a man weighiing 7-kg lies in a hammock whose ropes make angles of 20 and 25 degrees with the horizontal. What is the tension in each rope?

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  1. the horizontal components are equal in magnitude or the guy would accelerate.
    A cos 20 = B cos 25
    .940 A = .906 B
    A = .964 B

    A sin 20 + B sin 25 = 7*9.8
    .964 B (.342) + B (.423) = 68.6

    I think you can solve for B and go back and get A

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  2. Rofl, I think you're in my calculus class. I have the same question for homework too.

    I'm not sure if you're solving the problem using geometric or catersian vectors.

    It's always best to draw out the problem. . .

    Assuming you meant 70kg instead of 7kg. . .

    I'll do this problem using geometric vectors.

    You have a triangle with the following angles: 65, 45 and 70.

    70 corresponds with |T1| (Tension 1)
    65 corresponds with |T2| (Tension 2)
    45 corresponds with 686 N (9.8 x 70)

    Use sine law to solve for the tensions.

    |T1|/sin70 = 686/sin45
    |T1| = 911.6 N

    |T2|/sin65 = 686/sin45
    |T2| = 879.3 N

    Therefore, the tensions are 911.6 N and 879.3 N

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  3. thx, i just have problems drawing the diagrams

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