Force — Question 10

Question 10

A vector sum is a translational description

Two forces on an extended bar are 5 N east at its upper end and 5 N west at its lower end. The ends are separated vertically. A second arrangement applies both opposite forces at the same point.

1. Find the resultant in each arrangement.

2. Does this calculation distinguish their tendencies to turn the bar?

3. What information is lost by retaining only the force sum?

Question 10 - Solution

A vector sum is a translational description

(a) Add the vectors.

$$\boxed{\vec R=(5,0)+(-5,0)=(0,0)\,\mathrm N}\quad\text{in both cases}$$ Both arrangements have zero resultant: $(5,0)+(-5,0)=(0,0)$ N.

(b) Keep the domain of the calculation.

$$\tau_z=-(d/2)(5)+(d/2)(-5)=-5d\,\mathrm N,\quad\tau_{z,\mathrm{coincident}}=0$$ Let $d$ be the vertical separation, with counterclockwise torque positive. The net-force sum alone cannot distinguish their rotational effects. Separated opposing lines of action can turn a bar; coincident opposite forces cancel without that same turning effect.

(c) Retain application geometry when needed.

$$\vec\tau=\sum_i\vec r_i\times\vec F_i,\qquad\sum_i\vec F_i=\vec0\ \not\Rightarrow\ \vec\tau=\vec0$$ Points and lines of application are lost by replacing the forces only with a sum. A particle translation model uses the resultant, while extended-body rotation needs additional information and the later torque framework. Zero resultant alone does not guarantee complete equilibrium of an extended body.