Question about static equilibrium

  • #1
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1. Homework Statement


problem is above

Homework Equations


sum of torques = I * angular acceleration = 0 for rotational equilibrium

The Attempt at a Solution



OK, here's what I tried to do.

I wrote sigma Fy = Fleft pier up + F secondleftpierup - Fg,beam = 0
I assumed that the two piers have equal forces, so then I said that
2 * Fsinglepier = Fg,beam
2 * Fsinglepier = 100 kg * 9.8 m/sec^2.
Fsinglepier = 50 * 9.8.

Then I did my rotational equilibrium equation, and I assumed that there was no man in the pier initially. My axis of rotation is at the very left end of the pier:

sum torques = +Fsecondpier [which is 50 * 9.8] * 3 meters - [ 100 kg * 9.8 * 5 meters] (for the center of mass of sphere and the force of gravity acting through it

which gives = -3430 N-m.

This obviously implies that the system is already experiencing a net torque without the man, so the man can not even be on the system to balance it out in any fashion.

The solutions [which might be wrong, due to crummy question authors] say that 3 is the answer without providing any other guidance. Please help, and thanks in advance!
 

Answers and Replies

  • #2
haruspex
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To get the given answer, the second support has to be 4m from the end with the first support.
 
  • #3
billy_joule
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OK, here's what I tried to do.

I wrote sigma Fy = Fleft pier up + F secondleftpierup - Fg,beam = 0
I assumed that the two piers have equal forces, so then I said that
There is no basis for that assumption. It's incorrect.

The solutions [which might be wrong, due to crummy question authors] say that 3 is the answer without providing any other guidance. Please help, and thanks in advance!
Yes, that solution is wrong. As it is, there is no way the man can balance the beam.
 
  • #4
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So, it is not my fault, but in here, the solutions are incorrect?

To get the given answer, the second support has to be 4m from the end with the first support.
How do you know this for the four meters?
 
  • #5
haruspex
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How do you know this for the four meters?
I worked backwards from the given answer. Try taking the support as being 1m from the centre of the beam (towards the person) instead of 3m.
 
  • #6
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Try taking the support as being 1m from the centre of the beam (towards the person) instead of 3m.
Could you show me graphically what you mean? I am having a hard time following you with this. Thanks!
 
  • #7
haruspex
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Could you show me graphically what you mean? I am having a hard time following you with this. Thanks!
Just take the question as you originally posted it, but change the 3m to 1m.
 
  • #8
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There is no basis for that assumption. It's incorrect.
Why can I not assume like this? If so, how can I assume correctly?
 
  • #9
haruspex
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Why can I not assume like this? If so, how can I assume correctly?
As Billy posted, you have no basis for assuming the forces from the piers are equal. Nobody needs to give you a reason for saying that; the onus is on you to provide a reason for such an assumption.
Indeed, when the system is about to tip the force from one of them must be going to zero.
 
  • #10
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If I did not make such an assumption, then I would have too many variables to deal with as of now.

I am trying to find "x" the distance of the man, but if I have an unknown Pier1Force and Pier2Force, then that changes everything.

I actually started resolving it with the new distance that you gave me, but your post about not being able to make such an assumption changed that. I am stuck now. Thanks for the help!
 
  • #11
haruspex
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If I did not make such an assumption, then I would have too many variables to deal with as of now.
I did give you a different (and justifiable) assumption to make.
 
  • #12
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Indeed, when the system is about to tip the force from one of them must be going to zero.
So I can assume that the force from the left most pier becomes zero while the second pier's force is nonzero?
 
  • #13
haruspex
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So I can assume that the force from the left most pier becomes zero while the second pier's force is nonzero?
Yes.
 
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  • #14
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OK here's what I have

sumtorques = P2 * 4 - (100 * 9.8 * 5) + (60 * 9.8 *x) = 0

P2 is pier two's force.
the other two terms are the gravitational forces of the pier and the guy respectively. x is the guy's distance from the left point of the structure.

pivor point is left point of the structure.

My question is how can I substitute something for pier2 in now to solve for x.

Thanks!
 
  • #15
haruspex
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how can I substitute something for pier2 in now to solve for x.
Two ways to proceed:
- You have only used balance of moments. What other equations are available?
- Choose a reference point for taking moments that makes the force from the pier vanish from the equation.
 
  • #16
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Summing my equations in the y direction, I have Fpier2 = (force of gravity on pier and on the guy combined) = 9.8 * (100 + 60) = 1568 N.

sumtorques = P2 * 4 - (100 * 9.8 * 5) + (60 * 9.8 *x) = 0

Plugging the calculated value for P2 in and solving for x, I get x = -2.33 meters which does not physically make sense.
 
  • #17
haruspex
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sumtorques = P2 * 4 - (100 * 9.8 * 5) + (60 * 9.8 *x) = 0
How do you get that? In the first two terms looks like you are taking moments about the left hand end, with anticlockwise positive. The third ...?
 
  • #18
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Ahh, yess, my bad. Thanks for catching that sign mistake.

I meant:
sumtorques = P2 * 4 - (100 * 9.8 * 5) - (60 * 9.8 *x) = 0
 
  • #19
haruspex
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Ahh, yess, my bad. Thanks for catching that sign mistake.

I meant:
sumtorques = P2 * 4 - (100 * 9.8 * 5) - (60 * 9.8 *x) = 0
Right. What does that give you for x?
 

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