Finding the Minimum Distance for Static Equilibrium with a Hanging Weight

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Homework Help Overview

The problem involves a uniform rod supported by a cable and resting against a wall, with a focus on determining the minimum distance from a pivot point at which an additional weight can be hung without causing the rod to slip. The subject area pertains to static equilibrium and torque analysis.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of torques and forces acting on the rod, questioning the placement of the pivot and the conditions for equilibrium. There are attempts to express the net torque and forces in terms of the weights and friction involved.

Discussion Status

Participants are actively engaging with the problem, exploring different aspects of equilibrium and torque calculations. Some have provided guidance on setting up equations for vertical and horizontal forces, while others are questioning the assumptions made about the pivot point and the relationships between forces.

Contextual Notes

There is an ongoing discussion about the conditions for equilibrium and the implications of placing the pivot at different points. Participants are also considering the role of static friction and the coefficients involved in the problem.

unreal89
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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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unreal89 said:
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.)
 
okay so 0 = 2Fg + 4Ff - xFg

N = Tension in X direction

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

I don't get how to calculate tension =/
 
unreal89 said:
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?
 
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?
 
unreal89 said:
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.
 
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?
 
unreal89 said:
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.
 
unreal89 said:
Isnt the pivot at the right end, ...
Since this is a static problem, you can put the pivot wherever 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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