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Newton's second law with two objects

  1. Feb 11, 2007 #1
    My textbook is Essential University Physics by Richard Wolfson. I'm attending Illinois Central College, and my course is Engineering Physics 1: Mechanics. It's calculus based.

    1. The problem statement, all variables and given/known data
    From the Text:

    "A florist asks you to make a window display with two hanging pots as shown in the diagram. The florist is adamant that the strings be invisible as possbile, so you decide to use fishing line but you want to use the thinnest line you can. Will fishing line that can withstand 100N of tension work?"

    Original given diagram and my free body diagrams are attached.

    (edit: I had an ascii diagram worked out that didn't work in the preview, but when i scrapped it i forgot to change my angle notation, so I'm just arbitrarily naming the angles 1, 2, and 3)

    Mass of A = 3.85 kg
    Mass of B = 9.28 kg
    Angle 1 = 13.9 degrees
    Angle 2 = 54.0 degrees
    Angle 2 = 68.0 degrees
    Force of Gravity = -9.8 m/(s^2)
    Tensions : T(a1), T(a2), T(b1), T(b2) = ??

    I know this problem is of net forces to find string tensions. I started first trying to find the x and y components of each string tension assuming that the objects weren't in motion so the sum of y components would equal the inverse of gravity, but I figured that I couldn't assume that because I don't know if the string is actually breaking or not (if it were breaking, gravity's force would be larger, no?), and if I did assume that, I don't know the distribution of gravity between the y components (if the angles were equal, the y components would balance the forces equally, making this problem infinitely easier). Maybe I'm thinking too much about it (I tend to do that), or maybe I'm missing something from the basic concepts, or maybe i'm just really bad at remembering trigonometry.

    2. Relevant equations

    So far:


    Movement isn't considered, so (for both objects)
    T(1) + T(2) + G = 0

    Tension of the string between objects A and B
    T(b2) = -T(a2)

    This is where i start to get fuzzy

    Y components (for both objects)
    T(1y) + T(2y) + G = 0

    X components (for both objects)
    T(1x) + T(2x) = 0

    3. The attempt at a solution

    I was still working out an equation. I tried some trig functions, but they didn't seem to work out in my mind because of the conceptual problems I stated earlier. There was a similar, simpler example problem in the text that had a single mass hanging from a rope and equal angles that it was hanging from, and it gave me ideas, but the different angles in my problem keep screaming at me that there's something different at play, something I'm missing.

    4. Closing remarks

    What I'm asking here is not the answer. I just want a little help getting on the right track. Am I missing anything? Do I have too much of anything that is tripping me up? And for goodness sake, what is the secret gravity is hiding?

    Many thanks to any help whatsoever.

    Attached Files:

    Last edited: Feb 11, 2007
  2. jcsd
  3. Feb 11, 2007 #2


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    Staff Emeritus
    Science Advisor

    This is the acceleration of a particle due to gravity; the force on an object due to gravity is equal to mg
    Let me just get the question right in my head; the angles and masses are specified, and we want to calculate the tensions in the strings? So, we calculate the tensions, then if one is greater than 100N we know the string will break.

    You seem to be going along the right lines. Plug some numbers in; you've got four unknowns, and four tensions, so you can solve the problem.
    I dont know what you mean here. The force due to gravity on each basket is simply mg.
  4. Feb 11, 2007 #3

    Sorry, I wasn't clear on my earlier post... And the Force of gravity being 9.8... that was an error while I was typing, didn't think about it.
    I was wondering about the distribution of forces between the two strings (on each seperate mass). Is there some way I have to go about finding out that proportionality of tensions? Or is each hypotenuse just half the object's weight, and i use trigonometry to find their values? I should have explained that part better because this is what I was talking about with the "secret of gravity." I got a little too dramatic with that:-)

    I guess what I'm trying to say is - what aspect of the tension is equal and opposite to gravity? If it's just the sum of the tensions themselves (hypotenuse), then the problem is very easy, and it doesn't seem so simple.

    BUT, I just figured it out.

    Object B weighs about 90 N, which must be equal and opposite in the upward direction, so it's equal to the Y component sum of the tensions. Therefore, about 90 * sin(angle) will yield the rope's tensions. On object A, it's the same deal, except I already know the tension on one of the lines, and the other line is found by adding the Y component of the known line to the Y component of the unknown and solving for the actual tension using the given angle. I'm terribly sorry to waste your time:(
  5. Feb 11, 2007 #4
    Nevermind, I'm wrong again... I keep going back and forth:(
  6. Feb 11, 2007 #5


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    Staff Emeritus
    Science Advisor

    Let's go back to your handwritten work. Consider the drawing of the forces acting on mass B. Noting that here the force due to gravity is equal to 9.28g, try and write equations to balance the forces in the horizontal and vertical directions. e.g. do you know how to work out the component of TB2 in the vertical direction, given that you know the angle θ3?
  7. Feb 11, 2007 #6
    i worked on this problem for close to 8 solid hours now. I came up with an answer, and it seems viable, but i'll figure it out tomorrow in class. I 'm just having a problem getting around working the trig to work with the net forces. And no, i don't think i found how to get the force for Tb2 correctly:( Thanks for your help though.
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