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Determine force angles so that acceleration is least

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Figure below is an overhead view of a 12kg particle that is to be pulled by three ropes. One force (F1, with magnitude 50N) is indicated. Orient the other two forces F2 and F3 so that the magnitude of the resulting acceleration of the particle is least, and find that magnitude if

    (a) F2 = 30N, F3 = 20N
    (b) F2 = 30N, F3 = 10N
    (c) F2 = F3 = 30N.

    Figure: 0----->F1

    2. Relevant equations

    Ftx = F1 + F2x + F3x = ma

    a = (F1 + F2x + F3x) / m

    F2x = F2 * cos(θ)

    F3x = F3 * cos(θ)

    3. The attempt at a solution

    Initially, I said to myself, "well the least the acceleration can be is zero" so I assumed a = 0, Ftx = 0 and it follows that F1 = -F2x - F3x. Therefore:

    F1 = -F2 * cos(θ) - F3 * cos (θ)

    Solving for θ;

    θ = Arcos[F1 / (-F2x - F3x)]

    This gives you the angle of F2 and F3 and you've already made the assumption that a = 0

    This worked out for parts (a) and (c) of the problem, but it wasn't until I dug in to part (b) that I saw a few flaws in my logic:

    1) The problem doesn't say that acceleration is zero, it says to find the angle for the forces so that the acceleration is least. The only problem is, how do I set up my formula so that mathematically, the difference between the sum (F2x + F3x) and F1 is as small as possible?

    2) My final formula for θ assumes that both F2 and F3 have the same direction (in other words, θ2 = θ3), which may not be true. But if I introduce θ2 and θ3 in to my equations, they become very difficult to solve.

    Thanks in advance, MrMoose
     
  2. jcsd
  3. Sep 22, 2013 #2

    arildno

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    Remember that your acceleration is in TWO directions, the x and y directions. Thus,the magnitude of the acceleration is the "hypotenuse" of the "acceleration triangle".

    Let us take set up the problem:

    We have:
    F_1+F_2cos(a)+F_3cos(b)=ma_x
    F_2sin(a)+F_3sin(b)=ma_y

    We now square both equations, sum them together, in order to find the equation for the squared magnitude of the acceleration:
    We gain (remember sin^2t+cos^2t=1 for all t!):
    F_1^2+F_2^2+F_3^2+2F_2F_3(cos(a)cos(b)+sin(a)sin(b))+2F1(F_2cos(a)+F_3cos(b))=m^2(a_x^2+a_y^2)
    -------------------------------
    Now, this problem IS solvable, using partial derivatives, but I do not think that was the intention.
    Rather, I believe either you are the book has left out the critical information that a_y=0!!


    Am I correct about that?
     
  4. Sep 23, 2013 #3

    arildno

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    Okay, I made this too difficult.
    In a) and b), the answers must necessarily be to place the two other vectors in opposite direction to F_1

    But, in c), no acceleration in y-direction yields the condition:
    30sin(v_2)+30sin(v_3)=0, meaning v_2=-v_3

    Insert these values in the x-equation
     
  5. Sep 24, 2013 #4
    Yup, now I see what you did. Conceptually, you ask yourself, "how can I make the total force, Ft, as small as possible?" For (a) and (b), the sum of the forces f2 + f3 < f1. They will always produce the smallest Ft on the -x axis. However, in (c), f2 + f3 < f1, and you know that the x components of f2 and f3 will decrease as their angles increase. Now you have an opportunity to change the angles of f2 and f3 so their X components cancel out f1, yielding the smallest Ft, which is zero. In that case, I can use my assumptions from the original attempt (especially because f2 = f3 and you know θ2 = θ3). Thanks!
     
  6. Sep 24, 2013 #5

    arildno

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    Actually, v_2=-v_3, but it doesn't matter for the even cosine functions.
    If I did this right, you ought to get the equation cos(v_2)=-5/6, for zero acceleration in x-direction as well.
     
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