The Requirements of Equilibrium — Question 2

Question 2

Three nonparallel support forces must meet on a common line intersection

A planar rigid body is in equilibrium under exactly three nonzero forces and no applied couples. Two of their lines of action intersect at a finite point $P$, and are not parallel.

1. Prove that the third line must also pass through $P$.

2. Is concurrence alone sufficient? Test three concurrent forces $(3,0)$, $(0,4)$ and $(-3,-3)\,\mathrm N$.

Question 2 - Solution

Three nonparallel support forces must meet on a common line intersection

(a) Derivation and calculation

Let the third force act at $\vec r_3$ relative to $P$. The first two force lines pass through $P$, so $$\tau_{1P}=\tau_{2P}=0,\qquad\sum\tau_P=\vec r_3\times\vec F_3=0.$$ Because $\vec F_3\ne0$, its perpendicular lever arm must vanish: $$F_3d_\perp=0\quad\Longrightarrow\quad\boxed{d_\perp=0}.$$ Thus its line of action also passes through $P$.

(b) Result and interpretation

Concurrence guarantees zero torque about $P$ but not force balance. For the example, $$\sum\vec F=(3,0)+(0,4)+(-3,-3)=\boxed{(0,1)\ \mathrm N}\ne\vec0.$$ The force polygon fails to close. Static equilibrium additionally requires $$\sum F_x=0,\qquad\sum F_y=0.$$

Physical check

The argument excludes parallel force lines and applied couples, as stipulated.