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Very basic force pulleys problem

  1. Jun 23, 2015 #1
    1. The problem statement, all variables and given/known data
    There's a 200g piece of weight hanging by a string. That string is then split into two strings, each at θ from the horizontal and Φ from the horizontal. These two strings are each connected to force sensors which measure their tension in newtons. I know how to do this problem, as it's a simple trigonometric or pythagoras problem. The problem I'm having, is the fact that my answers are completely different to what my teacher got. She actually used real sensors and wrote down what she saw when she tested each θ and Φ.
    Maybe I don't know what I'm talking about, or maybe my teacher's force sensor is faulty. You decide.
    Here's a diagram: http://i.imgur.com/439esTg.png

    2. Relevant equations
    F=mg
    sin(θ)x0.2kgx9.8=NF1
    sin(Φ)x0.2kgx9.8=NF2
    Triangles, sin cos tan


    3. The attempt at a solution
    F1: sin θ x 0.2kg x 9.8
    F2: sin Φ x 0.2kg x 9.8
    Here's an image of my method. Please let me know whether it's correct or not. http://i.imgur.com/LTbo8Q9.jpg I cannot reproduce them as words here, as they are almost all diagrams.
    Note: The columns "Force F1 measured", "Angle θ", "Force F2 measured" and "Angle Φ" were already pre-filled when I was handed this worksheet. The only things I filled in were the "Force F1 calculated" and "Force F2 calculated". You will notice how incredibly 'off' my calculated measurements are from her measured measurements.
    So, am I going insane? Or does my teacher need to invest in a new force sensor?
    Thank you
     

    Attached Files:

  2. jcsd
  3. Jun 23, 2015 #2

    CWatters

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    Those aren't correct. What would happen if you make both strings vertical so that θ = Φ = 90? Try it and you will see it's wrong.

    The 200g mass is stationary (not accelerating) so what does that say about the net horizontal and vertical forces acting on it?
     
  4. Jun 23, 2015 #3
    But that's exactly the steps taken to solve 2 very similar (practically identical) problems in my physics book?? Look: http://i.imgur.com/nqK5TTY.png what am I missing here?


    The sum of horizontal forces are equal to zero, and the sum of the vertical forces equal zero. Right?
     
  5. Jun 23, 2015 #4
    In the sample problems, the weight was divided equally between the two strings. Why is that not the case in this problem?
     
  6. Jun 23, 2015 #5
    Edit: nevermind. There is no time to wait for a reply. Admins, please delete this thread.
     
    Last edited: Jun 23, 2015
  7. Jun 23, 2015 #6

    SammyS

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    What you are calling F1 is actually the component of the weight that's in the direction of string 1. Similar for F2 and string 2.

    What you actually need is the vertical component of F1 added to the vertical component of F2 to equal the weight.

    And as you stated earlier you need the following to be true.
    That is the horizontal component of F1 needs to be equal in magnitude to the horizontal component of F2 .
     
  8. Jun 23, 2015 #7

    haruspex

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    The thread may still have value for others, if only in showing how careful you have to be in assuming the same equations for set-ups that are not identical.
    Consider the junction between the three strings. There are three forces acting on this point. Apply the usual statics equation in each of the horizontal and vertical directions, ##\Sigma F = 0##.
     
  9. Jun 23, 2015 #8

    SammyS

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    By the way, your teacher's force sensors look to be working pretty well.
     
  10. Jun 23, 2015 #9
    F1 = ##\frac{1.96cos(θ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)} ##N

    and

    F2 = ##\frac{1.96cos(Φ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)} ##N

    Correct?
    My answers are much closer to my teacher's now. Which means it is correct.
     
  11. Jun 24, 2015 #10

    haruspex

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    Yes, those look right. You can simplify the denominator a bit.
     
  12. Jun 25, 2015 #11
    Are F1 and F2 reversed from your worksheet?
     
  13. Jun 25, 2015 #12

    haruspex

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    Good catch.
     
  14. Jun 25, 2015 #13
    Reversed? Well, F1 is pointing to the left and F2 is pointing to the right. As shown in the diagram.
     
  15. Jun 25, 2015 #14
    Okay, then phi and theta are reversed in your numerators. Re-do a calculation using your above formulas.
     
  16. Jun 25, 2015 #15
    Yes, they are.

    What? What's that supposed to mean? I already did the calculations, using my new equations, and got answers that were very close to my teacher's. What are you telling me to change and 're-do'?
     
  17. Jun 26, 2015 #16
    When I use your equations, I get the results for F1 and F2 reversed. When I switch phi and theta in the numerators in your equations, I get the correct answers.
     
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