# Moment Couple Problem

1. Sep 29, 2005

Okay, having some problems figuring out how to start this problem on a moment couple:
http://img347.imageshack.us/img347/6939/problem55hb.jpg [Broken]

And the same thing for this one:
http://img347.imageshack.us/img347/2488/problem64sx.jpg [Broken]

Any tips on how to start each one? I think I just need a jumpstart....

Last edited by a moderator: May 2, 2017
2. Sep 29, 2005

### hotvette

Fundamental methodology for statics:

1. Draw free body diagrams of objects w/ forces and moments

2. $\Sigma F_i = 0$

3. $\Sigma M_i = 0$

3. Sep 29, 2005

Okay, that makes sense, and I just finished the first 4 of my 6 problems correctly, BUT I'm still a little stuck on just starting the last 2.

For the first one there, what exactly should I first do? I understand the situation, and what's happening, just trying to get what needs to be calculated.

Can you be a little more descriptive?

4. Sep 29, 2005

### hotvette

The last 2 meaning the ones you posted links for? We'll, I'm not sure what part you are stuck on:

1. Drawing the free body diagrams for each object? I guess the only thing I can think of to add is a hint that touching objects have equal and opposite forces on each other when you split them apart into free body diagrams

2. Sum of forces = 0? The only hint I can think of is to sum the forces separately in each coordinate direction. Treat each object separately.

3. Sum of moments = 0? To clarify, moment about the center of mass. If a circular object of radius r has a tangengial force f, the moment about the center of mass is rf. Make sure to keep the directions straight (i.e. CW vs CCW). Treat each object separately.

The equations you get from 2 & 3 will generally be n equations in n unknowns that you can solve.

Hope this helps.

5. Sep 30, 2005

Okay, well, it's still not clicking.

1) To find the force N, how should I be setting up the moments? I completely understand how the moment couple is equal and opp., but do I start by using the radius of one of the rollers and the given tension force as the moment?

Like this?: Ma=(r)(Fsin(theta))=(0.025m)(75N)sin(90)=0.375Nm

Is that a step I would take for the first posted problem?

6. Sep 30, 2005

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Last edited: Oct 1, 2005
7. Oct 1, 2005

Again...

#1) Should I find a moment on one of the rollers using the tension force and radius? If so, then what?

#2) If I find the moments acting on each handle, then how to combine them and get the direction/space angles of the resultant?

I have lots of examples, but none like either of these problems.

8. Oct 1, 2005

Can anyone just give me a first line for the first equation/step in the first one?

I get the "theory" points, but am confused because of the system in the problem.

9. Oct 1, 2005

### hotvette

This is strange. I posted a reply last night but don't see it. I don't think I was dreaming.....

Anyway, I'm not sure this is valid, but I think it is.

1. Consider the two rollers as a single object. At the top of the object is a normal force and a tangential one. Same with the bottom.

2. Summing forces in both coordinate directions should be easy.

3. Summing the moments can be a bit tricky. The moment to be summed is about the center-of-mass. Think about where the center of mass is of the combined rollers.

Hope this helps get you in the right direction.

10. Oct 1, 2005

Okay. So "Summing forces in both coordinate directions" means adding the x (T) and y (N) forces at both the top and bottom to get the total force acting on each roller? Like T+N=F in the equation M=rFsin(theta)?

And the center of the mass should be right where the two tangentially touch one another?

Last edited: Oct 1, 2005
11. Oct 1, 2005

Like this?
http://img392.imageshack.us/img392/8307/problem5start5sm.jpg [Broken]

Oh no, wait....would I keep the r as just r, and then plug that in when finding the total moment of the system--the center point of the system?

Last edited by a moderator: May 2, 2017
12. Oct 1, 2005

Okay, went over it again:
M=(r)(F)

At top:
(F)=(-Ti-Nj)
so MA=(-r)(Ti+Nj)

At bottom:
(F)=(Ti+Nj)
so MB=(r)(Ti+Nj)

Looking right so far? Then, to add them together, I have to use r= the center point of the whole system? What about the angle?

