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  1. #1
    Taylor Gang 140wrestlerK's Avatar
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    physics homework help

    1. A 400 N child and a 300 N child sit on either end of a 2.00 m long seesaw. where along the seesaw should the pivot support be placed to ensure rotational equilibrium?
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    Registered User KGA617's Avatar
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    Fk physics. I hate this class.
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    Registered User Dden's Avatar
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    I dunno the exact nnumbers but your answer should be over 1.0 m and under 2.0 m from the 300n child. Im sure that doesnt help what so ever lol
    Last edited by Dden; 01-09-2011 at 05:45 PM.
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  4. #4
    gooby pls Father of Cajun's Avatar
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    Originally Posted by 140wrestlerK View Post
    1. A 400 N child and a 300 N child sit on either end of a 2.00 m long seesaw. where along the seesaw should the pivot support be placed to ensure rotational equilibrium?
    sum torque clockwise= sum torque ccw
    torque=fd

    so 400d1=300d2
    d1+d2=2
    d1=2-d2
    400(2-d2)=300d2
    800-400d2=300d2
    800=700d2
    d2=1.14m from the 300N child.

    Anyone wanna check my work for me?
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  5. #5
    Taylor Gang 140wrestlerK's Avatar
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    thankyou so much, repped. that actually makes a lot of sense. i hate how physics makes as much sense as reading a language you dont know until you have it broken down and explained.

    if anyone else is willing to explain another problem id really appreciate it.
    a bridge 20m long weighing 4x10^5N is supported by two pillars located 3m from each end. if a 1.98x10^4N car is parked 8m from one end of the bridge how much force does each pillar exert?
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  6. #6
    gooby pls Father of Cajun's Avatar
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    Originally Posted by 140wrestlerK View Post
    thankyou so much, repped. that actually makes a lot of sense. i hate how physics makes as much sense as reading a language you dont know until you have it broken down and explained.

    if anyone else is willing to explain another problem id really appreciate it.
    a bridge 20m long weighing 4x10^5N is supported by two pillars located 3m from each end. if a 1.98x10^4N car is parked 8m from one end of the bridge how much force does each pillar exert?
    ...............8m
    ..............o--o
    ---------------------------------
    /.../...../....../...../....../......../ |
    -------------------------------
    3m..|....20m........................|3m
    .....F1................................F2

    force down of 400,000 N
    Car weighing 19,800 N(force down) is 8m from one end of the bridge.


    sum force down = sum force up
    400,000N+19,800N= F1+F2(the two pillars)

    419800N=F1+F2


    sum torque cw=sum torque ccw

    torque=fd

    ccw forces: bridge,car

    (400,000N)(7m)+(19,800N)(5m)=F1(0)+F2(14m)

    14F2=2,800,000N/M+ 99000N/M
    F2=207,071.43N
    Plug into the equation from the force up/force down part,

    419,800N=F1+207071.43
    F1=212,728.57N
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  7. #7
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    1.14m from the 300N child.

    done.

    EDIT: fatherofcajun beat me to it
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    Originally Posted by Father of Cajun View Post
    sum torque clockwise= sum torque ccw
    torque=fd

    so 400d1=300d2
    d1+d2=2
    d1=2-d2
    400(2-d2)=300d2
    800-400d2=300d2
    800=700d2
    d2=1.14m from the 300N child.

    Anyone wanna check my work for me?
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  9. #9
    Registered User PenalCramp's Avatar
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    Sorry wrong thread
    Last edited by PenalCramp; 01-09-2011 at 11:44 PM.
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