Force — Question 9

Question 9

Components are not additional interactions

A 20 N force points $60^\circ$ above horizontal. A draft diagram includes that arrow plus separate rightward 10 N and upward $10\sqrt3$ N arrows, all labeled as physical forces.

1. Find the correct components of the 20 N force.

2. What resultant would the erroneous three-force inventory produce?

3. How should a component diagram distinguish representation from interaction?

Question 9 - Solution

Components are not additional interactions

(a) Resolve the single interaction.

$$F_x=20\cos60^\circ=10,\quad F_y=20\sin60^\circ=10\sqrt3\quad\mathrm N$$ $F_x=20\cos60^\circ=10$ N and $F_y=20\sin60^\circ=10\sqrt3$ N.

(b) Identify the double count.

$$\vec R_{\mathrm{draft}}=\vec F+F_x\hat i+F_y\hat j=2\vec F,\quad\boxed{R_{\mathrm{draft}}=40\,\mathrm N}$$ The two component arrows already sum to the original 20 N vector. Adding all three gives twice that vector, magnitude 40 N at $60^\circ$.

(c) Choose one accounting method.

$$\vec F\equiv F_x\hat i+F_y\hat j\quad\text{(one interaction)}$$ Use the physical force arrow with dashed component projections, or replace it algebraically by its components. Components are a decomposition of one force; they are not two additional agents acting on the body.