM}>2.7$ m

$\therefore$ $\frac{My+m(y_1+y_2+...y_n)}{M+nm}>2.7$

OR $\frac{\left(7500\right)\left(1.45\right)+73.5(2.95+3.45+...y_n)}{7500+n\left(73.5\right)}>2.7$

Sum of n terms of AP $S_n=\frac{n}{2}(y_1+y_n)=\frac{n}{2}(2.95+2.45+0.5n)$ (using $y_n=2.45+0.5n$)

OR $\frac{10125+73.5(2.95+2.45+0.5n)\frac{n}{2}}{7500+n\left(73.5\right)}>2.7$

$\therefore$ Solving we get $n^2>551.02$ $⟹n>23.47$

Hence, $24$ capybaras must be added so that COM of the system lies outside the elephant.