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Chandelier Tension Problem

  1. Oct 14, 2007 #1
    I actually have an answer from school but I couldn't follow some of the steps. And I have a big test tomorrow and I'm probably already down to a B because I screwed up a test and I need help on this please.

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

    Find F1 and F2

    2. Relevant equations

    Weight of the chandelier = 200kg
    Then 60 degrees on F1

    3. The attempt at a solution

    I drew it on coordinate plane.
    So far, I have

    For F1 :
    F1x = -F1(cos60) (not too sure about the negative)

    F1y = -F1(sin60) (not too sure about the negative)

    For F2 :
    F2x = F2

    F2y = 0

    For W :
    Wx = 0

    Wy = -1960 N (I did 200kg x 9.8m/s^2)

    Then I vaguely remember the teacher saying about this thing that looks like a 3.
    The zamation or something. It's the thing you click. Like a sideways M (clockwise rotation). I tried finding it on the list of symbols in this text box but it got confusing.

    So it's

    "M"Fx = 0
    = -F1(cos60) + F2 + 0
    = -F1(.5) + F2
    (I lost track after this. The teacher said the values had to be equal.)
    Here's what it said on the notebook:
    = (2253N)(.5) = F2
    =1127N = F2

    "M"Fy = 0
    = -F1(sin60) + 0 -1960
    = -F1(.866) - 1960
    (My notes don't make sense after that)

    "M"W : Wx = 0
    Wy = -1960N
  2. jcsd
  3. Oct 14, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good. You've found the components of all the forces.

    [itex]\sum[/itex] is the Greek letter Sigma, which is used to represent "Sum", as in the sum of the forces must be zero for equilibrium:

    [tex]\sum F_x = 0[/tex]

    [tex]\sum F_y = 0[/tex]

    To solve any equilibrium problem (like this one) you set the sum of the horizontal forces equal to zero:

    [tex]\sum F_x = 0[/tex]

    [tex]-F_1 \cos(60) + F_2 = 0[/tex]
    [tex]F_1 \cos(60) = F_2[/tex]

    That's the first equation that you need.

    Now analyze the vertical forces:

    [tex]\sum F_y = 0[/tex]

    [tex]F_1 \sin(60) - W = 0[/tex]
    [tex]F_1 \sin(60) = W[/tex]

    That's the second equation that you need. Since you know that W = mg, you can solve for F_1. Then use the first equation to solve for F_2.

    Make sense?
  4. Oct 15, 2007 #3
    Yes, it does kind of. I just had to look at someone else's notes.

    On my attempt at a solution, isn't F1y supposed to be


    instead of


    Because I tried it like the second one and I got the same answer as my classmate. But I don't know, we all take inaccurate notes. =S It it supposed to have a negative? Because the F1y is above the 0 vertically so it shouldn't be negative?

    Also, I should have done the Fy first because it was going to give me F1 which I could use in the second equation.

    But THANK YOU for the help. Oh, for the test, I got a 7/10. Two problems, five points each. I didn't do too well on the second problem which is like this except F1 and F2 are at 35 degrees angles.

    And the weight is 200.

    Do you think I could do the problem here and you can pick and tear my errors? Because I got a 4 but I reached the end enough so she gave me a 5. Which I feel bad about. I was SO sure.


    For F1:

    Fx = cos35 (-F1)
    Fy = sin35 (F1)

    For F2:

    Fx = cos35 (F2)
    Fy = sin35 (F2)

    For W:

    Wx = 0
    Wy = -1960


    = cos35 (-F1) + cos35 (F2) + 0
    = .82 (-F1) + .82 (F2)
    = .82(-F1 + F2)
    = 0


    ____ = sin35 (F1) + sin35 (F2) - 1960
    1960 = .57(2F)
    3438 = 2F
    1719 = F

    F1 and F2 have the tension 1719.

    Did I do it right?

    And if you don't mind, could I post the second question? >_> Because that's the one I screwed up in.
  5. Oct 15, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Absolutely. It should be positive, not negative. (I should have pointed that out.)

    The best strategy might be to write down both equations and then pick which one to use first.

    Looks good, assuming the angle is 35 degrees (your diagram says 30).

    Good. I would write it this way:

    [tex]\sum F_x = 0[/tex]

    -cos35 (F1) + cos35 (F2) + 0 = 0
    -F1 + F2 = 0
    F1 = F2

    Good. My version:

    [tex]\sum F_y = 0[/tex]

    sin35 (F1) + sin35 (F2) - 1960 = 0
    sin35 (F1) + sin35 (F2) = 1960
    F1 + F2 = 1960/sin35
    2F = 1960/sin35 [this comes from making use of the x-component equation]

    Yes you did. (In general, don't round off intermediate values. For example, sin35 is not exactly equal to .57 so keep a few extra digits and just round off your final answer.)

