Physics Help Forum Statics help

 Sep 7th 2013, 12:02 PM #1 Banned   Join Date: Sep 2013 Posts: 4 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.
Sep 7th 2013, 01:52 PM   #2

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,856
 Originally Posted by suvadip 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
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Sep 9th 2013, 08:58 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 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.
 Nov 17th 2013, 09:45 PM #4 Junior Member   Join Date: Nov 2013 Posts: 1 Thanks to provide good information and i am searching from many times. __________________ http://www.nutribullet-recipes.net/r...mond-smoothie/
 Jan 10th 2014, 05:08 AM #5 Junior Member   Join Date: Jan 2014 Posts: 3 Moment This seams quite easy but i am not sure about my result. Can any body help he out? .................................................. .................................................. ............ 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.
 Jan 10th 2014, 08:25 AM #6 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 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.
 Jan 10th 2014, 09:08 AM #7 Junior Member   Join Date: Jan 2014 Posts: 3 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)
 Jan 10th 2014, 10:20 AM #8 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 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.
Jan 10th 2014, 10:57 AM   #9
Junior Member

Join Date: Jan 2014
Posts: 3
 Originally Posted by ChipB 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

Jan 10th 2014, 12:03 PM   #10
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,354
 Originally Posted by Lemming22 you mean: (Mo/a)L - S(b-L) = 0
Yes, right.

 Tags statics

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post Samagra Kinematics and Dynamics 1 Dec 16th 2015 01:20 PM DustyGeneral Equilibrium and Elasticity 2 Sep 24th 2012 06:34 AM Preston019 Kinematics and Dynamics 1 Sep 11th 2012 06:04 AM dobe Equilibrium and Elasticity 3 Feb 6th 2010 07:07 AM Tulip007 Equilibrium and Elasticity 0 Oct 12th 2008 12:34 AM