Problems in Statics Involving Moments related to rigid bodies


by jnbfive
Tags: bodies, involving, moments, rigid, statics
jnbfive
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#1
Sep21-08, 05:14 PM
P: 47
I have a couple problems that I can't seem to get.

The first I'm really close, I can tell, because my professor did one that was similar in class. The answer the book gives is 250 lbs, I'm coming up with 225.

The second I think is similar to another problem he did in class but I'm not quite sure. I'll post it in the next reply.
Attached Thumbnails
Problem 3.10.1.jpg   Problem 3.10 work.1.jpg   Problem 3.71.1.jpg  
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#2
Sep21-08, 05:16 PM
P: 47
This is the problem that I think is similar to the one labeled 3.71. I just want to know if I'm correct in my assumption the two problems have a similar premise.
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Problem 3.71_similar.1.jpg  
PhanthomJay
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#3
Sep21-08, 08:15 PM
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Quote Quote by jnbfive View Post
This is the problem that I think is similar to the one labeled 3.71. I just want to know if I'm correct in my assumption the two problems have a similar premise.
The premise seems similar (Moment = force times perpendicular distance). For the force on the nails in part 1, one of your arrows points in the wrong direction.

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#4
Sep22-08, 02:09 PM
P: 47

Problems in Statics Involving Moments related to rigid bodies


I need help on part c of 3.71. I have the angle from b, which I believe I need to use. I originally thought that the way to set it up would be

86.2 = x(22.36)*sin(53.1)

^22.36 coming from the sqrt of 17.6^2+ 13.8^2

Any help on what I'm doing wrong would be appreciated.
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#5
Sep22-08, 04:18 PM
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Quote Quote by jnbfive View Post
I need help on part c of 3.71. I have the angle from b, which I believe I need to use.
how did you determine the angle that requires the min tension?
I originally thought that the way to set it up would be

86.2 = x(22.36)*sin(53.1)
check angle and position vector

^22.36 coming from the sqrt of 17.6^2+ 13.8^2
looks like it should be 15.2^2 + 13.8^2

Any help on what I'm doing wrong would be appreciated.
Note that if you are trying to minimize the tension, you want to maximize the perpendicular distance from the line of action of the force. Forget the r*T*sin theta approach.
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#6
Sep22-08, 06:23 PM
P: 47
*I took the inverse tangent of 11.4/15.2

*The angle of 53.1 is correct; thats the answer the book got.

*Why only 15.2?

*Thanks
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#7
Sep22-08, 06:36 PM
P: 47
If I could actually ask for help in another question, any help would be appreciated.

Excuse my chicken scratch work. I solved for the angle at which the force is acting at, which is 30 degrees in the xz plane. I solve for the actually distance from the x axis that said force was, which was about 387 mm or 100*sqrt(15). Given this I solved for the distance from the origin, 100*sqrt(19). I found my force, the set up a cross product. My answers are listed as such. I have the i vector correct, j is off by 8 (should be 22) and k is off by 1 (should be 39.1). I feel as if I'm really close to this one, was just wondering what I'm doing wrong here.
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Problem 3.94.1.jpg   Work for 3.94.jpg  
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#8
Sep22-08, 08:19 PM
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Quote Quote by jnbfive View Post
*I took the inverse tangent of 11.4/15.2

*The angle of 53.1 is correct; thats the answer the book got.

*Why only 15.2?

*Thanks
I was incorrectly thinking that the angle was 45 degrees, but it appears that the 53.1 is the correct angle that the tension force makes with the horizontal. But that is not the angle in between the force and the position vector for the minimum tension. . That angle is 90 degrees. Also, although I could easily be wrong, the distance to use is root((15.2)^2 + (11.4)^2)) + 2.4 = 21.4. So I get T_min = 86.2/21.4 = 4N.
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#9
Sep22-08, 08:26 PM
P: 47
I'm amazed at my ability to overlook the simple things in problems. Many thanks for your help with that.
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#10
Sep22-08, 11:19 PM
P: 47
In response to my above comment, I really realize that I overlook things too much. 3.94 I have the j and k of the force couple system (use the force that I obtained when breaking down the 220 N in the cross product. Duh.) Somehow I cant seem to get i. Any help would be appreciated.


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