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FBD troubles

  1. Dec 25, 2007 #1
    [SOLVED] FBD troubles

    1. The problem statement, all variables and given/known data [​IMG]

    I am looking at 3-15 (with diagram). I am having trouble since there is no angle given.

    I have so far [itex] -F+F_{ax}+F_{cx}=0[/itex] and in y direction [itex]F_{ay}=F{cy}[/itex].

    Since the y components are equal, the springs have equal unstretched lengths, can I assume that the angle they make with the horizontal must be [itex]45^\circ[/itex]?

    Or that since the vertical distance between the B and A must be 3 meters?
    Or is this not the way to be looking at it?

  2. jcsd
  3. Dec 25, 2007 #2


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    Pythagoras is the simple answer here. you can split the diagram into 2 right angled triangles with side lengths d and 3m.
  4. Dec 25, 2007 #3


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    You are given enough of the sides of the triangles formed by the pulley's (d is given to you in the problem) to find the angles and other sides of the triangles. You should be able to find all the angles you need just using trig. Take another look at the figure, and try to use trig and the Pythagorean theorem to find your angles and unknown sides.

    (Kurdt, you beat me to it! It was a close one though...)
    Last edited: Dec 25, 2007
  5. Dec 25, 2007 #4
    Hint: Let be [tex]\theta[/tex] the angle that the springs make with the horizontal. Then:


    Last edited: Dec 25, 2007
  6. Dec 25, 2007 #5
    Thanks guys. Could you answer my original question too? Is this why I have enough info^^? Since y components are equal and the springs are identical implies that the angle is bisected?

  7. Dec 25, 2007 #6
    Yes, since the two springs have the same stiffnes and the same unstretched lengths, when you apply them a force, they stretched to the same length. Therefore they form an isoceles triangle.
  8. Dec 25, 2007 #7
    I am not sure what I'm doing wrong here. The hypoteneuse=3.354m then I should have the magnitude of the spring force as [itex]F_s=-ks=-500(3.354-6)=1323 N[/itex].
    Then I have in the x direction for the applied force
    [itex]F_A=F_s\cos\theta +F_s\cos\theta=2*1323\cos63^\circ=1.2kN[/itex]
    But my answer key says 158 N

    I got the angle from [itex]\arctan\frac{3}{1.5}=63.43^\circ[/itex]
  9. Dec 25, 2007 #8

    The lenght of each spring is 3m.
  10. Dec 25, 2007 #9
    It says in the problem statement each spring has an unstretched length of 6 m.

    Where are you getting 3? Thanks for the help so far by the way. :)
  11. Dec 25, 2007 #10


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    Unstretched length of each spring is 3 m. Thereforer stretching is 3.354 - 3 = 0.354m. Now calculate the force.
  12. Dec 25, 2007 #11
    But the diagram shows that each spring has an unstretched length of 3 m. Maybe the statement is ambiguous...?

    Moreover, in this case: [tex]l_f>l_o[/tex]
    Last edited: Dec 25, 2007
  13. Dec 25, 2007 #12
    Okay, maybe I was brainwashed or something but I'll ask again: WHERE do you see something that says the 'unstretched spring is 3 m'???

    OP says length of unstretched spring=6 m

  14. Dec 25, 2007 #13
    I see your point chuy. I don't know why the statement says that, but this does get me the correct answer.

    I'll take it up with the author!
  15. Dec 26, 2007 #14


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    WHERE do you see something that says the 'unstretched spring is 3 m'??? In the quote it is given that unstretched length of AB and BC is equal to 6 m. Then ,from the diagram, obviuosly unstretche length of each spring is 3 m.
  16. Dec 26, 2007 #15
    AH HA! Thank you, was taking that to mean EACH, but 6m means the COMBINED length of the TWO.

    Thanks rl.bhat. Thanks chuy.
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