The Center of Gravity — Question 7

Question 7

One orientation fixes the weight line but not a unique point on it

A body has total downward weight $150\,\mathrm N$ and gravity torque $-45\,\mathrm{N\,m}$ about the origin in a vertical plane. Its orientation is held fixed.

1. Find the horizontal coordinate of the resultant weight line.

2. Can these two measurements determine the vertical coordinate of the center of gravity?

Question 7 - Solution

One orientation fixes the weight line but not a unique point on it

(a) Derivation and calculation

A downward resultant $\vec W=-150\hat y$ N acting at $(x_G,y_G)$ has moment $$\tau_z=x_G(-150)-y_G(0)=-150x_G.$$ Hence $$-150x_G=-45\quad\Longrightarrow\quad\boxed{x_G=.300\ \mathrm m}.$$

(b) Result and interpretation

The measurement fixes the line $$\boxed{x=.300\ \mathrm m},\qquad y\ \text{arbitrary}.$$ Translation of the application point along this vertical line leaves the moment unchanged: $$\Delta\vec r\times\vec W=(\Delta y\hat y)\times(-150\hat y)=\vec0.$$ A second independent orientation or geometric constraint is needed to recover a particular height.

Physical check

The clockwise moment is consistent with a downward force to the right of the origin.