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Homework Help: Finding the unknown magnitude of force

  1. Jun 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Here's a link to the website that has the problem. It's right under example 1.

    2. Relevant equations

    3. The attempt at a solution
    This is just for the first part; finding Curly's force.
    For the sled not to move up or down, both Curly and Moe will each have to have a y component force that cancels out the other one. I calculated Moe's y component force to be 17.32 N, which means Curly would also have to have 17.32 N of downward force to negate Moe's force. Then I calculated Curly's x component of force from having his y component of force and got 29.999 N of force.

    What they did is completely foreign to me. First, they got 8.660 N as Moe's y component of force because they multiplied .866, the sine, times the x component. I don't know why they did that.
    Then they just divided the 8.660 they got by Curly's sine to get Curly's x component of force. I don't see why they did that and they don't explain why they were doing it.

  2. jcsd
  3. Jun 18, 2010 #2


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    Gold Member

    Hi there!

    How did you get that result? The y component cannot be larger than the magnitude, which is 10 N. You have to use

    y component = magnitude times sin(angle measured with respect to the x axis)

    Not quite. They multiplied the sin by the *magnitude* of Moe's force. And that's in agreement with what I wrote above.

    They do not divide 8.660 by any sine. So I am not sure what you are refering to.

    Hope this helps!
  4. Jun 18, 2010 #3
    I assumed the magnitude was their x component force. So their magnitude was the total force they were exerting? Which would be the hypotenuse?
    Oh, ok.
    I was just kinda reading what the last equation was that they did. I'm going to try this again and see if I can do it....

    Ok thanks. I got 17.32 as his force, which is what they got. What confused me is the way they did it at the end. That and I thought the force they were talking about was the force just in the x direction. Don't know why.

    Thanks again.
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