Static equilibrium boy walking on beam

At that location, the torque about B is 300N times that distance.To check your understanding, if you multiply 300N times 0, what do you get? Yes, zero. If you multiply 300N times 9m, what do you get? 2700Nm, which is the torque that would be acting around B if the support were at the right end of the beam.
  • #1
trajan22
134
1
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 I am 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 I am 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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  • #2
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 I am 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 I am 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
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 can't explain why i don't understand this in more detail.
 
  • #4
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 can't explain why i don't 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.
 

Related to Static equilibrium boy walking on beam

1. What is static equilibrium?

Static equilibrium is a state in which an object is at rest and all the forces acting on it are balanced, resulting in no net force or movement.

2. How does a boy walking on a beam achieve static equilibrium?

A boy walking on a beam can achieve static equilibrium by evenly distributing his weight on both sides of the beam, keeping his center of mass directly above the beam, and adjusting his body and movements to maintain balance.

3. Why is it important to maintain static equilibrium when walking on a beam?

Maintaining static equilibrium when walking on a beam is important to avoid falling off the beam and potentially getting injured. It also ensures that the beam itself remains in a state of equilibrium and does not break or bend under uneven weight distribution.

4. What factors can affect the static equilibrium of a boy walking on a beam?

The factors that can affect the static equilibrium of a boy walking on a beam include his weight, the placement of his feet and body, the width and stability of the beam, and any external forces such as wind or vibrations.

5. How can static equilibrium be disrupted while walking on a beam?

Static equilibrium can be disrupted while walking on a beam if the boy's weight is not evenly distributed, if he makes sudden or jerky movements, or if external forces such as a gust of wind or a loose beam cause an imbalance.

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