The Center of Gravity — Question 6

Question 6

A vertical gravity gradient changes the resultant line of action

Two equal masses $m$ are fixed at $(x,y)=(-1,0)$ and $(1,2)\,\mathrm m$. The downward field has magnitude $g(y)=g_0[1+0.10y/(1\,\mathrm m)]$. Treat the masses as point particles.

1. Find the horizontal line-of-action coordinate of their total weight and compare it with $x_C$.

2. Can one replace the weights by $2mg(y_C)$ acting at the mass center and reproduce both force and torque?

Question 6 - Solution

A vertical gravity gradient changes the resultant line of action

(a) Derivation and calculation

The local weights differ: $$W_1=mg_0,\qquad W_2=1.2mg_0,\qquad W=2.2mg_0.$$ The vertical resultant line satisfies $$x_G=\frac{(-1)mg_0+(1)(1.2mg_0)}{2.2mg_0} =\boxed{\frac1{11}\ \mathrm m}.$$ By contrast, $x_C=(-1+1)/2=0$.

(b) Result and interpretation

The mass center has $y_C=1$ m, so the field-at-center approximation predicts $$2m g(y_C)=2m(1.1g_0)=2.2mg_0,$$ which gives the correct force. Its torque, however, is zero at $x_C=0$, whereas the actual torque is $$\tau_z=-mg_0(-1)-1.2mg_0(1)=\boxed{-.2mg_0\ \mathrm m}.$$ Correct net force does not guarantee the correct line of action.

Physical check

Parallel nonuniform gravity weights positions by local weight, not simply by mass.