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Summersnow8

  • one year ago

Please help! A bridge with mass 9000. kg supports a truck with mass 3000. kg that is stopped in the middle of the bridge. What mass (in kg) must each pier of the bridge support? = (9000 kg + 3000 kg) / 2 = 6000 kg If the truck in the preceding problems stops 9.00 m from the northern end of the 32.0-m bridge, what mass (in kg) must the northern pier of the bridge support? F1 + F2 = (9000 kg * 9.8 m/s^2) + (3000 kg * 9.8 m/s^2) F1 + F2 = 88200 + 29400 F1 + F2 = 117,600 N F1 (32.0 m) = (88200 N) (????? m) + (29400 N) (9.00 m) @nikato @michele_Laino @lightgrav @radar @Greg_D @peachpi the answers are suppose to be 5220 kg, 6580 kg

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  1. anonymous
    • one year ago
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    I tried this last night. Got the same numbers you did. 5220 kg doesn't even make sense for the first part because it's not enough support. The part you're missing for the 2nd part is 16 m. Weight acts at the center of the bridge, then you can solve for F1.

  2. Summersnow8
    • one year ago
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    @peachpi, yeah the answer for the first problem is 6000 kg, but there are 2 asnwers for the second problem... I need help with the second one only

  3. anonymous
    • one year ago
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    Is there another question for the 2nd problem?

  4. anonymous
    • one year ago
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    They only ask for mass as far as I can tell

  5. Summersnow8
    • one year ago
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    they are asking for the mass of the northern pier of the bridge.... so I don't know which answer it is or how you get the answers

  6. anonymous
    • one year ago
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    |dw:1437603862116:dw|

  7. Summersnow8
    • one year ago
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    how do I figure out which one is north and which one is south?

  8. anonymous
    • one year ago
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    North is the one closest to the truck. That's why I switched to N and S as subscripts to make it a little easier If you take the moment about the north side it's \[\sum M_N=-(9~m)(29400~N)-(16~m)(88200~N)-(32~m)F_S=0\] Solving that gives the force on the south support

  9. anonymous
    • one year ago
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    That should be + 32 Fs

  10. anonymous
    • one year ago
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    For the north you can do the same, take the moment about the south end \[\sum M_S=(23~m)(29400~N)+(16~m)(88200~N)-(32~m)F_N=0\]

  11. Summersnow8
    • one year ago
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    I got this: F1 (32.0 m) = (88200 N) (16 m) + (29400 N) (9.00 m) F1 (32.0 m) = 1675800 F1 = 1675800 / 32 m F1 = 52368

  12. anonymous
    • one year ago
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    yeah that's what I got, then when you divide by 9.8 you'll get 5343.75 kg

  13. anonymous
    • one year ago
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    that's for the south pier

  14. anonymous
    • one year ago
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    for the north pier you can use the equation above or just subtract from 12000

  15. Summersnow8
    • one year ago
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    but the answers are 5220 kg & 6580 kg , not what we got

  16. anonymous
    • one year ago
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    that's what I was saying above, those answers don't make sense. They don't even add up to 12,000 kg, which is the amount of mass that has to be supported

  17. Summersnow8
    • one year ago
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    well, those are the answers in the back of the book.... are you sure our equation is right? why do we divide by gravity? I am not understanding it

  18. anonymous
    • one year ago
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    I divided by gravity because they asked for mass, not force. The equation is right. The answer makes no sense. The mass of the bridge and truck is 9000 + 3000 = 12000 kg. The answers for the supports are 5220 + 6580 = 11800 kg. This would mean the mass on the bridge is 200 kg more than it can support. Those answers cannot be right, unless something is missing from the problem

  19. Summersnow8
    • one year ago
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    I put the wrong answer, in the back of the book the problem is asking he mass must each pier of the bridge support, which the correct answer is 6000 kg

  20. anonymous
    • one year ago
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    I'm confused. that's for the 1st question, right?

  21. Summersnow8
    • one year ago
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    sorry, i am confusing myself, yea that is for the 1st, but yeah the book says 5220 kg, 6580 kg for the second question...... I don't get it.....

  22. Summersnow8
    • one year ago
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    so did you get 5543.75 (south), and how do you solve for north?

  23. anonymous
    • one year ago
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    You can set up a moment equation from the south side, like I did above. Or you can just subtract the south from 12000

  24. anonymous
    • one year ago
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    since we know they have to add up to 12000

  25. Summersnow8
    • one year ago
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    so 6456.25 (north)

  26. anonymous
    • one year ago
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    yes

  27. Summersnow8
    • one year ago
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    okay, so the answer is 6460 ? I hope we solved this correctly :/

  28. anonymous
    • one year ago
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    yeah ok if you round

  29. Summersnow8
    • one year ago
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    hmmm, okay. thanks for your help

  30. mtimko
    • one year ago
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    The correct answer (at least for my purposes) ended up being 6655 +/- 10

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