Statics: Moment about a point

[Northbysouth]

Homework Statement

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.

Homework Equations

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 M_B 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

M_B = 250N*261.88cos(360-20.6841) + 250N*261.88sin(360-20.6841)
M_B = -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?

 

[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 definately 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.

 

[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?

 

[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.

 

[rezeew]

It seems like you have the right idea, but there are some issues with your calculations. Let's break it down step by step:

1. Finding R:

You are correct in finding the components of R, but you have made a small error in your calculations. The angle you should be using is the angle at B, not A. This is because we are trying to replace the force at A with a force-couple system at B. Therefore, the angle at B is the relevant angle for our calculations, not the angle at A. So the correct calculations for R are:

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

2. Finding the angle at B:

You are correct in using the angle at B, but the value you have calculated is incorrect. The angle at B can be found by using arctan(185/245) which gives us an angle of 37.0565 degrees, not 20.6841 degrees. This is the angle you should use in your calculations for MB.

3. Finding MB:

You are on the right track, but again there are some errors in your calculations. The first term in your calculation is correct, but the second term should be negative because the angle is measured counterclockwise from the positive x-axis. Also, the angle you should be using is the angle at B, not A. So the correct calculations for MB are:

MB = 250N*261.88cos(360-37.0565) - 250N*261.88sin(360-37.0565)
MB = -205.69

Overall, your approach is correct, but just be careful with your calculations and using the correct angles in each step. The final answer should be a negative value because the couple is clockwise, as you correctly stated in your response. Good job!