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Homework Help: Acceleration help

  1. Oct 1, 2004 #1
    A bucket of water which weighs 11.5 Newtons is pulled vertically using two forces. The first of magnitude 9.7 Newtons acts at an angle 53 degrees from the vertical. The second acts at an angle 20 degrees from the vertical. Find the acceleration of the bucket in m/s2. (g = 9.81 m/s^2)

    I keep getting 14.03m/s^2 but it's wrong can someone please help.

    First I found the sum of Fx = (9.7sin53 + -Tsin20) = 22.65
    Then, sum of Fy = (9.7cos53 + Tcos20) = -6.212
    Fx + Fy = 16.438
    F=ma so, 16.438 divided by (11.5/9.81) = 14.02

    What am I doing wrong? ALSO, how do I find the net force acting on the bucket (due to the forces and the weight) in Newtons?
    Last edited: Oct 1, 2004
  2. jcsd
  3. Oct 1, 2004 #2
    Why did you take the linear sum of Fx and Fy? The magnitude is sqrt(Fx^2 + Fy^2)
  4. Oct 1, 2004 #3


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    Not only that, but your computations of Fx and Fy don't make sense to me. You are never given the magnitude of the second force, yet somehow your unknown (T) just disappears! Am I missing something?
  5. Oct 1, 2004 #4
    I found T according to the steps in my textbook.
    Since the magnitude for the second force is not given, how do I find that?
  6. Oct 1, 2004 #5


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    But that's precisely what I'm saying...I'm not sure how to solve the problem without the magnitude of the second force, and "T" IS the magnitude of the second force! So it totally baffles me how you could have solved for it. :confused:
  7. Oct 1, 2004 #6
    cepheid, that's what I was having trouble with too because the second magnitude wasn't given.
  8. Oct 1, 2004 #7


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    Since the bucket is pulled vertically, no accelerations are present in the horizontal, hence the sum of "horizontal force" is zero, find the missing tension from there.
  9. Oct 1, 2004 #8
    arildno is correct. Ax = O, so Fx becomes zero as well (fx = mAx).
    but we can still use that information to solve the problem.

    i'm going to assume F1 and f2 are along the -x and +x axis.
    so sum Fx = 0 = -9.7N(cos53) + F2cos20

    9.7Ncos53 = F2cos20

    F2 = 9.7Ncos53 / F2cos20 = 6.21N

    okay, cool, now we can work with Y components.

    sum Fy = 9.7N(sin53) + 6.21N(sin20) Here both are positive b/c they're both pointing up

    sum Fy= 9.87N

    F = ma

    a = F/m = 9.87N/1.17kg = 8.43m/s2 Is that the answer you have?
  10. Oct 1, 2004 #9
    The x and y components are CORRECT. Yet the Fx must be set equal to 0 because there is no force in the horizontal direction. So set the first equation equal to zero and solve for T. Then use this T for the y component and your problem is solved...

  11. Oct 1, 2004 #10
    Hey, this is incorrect. Recall that are angles are given with respect to the vertical axis, not the horizontal axis...

  12. Oct 1, 2004 #11
    Thanks Marlon, I get it now :smile:
  13. Oct 1, 2004 #12
    my pleasure...

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