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Addition of Forces

  1. Sep 15, 2004 #1
    Two adults and a child want to push a wheeled cart in the direction marked x in the figure (attached). The two adults push with horizontal forces F1 and F2 as shown in the figure (attached).

    A) Find the magnitude of the smallest force that the child should exert. You can ignore the effects of friction.
    B) Find the angle force makes with +x-direction. Take the angles positive measured counterclockwize from the +x-direction.
    C) If the child exerts the minimum force found in part (A) and in part (B) , the cart accelerates at 2.0 m/s^2 in the +x - direction. What is the weight of the cart?

    A) I was able to do this one fine by finding the x and y components of both forces and then finding the difference between the y components. No issue there.

    B) I am completely at a loss as to what exactly they want. Is it just me or is this question poorly worded, or is that just on purpose? Anyway this is where I need a hand. I've tried using the answer I found for A (a third y force, that of the child if you will) and taking the arctan of it/the total force in x. No luck there. I've tried subtracting the arctan(force of the child/Force 2 in x) from arctan(force of the child/Force 1 in x). No luck there. I've also tried adding both of the previous angles together, and still no luck.

    Can anyone offer any insight into how to approach this? I'd really appreciate it.

    As far as C is concerned I am fairly certain that once I get B I'll be able to figure it out, so that can wait... for now at least. :P

    EDIT: Forgot to attach the figure. Oops!
     

    Attached Files:

  2. jcsd
  3. Sep 15, 2004 #2

    Doc Al

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    Staff: Mentor

    Man, part B is one poorly worded problem. Are they talking about the force from part A? (Did you copy it exactly? If that's your text, you have my sympathy.) If I had to guess what they might mean in B, I would say: What is the angle that the force of the child (from part A) makes?

    In any case, the way I read the problem, parts A and C are clear enough to solve even without B.
     
  4. Sep 15, 2004 #3
    Yep. I know, it is horrid haha. I'm going to give part C a shot.
     
  5. Sep 16, 2004 #4
    As you figured out in part a, the smallest force should only require the boy to balance out the y component. In part b, I believe the question is asking for the angle this force makes with respect to the x-axis and since this force is downward, the angle should be 270. For part c, you just need to sum up forces in the x-direction and divide by the acceleration to find the mass.
     
  6. Sep 16, 2004 #5

    Doc Al

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    part b

    That's exactly how I interpreted it.
     
  7. Sep 16, 2004 #6
    B was correct, and makes perfect sense.

    C is incorrect though as the problem is asking for the weight, not the mass. It gives the acceleration, however, there is no mass given. Since the answer it wants is the weight (in N) that leaves no purpose for the acceleration, as far as I can tell.

    If we were given mass then yes, we would need to multiply by the acceleration to find the weight. Or like you said if we were solving for mass, then divide the forces by the acceleration. That isn't the case though...

    So with that in mind I've tried totaling F1 (100N) + F2 (140N) + F3 (16.6N since its only in the Y), however that answer is wrong. I've also tried totaling just the X values (F1_x = 100 cos 60 = 50N) + (F2_x = 140 cos 30 = 121.24N) however that is wrong as well.
     
  8. Sep 16, 2004 #7
    But once you find the mass, you can calculate the weight. The net force in the x-direction causes the acceleration given. You know the relation between force, mass and acceleration, so you can solve the mass.
     
  9. Sep 16, 2004 #8
    Hi, I'm having the same problem. I have figured out parts A and B. For C, I have F = m*a, where a=2m/s^2. So, to find m I divide F (=171.24N) by a (=2m/2^2) to get 85.62 mkg. So, to find the weight in N, I'm trying to divide by the earth's acceleration 9.8m/s^2 and get 8.74N. However, This is wrong. Any help?
     
  10. Sep 16, 2004 #9
    You would want to multiply by the Earth's acceleration, not divide, as you have 85.62 for your mass, and mass * acceleration = weight.
     
  11. Sep 16, 2004 #10

    Doc Al

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    Why are you dividing mass by an acceleration?




    Edit: underthebridge beat me to it!
     
    Last edited: Sep 16, 2004
  12. Sep 16, 2004 #11
    thanks, that makes sense now that I consider units.
     
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