Statics Problem - Moment on a Bridge

AI Thread Summary
The discussion revolves around a statics problem involving a uniform work platform and a construction worker walking on it. The goal is to determine the location S where the combined moment of the weights of the man and platform about point B equals zero. Participants suggest treating the platform as a seesaw, focusing on the forces acting at point B while disregarding reactions at point A. There is confusion regarding the signs used in calculations and whether to consider forces at point A, with some participants emphasizing the need to clarify the moments around point B. The conversation highlights the complexity of balancing forces and moments in static systems.
TimChoi89
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Homework Statement



The uniform work platform, which has a mass per unit length of 28kg/m, is simply supported by cross rods A and B the 90-kg construction worker starts from point B and walks to the right. At what location S will the combined moment of the weights of the man and platform about point B be zero?

Platform illustration:

__A____________B___man (going to the right)___

Mass per unit length of bridge: 28kg/m
Mass of man: 90kg

Distance from far left end of bridge to A: 1m
Distance between A and B: 4m
Distance from B to far right end of bridge: 3m
Distance from B to man: S

Homework Equations



M=d*F or M=r*F (cross product)

The Attempt at a Solution



What I did was that I set point B to be (0,0) so any distances to the left of B is (-) and to the right will be (+). Also, I set all the weights to be (-) since it's going down the (-) y-axis.

Next I assigned the F from B to Man to be (90+28S) and the F from B to the remaining distance from Man to far right end of the bridge to be 28(3-S).

For the forces on the left side:

left end to A: 28
between A and B: 112

Next, I used M=d*F and set the moment of B from the left and right side to equal each other which leaves me with:

588=-90S-28S^2+(3-S)(-8.4+28S)
840=78S-56S^2
0=78S-56S^2-840
0=9.8(78S-56S^2-840) <-- to calculate weight
0=548.8S^2-764.4S+8232

When I try to solve for S, I use the quadratic formula but I can't solve for S because the number in the sqrt is (-) thus I can't solve for S. Am doing something wrong with the signs or is my approach completely wrong? Thanks for your help!
 
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TimChoi89 said:
At what location S will the combined moment of the weights of the man and platform about point B be zero?
Do you know the solution? Maybe you should look at the platform as a seesaw without reaction in A. In that situation you would only have reaction in B plus the weight of the platform and the man. It would be easy to calculate the moment around B.
 
method_man said:
Do you know the solution? Maybe you should look at the platform as a seesaw without reaction in A. In that situation you would only have reaction in B plus the weight of the platform and the man. It would be easy to calculate the moment around B.


I don't have the solution sorry. What do you mean by reaction in B? If I disregard A does that mean that I'll just have a combined distance of 5 from the left side to point B and just calculate the total weight from that side? Thanks again.
 
I've attached a picture to explain what I mean. On my picture, force A i faced downward. That is a situation when this guy walks far enough to the right (over distance S as I have assumed). If this guy stands close enough to the point B, force A is facing upward. So I guessed that there has to be a situation when force A becomes zero before shifting it's orientation. My assumption is that that will happen when distance is S. I hope that somebody will answer with better idea.
 

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Oh ok, at this point I thought that I can just ignore the forces on A b/c the question was asking about B. I'm still a little confused...sorry. It's a little hard to explain this via text only...lol
 
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