
#1
Sep2005, 07:00 PM

P: 9

I'm taking a Statics course in college right now. However, I wasn't able to get a book due to financial issues early on. Naturally I feel behind. Anyways, I have a test coming up and the teacher gave us four problems to solve as a study guide, saying similar problems would be on the test.
Looking over the assignment, I relized that I REALLY don't know what I'm doing. Is there anyone that could help me figure out how to do the four problems so that tomorrow night, when I take my test I'll have an idea of what I have to do? If anyone can help, I'll try and post the problems. If someone could just guide me through the steps I would be VERY greatful! 



#2
Sep2005, 07:02 PM

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So POST the problems already!




#3
Sep2005, 07:04 PM

HW Helper
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Sure, i'll help, maybe you want to review your vector algebra?




#4
Sep2005, 07:09 PM

P: 9

I need urgent help in Statics
ok... not really sure how since there is a picture... anyways
1)A 400N force P is applied at point A off the bell crank shown. (a) Compute the moment of the force P about O resolving it's components. A B / / \ \ <92> / / \ \ / / \ \ / / 42 \ O / 40 42 and 40 degree's off x Axis  and the Force  P, comes off of point A at 20 degrees (60 off axis) The smaller lenght (42 degree side) is 120 mm, and the 40 degree side is 200 Did I forget anything? 



#5
Sep2005, 07:12 PM

P: 9

Wow... spaces dont work in this forum.... ummmm fom the center point, there is a line going 120 mm at 132 degrees, and a line moving 200 mm at 40 degrees and the force is 400N at 20 degrees off point A.
Point A is the end of the 200 mm line, B is the end of the 120mm line, and O is the axis. 



#6
Sep2005, 07:19 PM

P: 9

Would it be easier if I just scanned the 4 problems and sent the pictures?




#7
Sep2005, 07:30 PM

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But spaces do in fact work if you use code tags (but without the spaces): [ code ] (stuff) [ /code ] Example:




#8
Sep2005, 07:53 PM

P: 9

Heres the first two problems.




#9
Sep2005, 08:04 PM

P: 9

On the first problem, what throws me off is the force. I've solved problems where there were two forces on a pivot. Thats done by making the Triangle and using sin and cosine laws.
But I don't see how this one would be done the same. If it was a force of 120N and 200N, at those angles, I would be able to do it.  On the second problem, I'm sure you could use the vector triangles, but how would the triangles be set up since there are 4 forces on it. I hate to sound dumb, but I really just don't know what I'm doing. Statics is VERY new to me. 



#10
Sep2005, 08:35 PM

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The definition of a moment is [itex]\vec{M}=\vec{r}\times\vec{F}[/itex], right? So you're calculating the moment about O, which means that you can resolve both the displacement and the force vectors into rectangular components and compute their cross product. Have you tried that?




#11
Sep2005, 08:49 PM

P: 9

The rectangle would have to be perpendicular to the Force. Since the force is at 60 degrees, perpendicular would be 110 degree. If I made a rectangle it wouldn't be perpendicular. So I wouldn't know how to go about doing that. I considered projecting downward from Point A and forming a right triangle. A lot of our example problems are done by making a right triangle and labeling the two sides, using the object as a hypotnus. I don't think I have enough information to determine the lenght of the sides using that method though.
(I'm telling you all this so you know that I really am trying, and not just looking for answers. I need to learn to do this on my own.) You suggested a rectangle, and I'm sure you could easily do this problem. But since the pivot isn't a 90 degree angle, and the foce isn't perpendicular to the object, I don't understand how you would work it that way. Could you explain? 



#12
Sep2005, 09:09 PM

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P: 2,280

For the first problem:
Simply obtain the radius vector from O to A, and then decompose the radius vector and the P vector into vectors in the parallel to the axis. Finally, do the cross product as Tom suggested. For the second problem: Those forces are concurrent, therefore you can easily add them throught their components in order to obtain a resultant. 



#13
Sep2005, 09:26 PM

P: 9

...perhaps it's not to late to withdraw...



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