# One Moment please HELP

1. Dec 29, 2007

### Saladsamurai

One Moment please...HELP!!!

I am supposed to use the principle of moments for this. that is break down each force into x and y components and find their individual moments about O and then add em up.

I am having most of my trouble with the SIGNS.

So far I have broken both r into components and both forces. Now. Is there a sign on r?
Like for the $r_OB}$ the x and y components are negative. So I need to sign them as such right?

And I am also having trouble with the right-hand rule for components. Take again the x component of $r_{OB}$ I would hold my wrist where the tail of itex]r_{OB}[/itex] meets O, point my fingers along the Y component of F_B and curl them in the direction of rotation right. That is, for each component of each r, I need to relocate my hand.

2. Dec 29, 2007

### Saladsamurai

I don't know why my solution is incorrect.

I am using $M_O=-r_{OA}_yF_{Ax}-r_{OA}_xF_{Ay}-(-r_{OB}_yF_{Bx})-(-r_{OB}_xF_{By})$

I am supposed to get -2.42 kiP*m

3. Dec 29, 2007

### Staff: Mentor

If you use the component method to find the cross product ($\vec{r}\times\vec{F}$), then you don't have to worry about the right hand rule. The signs are handled automatically.

What did you use for the components of each vector?

(Be sure you are computing the cross product correctly; review it here: Cross Product.)

4. Dec 29, 2007

### Saladsamurai

I did not think I was using cross product. I thought that the component method was scalar?! I will post my exact components in a moment. . . no pun.

5. Dec 29, 2007

### Saladsamurai

I used $M_O=-r_{OA}_yF_{Ax}-r_{OA}_xF_{Ay}-(-r_{OB}_yF_{Bx})-(-r_{OB}_xF_{By})$
$\Rightarrow M_O= -10\cos 30^\circ (3/5)(250)-10\sin 30^\circ (4/5)(250)+6(300\sin 30^\circ)+3(300\cos 30^\circ)$

6. Dec 29, 2007

### Staff: Mentor

Not sure what all those negative signs are doing there. The resultant moment is just the sum of the individual moments:
$$M_O= (-r_{OA}_yF_{Ax} + r_{OA}_xF_{Ay}) + (- r_{OB}_yF_{Bx} + r_{OB}_xF_{By})$$

The sign of the third term is incorrect.

7. Dec 29, 2007

### Saladsamurai

Well, this relates to my OP. Why are the signs incorrect? That is where my trouble lies.
Take the x component of of r_OB. It is on the Quadrant III so is negative-->-3 .......hmmm I may see it now. The x component of r_OB is - but since the y component of F_B is counterclock it is positive. Is this right?

8. Dec 29, 2007

### Staff: Mentor

The x and y components of r_OB and F_B are all negative.

As I said before, if you are using the component method to calculate moments (which is a cross product), then you don't have to worry about the right hand rule or about clockwise versus counterclockwise.

$$\vec{r}\times\vec{F} = r_x F_y - r_y F_x$$

List the components of each vector. Then just plug them into the formula.

9. Dec 29, 2007

### Saladsamurai

This would be great, but it's not what I was asked to do.

Doc ... can we just analyze this term by term here? I'd appreciate your time.

The first term is $(-r_{OA}_yF_{Ax}$ now the negative out front comes from the force component right? The r component is +.

2nd term $r_{OA}_xF_{Ay}$ you say is positive. I see r as positive, but why is F? Is it not moving clockwise?

3rd term $- r_{OB}_yF_{Bx}$ I see r is negative, but why isn't the F_Bx also negative? Is it not clockwise??

4th term $r_{OB}_xF_{By}$ r_x is positive. But is not F_BY counter clockwise? Making the whole term negative?

If someone could just reply to these as 1,2,3,4 respectively. I don't know where my concepts are getting effed...but they are!

Last edited: Dec 29, 2007
10. Dec 29, 2007

### Staff: Mentor

Huh?

