Static equilibrium boy walking on beam

  • Thread starter trajan22
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quick question,
A uniform, aluminum beam 9.00 m long, weighting 300 N, rests symmetrically on two supports 5.00 m apart. A boy weighing 600 N starts at point A and walks toward the right.
How far beyond point B can the boy walk before the beam tips?
ans=1.25 m
How far from the right end of the beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip?

this second part is the part im stuck on.

my equations are
all of these are the sum of the forces in that direction
Fx=0
no forces are acting in this direction

Fy=(F(support A)+F(support B))-(F(gravity on bar)+F(gravity on boy))
force of gravity of bar is at its center of mass between the two blocks

Tz=(F(gravity on bar)*(distance)-F(gravity on boy)*(distance))
rotation point is set around point b

i was able to solve for the first part of this problem however for the second part it appears that i have 2 unknowns with 1 effective equation so im not really sure how to go.

i have included a figure but chances are it will take some time for it to come up.
 

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Answers and Replies

  • #2
PhanthomJay
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trajan22 said:
quick question,
A uniform, aluminum beam 9.00 m long, weighting 300 N, rests symmetrically on two supports 5.00 m apart. A boy weighing 600 N starts at point A and walks toward the right.
How far beyond point B can the boy walk before the beam tips?
ans=1.25 m
How far from the right end of the beam should support B be placed so that the boy can walk just to the end of the beam without causing it to tip?

this second part is the part im stuck on.

my equations are
all of these are the sum of the forces in that direction
Fx=0
no forces are acting in this direction

Fy=(F(support A)+F(support B))-(F(gravity on bar)+F(gravity on boy))
force of gravity of bar is at its center of mass between the two blocks

Tz=(F(gravity on bar)*(distance)-F(gravity on boy)*(distance))
rotation point is set around point b

i was able to solve for the first part of this problem however for the second part it appears that i have 2 unknowns with 1 effective equation so im not really sure how to go.

i have included a figure but chances are it will take some time for it to come up.
Your 2nd equation is a bit confusing. And I'm not sure how you solved the first part. In both cases, the beam is just about to tip when the reaction force at A is 0, because A can not support any upward force on it (assuming it isn't bolted or otherwise fastened to the beam).

Also in both cases, the sum of torques about any point must be zero, because the beam is in equilibrium as long as it is not tipping.

In part a, summing torques about B is easiest, because you know that the reaction at A is zero. Thus,
[tex] 600x -300(2.5) = 0[/tex]
[tex] x = 1.25 meters[/tex] which agrees with your result.
In part b, you are trying to locate the position of support B such that no tipping occurs when the boy reaches the far end of the beam. Again summing torques about B,
[tex]600x - 300(4.5 -x) = 0[/tex]
[tex]600x - 1350 + 300x = 0[/tex]
[tex]x = 1.5 meters[/tex]
Remember, the reaction at A is 0!
 
Last edited:
  • #3
132
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im confused about the last part of your explanation when you multiplied the 300 by (4.5-x) why by do you take half the length of the beam and subtract that from x...sorry i cant explain why i dont understand this in more detail.
 
  • #4
PhanthomJay
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trajan22 said:
im confused about the last part of your explanation when you multiplied the 300 by (4.5-x) why by do you take half the length of the beam and subtract that from x...sorry i cant explain why i dont understand this in more detail.
The beam is 9m long; the cg of its weight, 300N, is located dead center at 4.5 m from the right end. The support B is located x meters from the right end. Draw a sketch showing the location of these 2 distances from the right end. It should then jump out at you that the 300N force is located (4.5-x) m from support B.
 

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