Question:
Two unlike parallel forces $P$ and $Q$ $(P>Q)$ act at $A$ and $B$ respectively. If P and Q are both increased by R, show that the resultant will move through a distance $d=\frac{R}{P-Q}$.
My book's attempt:
Let the resultant of two unlike parallel forces $P$ and $Q$ act at $C$.
$$\frac{P}{BC}=\frac{Q}{AC}=\frac{P-Q}{AB}$$
$$\text{Each force is proportional to the distance between the points of application of the other two.}$$
From 2nd and 3rd ratios,
$$AC=\frac{Q}{P-Q}AB\tag{1}$$
Let when $P$ and $Q$ are increased to $P+R$ and $Q+R$, their resultant act at $D$, where $CD=d$.
$$\frac{P+R}{BD}=\frac{Q+R}{AD}=\frac{P+R-Q-R}{AB}$$
Each force is proportional to the distance between the points of application of the other two.
From the 2nd and 3rd ratios,
$$AD=\frac{Q+R}{P-Q}AB\tag{2}$$
$(2)-(1)$,
$$AD-AC=\frac{Q+R-Q}{P-Q}AB$$
$$d=\frac{R}{P-Q}AB\ (\text{showed})$$
My comments:
My book assumed that the forces $P$ and $Q$ acting at $A$ and $B$ respectively will have a resultant force $P-Q$ that will act at $C$. I take issue with this. I would've agreed with the book if P and Q were both acting at the same point; then the resultant would've been $P-Q$, which would've acted at the same point as $P$ and $Q$. However, that's not the case here. P and Q are acting at two different points in opposite directions. I think what will happen is that the body that the two forces are acting on will experience a net torque and start rotating.
My question:
- Isn't my book's attempt wrong?
Best Answer
There is nothing in that resultant force calculation that contradicts what you've said. Saying that the resultant force has size $P - Q$ and acts at a point $C$ does not somehow stop the body from rotating about a pivot - either a fixed point that is not free to move, or the body's center of mass if it is free - unless the point $C$ just happens to be that pivot. A non-zero resultant normal force acting at some distance away from the pivot will cause it to rotate.
The resultant force of several forces acting on a rigid body is a mathematical simplification. From the equations of motion, we find it is possible to treat all the force vectors acting at various points on a rigid body as if they were all acting at a single point, the "center of force". The behavior of the body as a whole will be the same for all the forces acting there as it is for each force acting in its real location. And once we move all the forces to act at the center of force, we can now just add them all together to get a single resultant force.
Your course has already covered this, but you apparently missed the significance. You can find the proof that it works there. The book expected you to already know what it was talking about with this calculation, so it didn't explain this part. $C$ is not some arbitrary point. It is rather the center of force for $P$ acting at $A$ and $Q$ acting at $B$. That is, it is the one point where the body would respond the same if $P$ and $Q$ were both acting at $C$.
Similar remarks for $D$ and $P+R, Q+R$ acting at $A, B$.