# Homework Help: Statics: Moment about a point

1. Sep 8, 2012

### Northbysouth

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

Points A and B are the midpoints of the sides of the rectangle. Replace the force F acting at A by a force-couple system at B. Give the general result and then answer for the specific case: F = 250 N, h = 185 mm, b = 490 mm. The couple is positive if counterclockwise, negative if clockwise.

2. Relevant equations

3. The attempt at a solution

I found R to be:

R = 250cos(37.065) + 250sin(37.0565)
R = 199.51i + 160.651j

I know my R is correct but I can't get the MB right.
If I draw a straight line connecting A and B, I find the angle at A in this triangle to be 20.6841 by using arctan.

At this point I have

MB = 250N*261.88cos(360-20.6841) + 250N*261.88sin(360-20.6841)
MB = -38.1249

It's negative because the force would clockwise. I've think I tried this answer previously and it was wrong. Does the y component go through the point B, thereby making it's value 0?

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2. Sep 9, 2012

### cmmcnamara

Hmmmm I do see that the program you are using for HW shows you getting the Ry of 160.651 as correct, really odd but I got Ry as 150.651....

For your moment, I'm not sure what you are actually doing here. If you attach the force at B, the original moment about O was already zero thanks to symmetry, but you now have to worry about the moment being created at O by placing the same force at B. If you look at the force components and perpendicular distances in symbolic form, you can see a bit easier I think:

$$\vec{F}=<\frac{Fb}{2\sqrt{b^2/4+h^2}},\frac{Fh}{\sqrt{b^2/4+h^2}}>$$

This I got simply by noting that F lies along line OA, so by finding the unit vector along OA and then scaling it by the force's magnitude I get the vector for force F. Additionally I can create position vectors for both points A and B, by noting their position relative to O and keeping in mind they are midpoints along each side:

$$\vec{OA}=<\frac{1}{2}b,h>;\vec{OB}=<b,\frac{1}{2}b>$$

All three of these vectors can now be used in moment equations about O. Originally I said that due to symmetrical geometry that the original position of the force creates no moment about O which you can test, but I'll skip it and just show the moment about O when the force is moved to B:

$$\vec{M_{OB}}=\vec{OB}\times\vec{F}=\frac{3}{4}Fbh(b^2/4+h^2)^\frac{-1}{2}\vec{k}$$

However, you need to counteract this moment with one placed at B, making the force couple system. This means your couple has to be opposite in direction to the created moment by the force to make a total of zero moment (the original amount when the force was at A).

If you plug all the variables in you should get a couple value of -55.4k.

About your last point, the y-component of the force definetly travels through B....that's the new point of application! But you are mixing up your reasoning here. A force creates no moment through a point which lies on its line of action, so yes the force would create no moment through B....however, B is not your point of interest here, it would be any other point on the object for which you would take the moment about.

3. Sep 9, 2012

### Northbysouth

Ry is 150.651. The 160.651 that I put was a typo, sorry about that.

I'll be honest and say that I'm not sure what I was doing with the moment.

With the F vector is that just the unit vector of the force? I'm a little unsure of what you did there. Could you explain it again?

4. Sep 9, 2012

### cmmcnamara

No the vector F was the full vector of the force. Basically the forces line of action is OA. So I made the position vector of A in relation to O. I then found the magnitude of OA and divided the position vector of A by the magnitude in order to create direction vector of A's position vector which also happens to be the position vector of the force. Any force can be written as the product of its magnitude and it's position vector which is how it is written above. I used it in that form to write a simple symbolic formula for M that I could just plug the values into.