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What net force does the scale read?

  1. Nov 10, 2008 #1
    I stayed after with my physics teacher today but ran out of time so I was not able to ask him how to do this one last question.

    1. The problem statement, all variables and given/known data

    This problem is a picture so I have included a scanned photo of it.


    The question is stated in the picture. On the answer sheet the following questions are asked:

    F1 =
    F2 =
    F3 =
    Funk = (unk is in the subscript and no it is not a typo)

    2. Relevant equations

    3. The attempt at a solution

    I originally tried doing some trigonometry but I do not think that is what needs to be done.

    F1 = Sin (30) = op/200 x 200 =
    0.5 x 200 = op = 100
    1. The problem statement, all variables and given/known data
  2. jcsd
  3. Nov 10, 2008 #2
    Try another trigonometric function. ;) Draw the triangle with 200N as the hypotenuse then look for what you're solving for.
  4. Nov 10, 2008 #3
    That is what I thought I did since Sin = op/h = op/200.
  5. Nov 10, 2008 #4
    The force opposite that of the angle is 100N and it faces directly to the right or the left. Do either of those impact the weight of the scale if they cancel?
  6. Nov 10, 2008 #5
    Yes?? I am not sure.
  7. Nov 10, 2008 #6


    User Avatar
    Homework Helper

    What you have are three forces.

    What you need to know is what the sum of all the vertical components are of each force.

    The middle one is easy. 200N

    The ones to either side have what vertical components of force?

    When you figure that out, add them all together.
  8. Nov 11, 2008 #7
    Can anyone tell me exactly what I am doing wrong?? I got an opposite of 100 for both angles. I added them up to get 400 N. I did

    F1 = Sin (30) = op/200 x 200 =
    0.5 x 200 = op = 100

    Someone please help asap I ahve a physics class in a few hours./
  9. Nov 11, 2008 #8

    D H

    Staff: Mentor

    Why sine? Think about it this way: The middle rope is obviously exerting a force of 200 N, yet this rope has at a zero degree deflection. If you used sine, that would suggest the middle rope would exert a force of 0 Newtons on the scale since the sine of 0 degrees is zero.
  10. Nov 11, 2008 #9
    I see what you mean. Could somebody tell me how to do this problem?
  11. Nov 11, 2008 #10


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    Homework Helper

    Divide the components of the tension into the x and y components.

    How do you do this? What trig function defines the x component in your picture and which defines the y component?

    When you can answer that, you can answer your question, as the rest has been explained.
  12. Nov 11, 2008 #11
    The x component of both is 100 N and the Y is 200 N. I cannot answer the question
  13. Nov 11, 2008 #12


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    Homework Helper

    Come on now. You know better than that.

    If that was true your tensions on the side ropes would be given by:
    (1002 + 2002)1/2N

    But you know the overall tension of that hypotenuse is itself only 200N.
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