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Find force

  1. Sep 20, 2007 #1
    A bar 10 ft long carries a weight of 20 lbs, 6 ft from the end. What force must be applied at each end to support the rod.

    P1 6 | 4 P2
    |____________________________________|
    B | A C
    20lb
     
    Last edited: Sep 20, 2007
  2. jcsd
  3. Sep 20, 2007 #2

    cristo

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    Staff Emeritus
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    Welcome to the forums. Please note that, for homework questions, you are required to show some work before we can help you. Also, in the future, please post in the relevant homework/coursework forum.

    So, how do you think you'll approach this question?
     
  4. Sep 20, 2007 #3

    Astronuc

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    Staff: Mentor

    Please post homework problems in the appropriate Homework forum - in this case Introductory Physics or perhaps Engineering.

    Also, we ask that students show some effort in solving the equation before asking for help.

    This appears to be a static beam problem where one must find the reaction forces using sum of forces = 0, and sum of moments = 0.
     
  5. Sep 20, 2007 #4
    AY 10(6)+10(4)=100 100/10= 10
     
  6. Sep 20, 2007 #5
    20(6)=120 120/10=12 B=12
    20(4)=80 80/10=8 C=8
     
  7. Sep 20, 2007 #6

    Astronuc

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    Staff: Mentor

    The greater force has to be applied at the shorter moment arm.

    One knows that P1 + P2 = 20 lbs, since the net forces must be zero.

    Then pick on end and determine the moments about that end. If one picks P1, then the 20 lbs is at 6 ft, and the moment is therefore 120 lb-ft.

    The force P2 is at the other end of the beam, at 10 ft, so its moment is _________, which must equal the 120 lb-ft moment in order to maintain static equilibrium.
     
  8. Sep 20, 2007 #7
    moment at 10' is 40 lbs.
     
  9. Sep 20, 2007 #8
    moment at 10' is 200lbs
     
  10. Sep 20, 2007 #9

    Astronuc

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    Well the moment of P2 at 10 ft is P2*10 and one has to solve for P2, by

    0 = (20 lbs)(6 ft) - P2 (10 ft), which gives P2 = _________


    The use P1 + P2 = 20 lbs.


    The moments have to cancel to obtain static equilibrium.
     
  11. Sep 21, 2007 #10
    [QUOTE=Mike McCue;the moment at 10ft would be 200lbs.
     
  12. Sep 21, 2007 #11
     
  13. Sep 21, 2007 #12
     
  14. Sep 21, 2007 #13
    (20lb(6ft)-P2 (10ft)=0
    120=10P2
    120/10=P2
    P2=12
     
  15. Sep 21, 2007 #14
    120-P2(10ft)=0
    120=P2(10)
    120/10=P2
    12=P2
     
  16. Sep 21, 2007 #15

    Astronuc

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    0 = (20 lbs)(6 ft) - P2 (10 ft), which gives P2 = 120 lb-ft / 10 ft or P2 = 12 lb, which one did correctly in the two preceding posts.

    Then P1 + P2 = 20 lb = P1 + 12 lb, so P1 = 8 lb.
     
  17. Sep 21, 2007 #16
    P1=8lbs
    P2=12lbs

    Thanks for your help Astronuc.
     
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