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Math Help - forces problem

  1. #1
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    forces problem

    Please help me out here. thanks
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  2. #2
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    Quote Originally Posted by Mr_Green View Post
    Please help me out here. thanks
    Hi, Mr_Green,

    in the attachment you'll find a your slightly modified diagram.

    You are dealing with a parallelogram of forces. Take the left half of the parallelogram. Since AD || BC you only have to calculate the side AD = BC. Use Sine rule:

    AD/70 N = sin(45)/sin(30)

    AD = 140*1/2*sqrt(2) N = 70*sqrt(2) N ≈ 98.994949.. N ≈ 99 N
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  3. #3
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    Quote Originally Posted by Mr_Green View Post
    Please help me out here. thanks
    This is the more "traditional" way to do this:

    This is a Newton's 2nd Law problem.

    Let the tension on the left be T1 = 70 N and the tension on the right be T2. Let the mass be m. I will use a coordinate system with +x to the right and +y upward.

    Now take Newton's 2nd Law in each coordinate direction acting on the point where the 3 strings meet:
    (Sum)F_x = -70*cos(45) + T2*cos(60) = 0 <-- Since this point has no mass (and is not accelerating either.)

    (Sum)F_y = 70*sin(45) + T2*sin(60) - mg = 0


    From the x equation we know that:
    T2 = 70*[cos(45)/cos(60)] N = 98.9949 N

    As a bonus:

    Inserting this into the second equation we get:
    70*sin(45) + (98.9949)*sin(60) - mg = 0

    m = 1/g*[70*sin(45) + (98.9949)*sin(60)] kg =
    13.7989 kg

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post
    ...
    m = 1/g*[70*sin(45) + (98.9949)*sin(60)] kg =
    13?.7989 kg

    -Dan
    Hi, topsquark,

    I've found the digit you've lost: ? = 5
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