- #1

zenterix

- 515

- 74

- Homework Statement
- Three forces are applied to a square plate as shown in the figure below. Find the modulus, direction, and the point of application of the resultant force, if this point is taken on the side ##BC##.

- Relevant Equations
- ##F=ma##

Here is what I came up with

First, let's compute the resultant force.

$$F_{R,x}=F+F\sqrt{2}\cdot\frac{\sqrt{2}}{2}=2F$$

$$F_{R,y}=F-F\sqrt{2}\cdot\frac{\sqrt{2}}{2}=0$$

At this point we already have the magnitude and direction of the resultant force. We need the point of application.

**My question is how to find this point.**I don't actually know how to do this.

What I thought at first was that the resultant force would have to be applied at a point such that the torque produced by it would be the same as the net torque on the rigid body.

Let ##l## be the length of a side of the square plate.

Net torque about the center is

$$\vec{\tau}=-F\frac{l\sqrt{2}}{2}\cdot 2\hat{k}+F\sqrt{2}\frac{l\sqrt{2}}{2}\hat{k}$$

$$=Fl(1-\sqrt{2})\hat{k}$$

and if ##\vec{r}=x\hat{i}+y\hat{j}## is the position where the resultant force is applied we have

$$\vec{r}\times\vec{F}_R=(x\hat{i}+y\hat{j})\times 2F\hat{i}=-2Fy\hat{k}$$

If we equate this to net torque and solve for ##y## we get

$$y=\frac{l(\sqrt{2}-1)}{2}$$

which seems to be the purple line below

It seems not to matter what the ##x##-coordinate is.

The solution manual says the following:

##F_R=2F## This force is parallel to the diagonal AC and is applied at the midpoint of the side BC.

Which is not in accordance with my attempt at a solution. Which isn't surprising given that I am very unsure about how to find the point of application of a resultant force on a rigid body.