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Old Sep 7th 2013, 12:02 PM   #1
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Statics help

A heavy rod AB is suspended from a point O by two strings OA and OB. Show that the plane OAB is vertical.

How to show it? Please help.
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Old Sep 7th 2013, 01:52 PM   #2
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Originally Posted by suvadip View Post
A heavy rod AB is suspended from a point O by two strings OA and OB. Show that the plane OAB is vertical.

How to show it? Please help.
Say that we orient the plane OAB such that we are looking into the page. ("Vertical" becomes "in the plane of the page.") There's a symmetry here: the plane OAB looks basically the same if we look at it from above the page or below it. So OAB has to be in the plane of the page since there is as much reason for it to be pointed out of the page or into it.

It's the correct argument but I'm not positive that I explained it well enough. Please let us know.

-Dan
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Old Sep 9th 2013, 08:58 AM   #3
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One way to show it is to consider what happens of the strings weren't vertical - the result of the force of gravity would be a force component perpendicular to the strings, causing the rod to swing toward the vertical. This is exactly what happens with a pendulum. This "restoring force" of gravity cause the pendulum to oscillate about the vertical. The rod would work the same way - it will oscillate about the vertical, and if made motionless would be stable only in the vertical plane.
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Old Nov 17th 2013, 09:45 PM   #4
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Thanks to provide good information and i am searching from many times.
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Old Jan 10th 2014, 05:08 AM   #5
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Moment

This seams quite easy but i am not sure about my result.
Can any body help he out?

Statics help-unbenannt.png

.................................................. .................................................. ............

The next question i can't solve at all. I know how graphical solution works but in this case it does not work out for me.

Statics help-unbenannt2.png
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Old Jan 10th 2014, 08:25 AM   #6
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Lemming:

It is best to not tag a new question onto an old thread, but rather submit a new thread instead. But in any event:

Regarding the first problem - you say you're not sure of your result: please tell us what you think the answr is, and what doubts you have about it. We'll help if you get stuck.

As for the second problem - think about moments about support A, which must equal zero. The only outside forces that affect moments there are M_3 and the reaction force at support B. From this you should be able to determine force B.
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Old Jan 10th 2014, 09:08 AM   #7
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ok thx for the reply

for Q1 i get close to answer E but in the denominator i have not L but (L-a)

what i did is:

RA*a=Mo ...... RA....Reaction at Section A-A

RA=Mo/a

(Mo/a)*(l-a)-S(b-l)=0

Mo=S(b-l)a / (l-a)
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Old Jan 10th 2014, 10:20 AM   #8
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It seems you are setting the moment about point L to zero, which is a fine approach. The problem is that RA = M0/a is not the reaction at a, but rather is the reaction at the pivot support. So sum of moments about the point L is:

(Mo/a)L - S(L-b) = 0

Hope this helps.
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Old Jan 10th 2014, 10:57 AM   #9
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Originally Posted by ChipB View Post
It seems you are setting the moment about point L to zero, which is a fine approach. The problem is that RA = M0/a is not the reaction at a, but rather is the reaction at the pivot support. So sum of moments about the point L is:

(Mo/a)L - S(L-b) = 0

Hope this helps.
you mean:

(Mo/a)L - S(b-L) = 0
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Old Jan 10th 2014, 12:03 PM   #10
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Originally Posted by Lemming22 View Post
you mean:

(Mo/a)L - S(b-L) = 0
Yes, right.
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