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 Equilibrium and Elasticity Equilibrium and Elasticity Physics Help Forum

 Nov 7th 2016, 12:16 PM #1 Junior Member   Join Date: Nov 2016 Posts: 9 Moments Help! Hello everyone, I can do part a easily, but I have no idea how to do part b. Can I have a step-by-step of how to do it - I know that you have to take moments about either A or B, and I know that the horizontal force doesn't have to be considered in the problem either (due to it being perp. to the weight). Can anyone help? Attached Thumbnails
 Nov 7th 2016, 01:06 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 Your free body diagram should include a horizontal force on the left end of the pole at A, and both horizontal and vertical forces at B. From sum of vertical forces = 0 you have Bv = W, and from sum of horizontal forces =0 you have Ah = Bh. Then take the moment about point B and set it to zero. You now have three equations in 4 unknowns; the 4th equation you need is from the fact that at the contact point B the reaction force acts perpendicularly to the pole - in other words Bh/Bv = tan (theta). Can you take it from here?
 Nov 7th 2016, 01:13 PM #3 Junior Member   Join Date: Nov 2016 Posts: 9 Thanks for the help - the bit that I'm having the most trouble with is the moments part itself. Would I put the length of the rod inside the skip as x, or is there a way of working it out?
 Nov 7th 2016, 01:34 PM #4 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 Regarding the moments about point B, the horizontal force at A has a moment arm of 2m x sin(theta), do you see why that is? And the weight of the pole acts vertically and has a horizontal moment arm from B to the center of mass of 2.5m x cos(theta) - 2m.
 Nov 7th 2016, 02:11 PM #5 Junior Member   Join Date: Nov 2016 Posts: 9 So does 2 sin (theta) = 0.5 cos (theta) ? Last edited by electrogeek; Nov 7th 2016 at 02:22 PM.
Nov 7th 2016, 04:39 PM   #6
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,352
 Originally Posted by electrogeek So does 2 sin (theta) = 0.5 cos (theta) ?
No.

$\displaystyle B_h \ \times \ 2 \sin \theta = W(2.5 \cos \theta -2)$

Maybe the attached will help you visualize this. Then use the fact that $\displaystyle B_h/B_v = \tan \theta$, so the $\displaystyle B_h$ on the left hand side can be replaced by $\displaystyle B_v \tan \theta = W \tan \theta$. Then it's a matter of some manipulation to solve for cos theta.
Attached Thumbnails

Last edited by ChipB; Nov 8th 2016 at 08:51 AM.

 Nov 8th 2016, 07:58 AM #7 Junior Member   Join Date: Nov 2016 Posts: 9 I understand - thank you so much for the diagram! Everything has jut clicked into place! Thanks for all the help. Last edited by electrogeek; Nov 8th 2016 at 08:20 AM.

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