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Homework Help: Help needed - Newtons first law - particles in equilibrium

  1. Apr 17, 2008 #1
    For some odd reason, i'm having no problems with the rest of the exercises, though this one is really annoying as i just can't seem to get it right.

    Here's a picture of the exercise

    h t t p://img2.freeimagehosting.net/image.php?4bc2e12442.jpg
    http://img2.freeimagehosting.net/image.php?4bc2e12442.jpg [Broken]
    Could someone please give me a detailed description on how to solve it ?

    I've used m=w/g to find the weight of the person (882.9 N )
    But when i try to use sin and cos with the angle, i don't get the right result
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Apr 17, 2008 #2


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    Why don't you show us what you have done?

    What angles did you use in the sin and/or cos functions?

    Please provide us with more information about what you have done.
  4. Apr 17, 2008 #3
    I tried dividing it up in two sets, where the rope to the left was A and the rope to the right was B

    Since it's in equilibrium, the sum of forces in the x direction (Fx) is 0, Fy is also 0

    Since Fx=0 : B * sin 10 + (-A)
    Since Fy=0 : B * cos 10 + (-w)

    I then used the second one to find the tension from part B, but i'm not allowed to sum them together (?), so i don't really know what to do.

    Basicly, i haven't got a clue right now
  5. Apr 17, 2008 #4


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    Hi Hemmelig,

    Newton's law in the x-direction, for the case of no acceleration in the x-direction is [itex]\sum F_x =0[/itex]. Writing this out for the case of your three forces, A, B, and w gives:

    A_x+ B_x + w_x =0

    Your got that w_x=0 (so the weight force does not appear in the x direction equation); however, it appears that you did not get the right x-components for A and B.

    In the y direction, the same thing is happening. Your y-component for the weight is okay ([itex]w_y=-w[/itex]) but the y-components for A and B are incorrect.
  6. Apr 17, 2008 #5
    hmm, what are the right components for x and y then ?
  7. Apr 17, 2008 #6
    I got the right result (well, in a way)

    If i call the rope C

    C * sin 10 + (-w)

    w/sin10 = C

    But for some reason, the answer i get is exactly twise the correct result

    Am i doing it right and i'm just supposed to divide the answer with two ?
  8. Apr 17, 2008 #7


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    For the part of the rope pulling to the right (force B), it is pulling both upwards and to the right. It's x-component is the horizontal component of the force and if you draw a triangle for the components of B you can see that the horizontal component is adjacent to the angle.

    So, for example, [itex]B_x = +B \cos\theta[/itex]. If you find [itex]B_y, A_x, A_y[/itex] you can then find the tension.
    Last edited: Apr 17, 2008
  9. Apr 17, 2008 #8


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    If you solve the equation for forces in the x-direction and plug it into the force equation for the y direction, you'll get the above formula except with a factor of 2.

    Solving the x-direction equation will also tell you something important about the tension along ropes in these problems that will show you why the above equation is so close.
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