The Center of Gravity — Question 9

Question 9

Ballast must keep the combined center inside the support interval

A $20\,\mathrm{kg}$ machine has center of gravity at $x=.70\,\mathrm m$. Its ground support interval is $-.50\le x\le.50\,\mathrm m$. A point ballast of mass $m$ can be mounted at $x=-1.0\,\mathrm m$ on a massless rigid arm. Gravity is uniform.

1. Find the full range of $m$ that keeps the combined gravity projection within the support interval.

2. Interpret both endpoints and distinguish marginal from strict stability against tipping.

Question 9 - Solution

Ballast must keep the combined center inside the support interval

(a) Derivation and calculation

With ballast mass $m$ in kilograms, the combined center is $$x_G=\frac{20(.70)+m(-1)}{20+m}=\frac{14-m}{20+m}\ \mathrm m.$$ Admissible support requires $$-.5\le\frac{14-m}{20+m}\le.5.$$

(b) Result and interpretation

Since $20+m>0$, the upper bound gives $$14-m\le10+.5m\quad\Longrightarrow\quad m\ge\frac83.$$ The lower bound gives $$14-m\ge-10-.5m\quad\Longrightarrow\quad m\le48.$$ Thus $$\boxed{\frac83\le m\le48\ \mathrm{kg}}.$$ Use strict inequalities to keep the resultant strictly inside the support footprint.

Physical check

Too little ballast permits right-edge tipping; excessive ballast can cause left-edge tipping.