Last edited: Oct 1, 2005
13. Oct 1, 2005

M=(r)(F)

At top:
(F)=(-Ti-Nj)
so MA=(-r)(Ti+Nj)

At bottom:
(F)=(Ti+Nj)
so MB=(r)(Ti+Nj)

Then:
Meq=MA+MB=[(r)(Ti+Nj)]+[(r)(Ti+Nj)]
Meq=(-r)[(Ti+Nj)]
=(-rTi)+(-rNj)
-rNj=Meq+rTi (Meq goes to zero?)
Nj=(rTi)/(-r)
Nj=(-Ti)
Nj=(-15N) Something's wrong, this can't be right, but where did I flub up? I think the radius to the center moment needs to be included, but couldn't see how, since it cancelled out, or so I thought...

14. Oct 1, 2005

Where'd I "miss"?

15. Oct 1, 2005

Please, anybody, I really have to get these done!

Anyone know or are they just weird problems?

16. Oct 1, 2005

### hotvette

I hope we're talking about the first problem, the one with the 2 rollers between the belts.

Re summing the forces, actually that's already done for you in the problem, and the appropriate coordinate system for that is horizontal (x) and vertical (y) because those are the directions the forces lie in. All this means is that the N at the top is the same as N in the bottom but in the opposite direction. So, $\Sigma F_y = 0$ just means N - N = 0, or N = N, which is already given. It might make more sense if the top one had been labeled N1 and bottom one N2. Anyway, they have to be equal to each other because there are no other forces in the y direction.

Now, take the x direction. $\Sigma F_x = 0$ just means that T - T = 0. Again, trivial in this problem. So, forget about summing forces because it's already done.

Now, for the moments. By the way, you are correct that the center of mass is where the 2 rollers touch. There are 4 moments acting at the center of mass. At the top you have N times it's 'lever arm' (don't know what terminology you use), T times its lever arm acting in the opposite direction as the first one. You also have the same situation at the bottom. Thus, all you have to do is calculate the lever arms based on the geometry, multiply those by the forces (keeping the CCW and CW directions correct), sum them, set the sum equal to zero, and solve for N (because T is expressed in terms of N). You have single equation in N.

If you aren't sure what I mean by 'lever arm', a moment is determined by multiplying a force by a distance that is perpendicular to the direction of the force and through the center of mass. In other words, the force and the lever arm are at right angles.

P.S. I hope your other post about nobody responding was referring to the 2nd problem, not this one.

Last edited: Oct 2, 2005
17. Oct 2, 2005

OH!! That makes sense! Thank you so much!

And yeah....got any clues on that one? I THINK I may have calculated the moments on each wheel correctly (but very likely didn't), but then am really confused as to how to get their resultant--is the resultant about the axes origin? What about the resultant angles? Since nothing else is given but the 60deg. angle from the y-axis, I'm wondering just what to do--sum them somehow? Here it is again:
http://img347.imageshack.us/img347/2488/problem64sx.jpg [Broken]

Last edited by a moderator: May 2, 2017
18. Oct 2, 2005

### Staff: Mentor

Each couple produces a vector moment. (What is the direction of those moments?) Add those two vectors.

Last edited by a moderator: May 2, 2017
19. Oct 2, 2005

Sorry about that....a little late night hysteria

Thanks, I'll see where this all gets me.

20. Oct 2, 2005

Okay, so for the first problem with the rollers, would the "lever" arm be the distance from point A to the center of the masses? Then point B to the center, etc.? I'm still a little confused on that one. Would the moments with N have a different r than the moments with T?

Like:
At top:
M1)=(-N)(r)
M2)=(-T)(r)

At bottom:
M3)=(N)(r)
M4)=(T)(r)

where r is the distance mentioned above, and then plugged in at the end when all the moments are summed? Do I need an angle for each?

Last edited: Oct 2, 2005