Torque Revisited — Question 5

Question 5

Uniform gravity acts at the center of mass for torque calculations

Three particles lie at $(0,0)$, $(2,0)$ and $(0,3)\,\mathrm m$, with masses $1$, $2$ and $3\,\mathrm{kg}$ respectively. Gravity is uniform, $\vec g=-10\hat y\,\mathrm{m/s^2}$.

1. Find the gravity torque about the origin by summing particles.

2. Locate the center of mass and verify an equivalent single resultant.

Question 5 - Solution

Uniform gravity acts at the center of mass for torque calculations

(a) Derivation and calculation

Each downward force is $\vec F_i=-10m_i\hat y$. Thus $$\tau_{z,i}=x_iF_{iy}=-10m_ix_i,$$ $$\boxed{\tau_z=-10[1(0)+2(2)+3(0)]=-40\ \mathrm{N\,m}}.$$

(b) Result and interpretation

The total mass and mass-center coordinates are $$M=6\ \mathrm{kg},\quad x_C=\frac{1(0)+2(2)+3(0)}6=\frac23\ \mathrm m,$$ $$y_C=\frac{1(0)+2(0)+3(3)}6=\frac32\ \mathrm m.$$ The resultant $\vec F_g=-60\hat y$ N at $C$ gives $$\boxed{\vec r_C\times\vec F_g=-40\hat z\ \mathrm{N\,m}},$$ consistent with $\sum\vec r_i\times m_i\vec g=\vec r_C\times M\vec g$ for uniform gravity.

Physical check

Only horizontal lever arms matter for the moment of vertical gravitational forces.