# Homework Help: Statics problem.

1. Dec 9, 2011

### -Dragoon-

1. The problem statement, all variables and given/known data
In the attached image, find d.

2. Relevant equations

N/A

3. The attempt at a solution
Essentially, I did solve this problem correctly by first using forces and Newton's second law and then finishing it with torque, but I ended up with the wrong answer only because I assumed that the lower beam, which has a force (FG)1 in the diagram, exerts a positive torque since it would try to rotate the system counterclockwise, would it not? Why does my textbook assume that the torque of the lower beam is negative and, thus, would cause a clockwise rotation of the system?

Also, another classmate of mine said this problem could be done much more easily with using the center of mass equations. If they are correct, how would I go about using that method?

attached image: http://imageshack.us/photo/my-images/690/torqueproblem.jpg/

Last edited: Dec 9, 2011
2. Dec 9, 2011

### Delphi51

It looks to me like it will rotate clockwise because most of the mass is to the right of the pivot point. What is the question? Show your work so we can offer help!

3. Dec 9, 2011

### -Dragoon-

I've essentially solved the problem, I just wanted to know why they assumed the portion to the right of the pivot of the lower beam is going to try to rotate the system clockwise and thus have a negative torque? Oh, and as I was working through the problem, I separated the two masses so that the upper beam would exert a torque while the lower beam would exert another, just like the book did it.

4. Dec 10, 2011

### Delphi51

Sounds good. Maybe "they" took clockwise to be positive. It is difficult to comment without seeing the solution you are writing about.

5. Dec 10, 2011

### krash91

Yea, clockwise is usually negative. But some authors take it as positive. In this case, check to see if your math and your method is correct. The concept is more important. On a test or HW, you can always define CW and CCW.

6. Dec 11, 2011

### -Dragoon-

Ok, I think posting my worked solution would be the best way to proceed:

Let m be the mass of the lower beam and M be the mass of the upper beam. Let r1 be the distance from the pivot to the application of force of the lower beam and r2 be the distance from the pivot to the application of force. Firstly, I choose the pivot to be the left end of the lower beam to make the problem easier to solve. Using Newton's second laws:

$\sum f_{y} = F_{P} - F_{G_{m}} - F_{G_{M}} = 0$
$F_{P} = F_{G_{m}} + F_{G_{M}} = mg + Mg = g(m+M)$

Now to use Newton's second law for rotational statics:
$\sum \tau = \tau_{P} + \tau_{m} - \tau_{M} = 0 => \tau_{P} = -\tau_{m} + \tau{M}$
$dF_{P} = -r_{1}F_{m} + r_{2}F_{M} => d = \frac{-r_{1}F_{m} + r_{2}F_{M}}{F_{P}} => \frac{-r_{1}F_{m} + r_{2}F_{M}}{g(m+M)}.$

This is my worked solution. The textbook assumed that the torque exerted by the lower beam is negative, so they had a sum rather than a difference in the numerator of their solution. That assumption is what bothers me, as I thought I was right in assuming the lower beam would attempt to rotate the system counterclockwise.

7. Dec 12, 2011

### -Dragoon-

Any one?

8. Dec 12, 2011

### netgypsy

AM I missing something? What are you asked to find? I see the diagram but not the problem?? OK Sorry - find d

Is it in equilibrium?? I assume so.

Last edited: Dec 12, 2011
9. Dec 12, 2011

### -Dragoon-

The answer is 1.4 m, I would have gotten the same answer had I made one different assumption: Assuming the gravitational torque exerted by the lower beam that is left of the pivot causes a negative torque and thus tries to rotate the system clockwise rather than counterclockwise, which just doesn't make any sense to me at all.

10. Dec 12, 2011

### netgypsy

OK that's what I got only I did alot less work to get it.

First of all it doesn't really matter which you call positive. Clockwise or counterclockwise. Place the axis of rotation at Point P so the only torques you have are from the two masses (weights) of the plank and the fat plank.

Call the distance from the long plank's weight to P a variable like X This means that the distance from the fat plank's weight to point P will be .5 - x.

You now have two torque you can equate since they act in opposite directions.

The torque from the long plank (about point P)will equal the tgrque from the fat plank (about point P) . Solve for X and add it to the 1.0 meter length of half the long plank and you have d.

Try it like this and see if it doesn't make sense.

11. Dec 12, 2011

### -Dragoon-

The problem was not solving it, but I see it now. I was thinking of the fulcrum as the pivot even though I selected the pivot to be at the left end of the system, in which case the torque of the lower beam would try to rotate the system clockwise just as the torque of the upper beam would.

Thanks for all the help.

12. Dec 12, 2011

### netgypsy

The problem usually isn't the math, it's visualizing how to set it up in the first place.

The beauty of a statics problem with torque is that you can put the axis of rotation at the most convenient point to get rid of an unknown, which in this case is the force normal and it makes the distances easier. Other solutions are just as correct, not just as easy.