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anonymous

  • 4 years ago

A bridge of weight 50,000N is supported on piers A and B 20m apart. The weight of the bridge acts at its midpoint. A lorry of weight 30,000N stands with its center of gravity 4m from pier A. Find the forces acting on each pier.

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  1. JamesJ
    • 4 years ago
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    (btw, emcrazy, did the answer to your earlier question on light underwater make sense?)

  2. anonymous
    • 4 years ago
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    Yeah. Thank you. :)

  3. JamesJ
    • 4 years ago
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    still here? If so, we can talk this problem through

  4. anonymous
    • 4 years ago
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    Yea.

  5. JamesJ
    • 4 years ago
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    So there's a torque on the piers from the weight of the bridge and the truck. But is there any force?

  6. JamesJ
    • 4 years ago
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    Any linear force, if you like; any force that isn't rotational?

  7. anonymous
    • 4 years ago
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    There will be an upward force on the bridge.

  8. JamesJ
    • 4 years ago
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    What do you mean. There are downward forces from the weight of the bridge and truck acting on the bridge. But what forces are acting on the piers, as per the question?

  9. anonymous
    • 4 years ago
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    See, the piers exert upward forces on the bridge but I don't know how to calculate them .

  10. JamesJ
    • 4 years ago
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    Well, suppose there was no truck on the bridge. The weight of the bridge must be reacted to by the piers, and it must be done symmetrically. Hence from the weight alone, each pier experiences a force of 25,000 N.

  11. JamesJ
    • 4 years ago
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    And the reason they are equal for that comes from looking at the torque on the center of the bridge, call that C. Call the force on pier A, \( F_A \) and on pier B, \( F_B \). Then the torque on C from A is \[ \tau_{CA} = r_{CA} F_A = -10 F_A \] negative because the displacement from C to A is -10 m. Similarly, the displacement from C to B is +10 m, thus the torque on C from B is \[ \tau_{CB} = r_{CB} F_B = 10 F_B \] We know that \( \tau_{CA} + \tau_{CB} = 0 \), hence \[ 10(F_B - F_A) = 0 \] i.e., \[ F_A = F_B \] ==== Now, use the same logic to find the torque from the truck on piers A and B, centered around the truck.

  12. JamesJ
    • 4 years ago
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    Make sense?

  13. JamesJ
    • 4 years ago
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    This diagram might help.

  14. anonymous
    • 4 years ago
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    Ok, so taking A as pivot; 30,000x4=20xF F=6000N (This is the force at B) Taking B as pivot; 30,000x16=20xF F=24000N (This is the force at A) So, adding up the forces due to weight of bridge and weight of truck, the force on pier A becomes 25000+24000=49000N. And force on pier B is 25000+6000=31000N. Is it right?

  15. JamesJ
    • 4 years ago
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    Yes. That's another approach to the problem: the bridge must not pivot around either pier hence the torque about them is zero. Finishing the approach I started with above, we've showing the the force on the two piers from the weight of the bridge is 25,000 N each. From the truck itself, we know the force on each pier must again equal the weight, Fa + Fb = 30,000 --- (*) If we calculate the torque about the truck, the torque from Fa is 4.Fa. The torque from Fb is -16.Fb. Hence 4.Fa -16.Fb = 0 i.e. Fa = 4.Fb Substituting into (*), we have 5.Fb = 30,000 thus Fb = 6,000 and Fb = 24,000 Adding up now the force we've calculated from the weight of the bridge and from the weight of the truck we have Total Fa = 25,000 + 24,000 = 49,000 N Total Fb = 25,000 + 6,000 = 31,000 N This the same result you found.

  16. anonymous
    • 4 years ago
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    Okay. Thanks a lot. :)

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