Static Equilibrium please help

  • Thread starter unreal89
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  • #1
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Homework Statement



One end of a uniform 4.00-m-long rod of weight Fg is supported by a cable. The other end rests against the wall, where it is held by friction, as in Figure below. The coefficient of static friction between the wall and the rod is s = 0.500. Determine the minimum distance x from point A at which an additional weight Fg (the same as the weight of the rod) can be hung without causing the rod to slip at point A.

attachment.php?attachmentid=16777&d=1228833076.jpg





Homework Equations



TORQUEnet = (F)1(d)1 + (F)2(d)2 + ...

The Attempt at a Solution



Assuming fulcrum is at the right tip of the bar
TORQUEnet = 2Fg + 4Ff - x Fg
 

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Answers and Replies

  • #2
Doc Al
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Assuming fulcrum is at the right tip of the bar
TORQUEnet = 2Fg + 4Ff - x Fg
Redo this: How far is the hanging weight from the pivot? In what direction do the torques act?

What other conditions for equilibrium must be satisfied? (And will give you additional equations.)
 
  • #3
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okay so 0 = 2Fg + 4Ff - xFg

N = Tension in X direction

Ff = N (Mu) = (tension in X)(Mu)

I dont get how to calculate tension =/
 
  • #4
Doc Al
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okay so 0 = 2Fg + 4Ff - xFg
As I said, you'll need to redo this.

What forces act on the beam? At the left end? At the right end?

What's the net force in any direction?
 
  • #5
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there's weight of beam, weight of sphere, tension, normal, friction.
Friction is at the left, tension at the right.
Isnt the pivot at the right end, so TORQUEnet = 0(Tension) - 2Fg + 4Ff - x Fg and TORQUEnet = 0 so that it is in equillibrium?
 
  • #6
Doc Al
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there's weight of beam, weight of sphere, tension, normal, friction.
Friction is at the left, tension at the right.
Good. Set up equations for the vertical and horizontal forces to equal zero.
Isnt the pivot at the right end, so TORQUEnet = 0(Tension) - 2Fg + 4Ff - x Fg and TORQUEnet = 0 so that it is in equillibrium?
x is measured from the left end.
 
  • #7
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so,

0 = -2Fg + 4Ff - (4-x)(Fg)

0 = Tx - N
0 = Ty + Ff - 2Fg

Are these correct? I have no idea where I'm going with this... should it end with a finite number or will it have variables in it?
 
  • #8
Doc Al
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so,

0 = -2Fg + 4Ff - (4-x)(Fg)
Good.
0 = Tx - N
0 = Ty + Ff - 2Fg
Good. Instead of using Tx & Ty, use some trig and the given angle to express it just in terms of T.

Hint: How does Ff relate to N?

By solving all these equations together, you should get a numerical answer for x.
 
  • #9
turin
Homework Helper
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Isnt the pivot at the right end, ...
Since this is a static problem, you can put the pivot whereever you want. I chose to put mine at the point where the rod touches the wall. Any idea why I chose it here? (There are actually two reasons.)
 

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