# Summing Torques about a square

1. Mar 26, 2014

### sreya

1. The problem statement, all variables and given/known data

A square metal plate 0.180 m on each side is pivoted about an axis through point O at its center and perpendicular to the plate (Figure 1) .

Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F1 = 25.0N , F2 = 15.5N , and F3 = 17.0N . The plate and all forces are in the plane of the page. Take positive torques to be counterclockwise.

2. Relevant equations

$\tau=rFsin(\phi)$

3. The attempt at a solution

$rF2+rF3-rF1*sin(45) = .675$, r=.09

Apparently that's incorrect though, could anyone provide insight as to why?

Solution

There are two ways to express torque:

$\tau = r(Fsin\phi)=rF_t$
$\tau = rsin\phi F =r_{\perp}F$

So, the solution would go like so:

$r_1=r_2=r_3=\frac{l}{2sin\theta}$ where $\theta=45$ and l = length of side
$r(F_1*sin(45)+F_3-F_2*sin(45))$

Since the moment arm ($t_{\perp}$) forms a 45 degree angle with F1 and F2

So unlike i tried in the beginning the real equation has the form:

$F_1r_{\perp}-F_2r_{\perp}+rF_t$ where $F_3=F_t$

Last edited: Mar 26, 2014
2. Mar 26, 2014

### paisiello2

Looks like you set it up fine in the first equation. You probably just did an arithmetic mistake somewhere.

3. Mar 26, 2014

### collinsmark

Not quite.

By the equation above, you have the sin(45) associated with F1. Which force is at the 45o angle? Is it F1 or F3?

(using r = 0.09 m, in this case).

Just try to be a little more careful which vectors are at an angle, and which are not.

I'm not sure I follow you on that 100%, but from what I assume you are doing, it might be a valid approach. There is one problem though regarding your signs. Two of the forces produce a counterclockwise torque (positive for this problem) and one of your forces produces a clockwise torque (negative for this problem). Which is which?

Last edited: Mar 26, 2014
4. Mar 27, 2014

### sreya

F1 and F3 produce a counterclockwise torque, F2 produces a clockwise, I thought that was evident by the signs associated with each torque:

$r(F1∗sin(45)+F3−F2∗sin(45))$

There are other angles in my solution that I couldn't draw on the picture, I didn't mean to assume that the angle theta given in the problem somehow relates to the other two forces. The reason there is no sine associated with the F3 is because the force is already perpendicular to r.

The reason the other two forces have sines associated with their forces is because they do not act on r at a perpendicular angle and since their components are itself (in the y) and 0 (in the x) you have to associate the sine with the moment arm (which is the component of r that is perpendicular to the force), all the angles just happen to be equal 45 degrees due to the nature of a square

Last edited: Mar 27, 2014
5. Mar 27, 2014

### paisiello2

From your diagram F2 and F3 produce a counterclockwise torque. F1 produces a clockwise torque.

But other than the signs your formula is correct.

I think it would be more straight forward to just take F2 and F1 with l/2 lever arms directly rather than convert them to the same lever arm as F3. Less work and less chance of making a mistake.

6. Mar 27, 2014

### sreya

Oh! Yes, you are absolutely right that's a typo, F1 should have a negative value and F2 should be positive