The r component is + and the F component is +, so the term is negative; the negative sign comes from the definition of cross product.

r is +, but F is -. Where did I say anything different?

(Forces are not clockwise, torques are!) Here both r and F are negative, so the term is negative due to the minus sign out front. (Compare this with what you did.)

Both components are -, making the term +.

11. Dec 29, 2007

### Saladsamurai

I don't understand your question??

Quote:
2nd term you say is positive. I see r as positive, but why is F? Is it not moving clockwise?

r is +, but F is -. Where did I say anything different?

Right there..... if r is + and F is - then how can r*(-F)=+ ? I don't get it.

12. Dec 29, 2007

### Saladsamurai

For the general case $r_yF_x$ should I just calculate the magnitude $r_yF_x$ and then determine the sign after by taking note whether this would cause the point where the force acts to tend to rotate about O clockwise or counterclockwise?

13. Dec 29, 2007

### tongpu

Find the magnitude of Moments (always positive) |rXF|= |f||r|sin90 always if you use components. Then use the right hand rule (rXF) to figure into-the-page (negative) or out-of-the-page (positive)

Last edited: Dec 29, 2007
14. Dec 29, 2007

### Staff: Mentor

(1) The torque produced by F_A is definitely clockwise and thus is negative.
(2) The way you are calculating the torque is by using components: The x-component of F_A is positive and the y-component of F_A is negative. Thus:

Torque_A = -r_y*F_x + r_x*F_y
The first term is -(+)*(+) = negative;
The second term is +(+)*(-) = negative.

Thus the torque is negative, which is consistent with (1).

No. There are two ways to calculate torque: (A) By components and the rule stated above, or (B) Using rF sin(theta).

When using method A, just plug in the components like a machine. What matters is whether the components are negative or positive. This is what you are trying to do.

Only when using method B do you have to worry about using the right hand rule to figure out whether the torque is clockwise (-) or counterclockwise (+). But you aren't using this method, so you don't need to worry about the right hand rule.

Of course you can always use the right hand rule to double check your work using method A. For example, we know from the right hand rule that the torque from F_A must be negative, which confirms what we already know from using the components.

If you wish, and perhaps this is what you are doing, you can view F_A as being composed of two components and then find the torque due to each component. The torque due to the x-component of F_A is clockwise (using the right hand rule) so must be negative = -|r_y|*|F_x|; similarly, the torque due to the y-component of F_A is clockwise (using the right hand rule) so must be negative = -|r_x|*|F_y|. Using this method we forget about the sign of the components, just using their magnitudes, and get the sign of the torque using the right hand rule.

15. Dec 30, 2007

### Saladsamurai

Yes. In the first line of my OP, I belive this is what I stated. Hence my telling you that plugging and chugging would be great and all, but it was not the method that I was told to use.

Now in this case, I just pay attention to magnitudes, and then lastly determine in what direction each component would cause the point to rotate about O.

Thanks for all of your help Doc!
Casey

16. Dec 30, 2007

### Staff: Mentor

D'oh! Sorry for not reading your first post carefully. :uhh: I hope the digression did not confuse you too much!

Anyway, have you found your sign error?

My comment in post #6 remains true. Your third term there should be negative, since that's the moment due to the x-component of F_B. That torque is clockwise, so it should be negative.

Last edited: Dec 30, 2007
17. Dec 30, 2007

### Saladsamurai

Not to worry! I learn the most when I get severely confused and then suddenly it all becomes clear.

And yes, i found my error. I just made a table in which one column was the magnitudes
r_yF_x and so on and then next to each I drew an arrow that showed the sense of each. I then assigned a - to any clockwise 'sense arrows' and a + to any counterclock. Then I summed them altogether.

This Varignon's Theorem (the principle of moments) correct?

Thanks again!
Casey

18. Dec 30, 2007

### Staff: Mentor

Good.

Absolutely.

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