    Of course.
  6. Oct 15, 2007 #5
    Oh, yay. Tension problem

    Okay, so the problem is that there's a meter stick that's supported at the center. On the 80 cm line, there's a weight that's 20N hanging. Balancing it on the other side is an unknown weight, but it's sitting on the 10cm line. What's the unknown weight?


    Here's my almost solution.

    So the middle of the stick should be 50cm so it's the center. (.5 M)

    Then, found the torque for the right side of the stick.

    T = force x lever
    T = 20N x .3 meter
    T = 6Nm

    Then for the left side of the meter stick.

    6Nm = force x lever
    6Nm = (9.8 x MASS) x .4
    15N = 9.8 x MASS
    1.53 = MASS

    Is this right?

    Because I swear that's what I did to my paper and I got a 2/5 on the problem. When the teacher was explaning, she said something about the lever being 10cm away from the edge and that's from where she measured.

    - - - - - - - - -

    Here's the problem before the test that I got wrong:



    John's lever = .5m
    Jason's lever = 1.0m

    John's Force = 150N
    Jason's Force = ?

    MY Solution:

    John's Torque:

    T = Force x Lever
    T = (.5m)(150N)
    JOHN'S TORQUE = 75Nm

    Jason's Force:

    75Nm = (1.0m)(9.8m/s^2)(MASS)
    75N = 9.8(MASS)
    7.65 = MASS
    N = 75N
    Jason must exert a force of 75N to prevent the door from closing.


    Wd = 1.0m
    F1 = 150 N
    L1 = 0.1 m (near center)
    L2 = 0.4 (from edge)

    Find T:

    T = ?
    T = F1 x L1
    T = (150N)(0.1m)
    T = 15Nm

    F(.4) = (150N)(0.1m)
    F(.4m)/.4m = 15N/.4m
    F = 37.5 N

    First of all, I don't understand why they measured from the edge of the door to the center. I understand that it's not the exact center but it could've been on the .6 line or on the .4 line. Anyway, why? Because the torque is the point where the rotation occurs which is the door hinge, not the center of the door. It just doesn't make any sense! So yes, I need help on this. Maybe some explanations because I don't understand the notes.
  7. Oct 15, 2007 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Your solution is perfectly correct. I would have given 15 N as the answer, since they asked for the weight not the mass. But that's not the issue.

    You certainly could have used the left end of the stick (or any other point) as your reference pivot for calculating torques. But you certainly don't have to. (In fact, choosing the endpoint as the pivot involves an extra step, so I like your choice better.)

    You could try explaining to your teacher that any point can be used for calculating torques. And that your point is better than hers!

    Kind of funny that there's a typo (extra sentence) in the problem statement.

    All this is good.

    You "got the right answer", but that stuff in the middle with weight = mg doesn't belong. (Weight acts down--nothing to do with the forces in this problem.)

    It's much simpler than that:
    John's torque = Jason's torque
    75 = (Jason's Force)(1.0)
    Jason's Force = 75 N

    This makes no sense at all to me. For one, unless she's got access to extra information not given in the problem, "near the center" means use 0.5m from the hinge and "near the edge" means use the edge = 1.0 m from the hinge.

    For some wacky reason, she thinks that John pushes not at the center, but "near" the center at a point 0.4m from the hinge or 0.1m from the center of the door. (Where in the world she gets this from I have no idea.) She also thinks that Jason pushes "near" the edge, which she takes to be at 0.9m from the hinge or 0.4m from the center. (Weird!) Then, given those groundless assumptions, she attempts to calculate an answer using the center of the door as the pivot point. She calculates it incorrectly since she neglects the force that the hinge exerts on the door! (That's one reason why it's better to choose the hinge as your pivot--you can ignore the hinge force, since it doesn't exert a torque about that point.) Using her secret data about where the forces really were applied, the correct answer would be 67 N. (You can verify that by using the hinge as your pivot--much quicker than using the door center as the pivot. And less likely to make mistakes, like she did. :wink:)

    I wonder if she used the same incorrect method to solve the meterstick problem and that's why she marked yours wrong!

    I think your teacher is losing it. Let me know if there's anything in what I wrote that you don't understand.
    Last edited: Oct 15, 2007
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