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Acceleration of couch pushed by non-horizontal force

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    1. An 82 kg couch is being pushed with a force of 56 N at an angle of 40o above horizontal. Friction is to be considered. Draw a free-body diagram representing this scenario.

    2. If the coefficient of friction between the couch and the floor is 0.051, what is the acceleration of the couch? Be sure to reference the free body diagram in your response.
    Given:
    Mass = 82kg
    Fapp= 56N
    theta= 40°

    2. Relevant equations


    3. The attempt at a solution
    I wanted some clarification, I'm not sure if my steps were correct, I'm still fuzzy about this concept because I haven't taken physics in a year. I based my solutions off another example.

    My solutions are posted in the picture attached.
    p.s sorry for they low quality, my camera is pretty bad.
    physics hw help.jpg
     
  2. jcsd
  3. Feb 25, 2017 #2

    Nidum

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    couch-2.jpg

    Best I could do with my image enhancement software . You really need to re-draw the FBD and make it much larger and clearer and also type out the calculations .
     
  4. Feb 25, 2017 #3

    gneill

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    Perhaps you can draw one larger picture of your FBD and type in at least your Relevant Equations?

    It looks like you have your "couch" on a ramp. The problem statement doesn't mention a ramp or slope of any kind. In fact it mentions a floor, which is usually taken to be horizontal. It does state that the applied force has a certain orientation with respect to the horizontal though.
     
  5. Feb 25, 2017 #4
    As gneill suggested, I was also wondering about your interpretation of the wording, John A. But looking at your work, it looks like you calculated the components of the 56 N force (the direction of the force being in question), but I didn't see where you used them anywhere.

    Also, in calculating the parallel and perpendicular components of the weight, you took sin and cos of the magnitude of the force (sin56 and cos56) instead of the sin and cos of the angle.

    Edit: I thought the wording of the problem was kind of ambiguous, but I thought it meant that the couch was on a flat surface, but the force was being applied at an angle 40 degrees above horizontal.
     
  6. Feb 25, 2017 #5

    gneill

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    Mod note: The thread title has been changed to describe the problem being asked.

    Please note that over-general titles or pleas for help are not considered to be unique enough or descriptive of the actual topic. Helpers should be able to recognize the type of problem and physics involved from the thread title. Thanks! :smile:
     
  7. Feb 25, 2017 #6

    kuruman

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    There is no mention of an incline in the statement of the problem. The couch is on a horizontal surface being pushed from below at a 40o angle. That's my interpretation and I agree with Tom Hart.
     
  8. Feb 25, 2017 #7
    Is this a better FBD?

    fbd.jpg

    Also I'm confused about the significance of the force being applied at an angle. I would just find the vertical and horizontal results of Fapp like I did previously, correct?
     
  9. Feb 25, 2017 #8

    kuruman

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    You show no force applied at 40o in your FBD. You need to redraw it.
     
  10. Feb 25, 2017 #9
  11. Feb 25, 2017 #10
    Yes, you would find the vertical and horizontal components of the 56 N force. In the 2nd FBD you posted, you show a vertical force called Fg, and you show a horizontal force called either Fa or F9 or Fg. (It's hard to tell.) What is that force? If that is the 56 N force, then you have it incorrectly shown as a horizontal force. But it looks like in your original FBD you have 2 forces called Fg - one is vertical and appears to be the weight and the other is in the direction of the ramp, and is . . . ? I don't know.
     
  12. Feb 25, 2017 #11
    I think your latest FBD in post #9 looks good.
     
  13. Feb 25, 2017 #12

    kuruman

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    I agree. Remember FN is not equal to the weight Fg. It is what is necessary to provide the observed acceleration in the vertical direction.
     
  14. Feb 25, 2017 #13
    Thank you all for your help, I'll attempt to solve it and reply if I get stuck. I'll post my final answer when finished.
     
  15. Feb 25, 2017 #14
    final.jpg

    I got 4.88 x 10^-4 for my acceleration, which is really low.
     
  16. Feb 25, 2017 #15

    kuruman

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    You say FN = Fg+Fsomething illegible y.
    Can you explain that? Remember, up is positive down is negative.

    On edit: Please provide better pictures or (learn to) use LaTeX. You may find that people will not be quick to respond if they can't read your posts.
     
  17. Feb 25, 2017 #16
    I found the equation online, it was Fn Fg + Fay (Fapp in y-direction)
    I'm not sure what LaTeX is but next time, I think I'll just draw up a diagram with a picture software to make it clearer. I think I may just move on to a different question because I'm stumped on this one.
     
  18. Feb 25, 2017 #17
    John A, you wrote: FN = Fg + Fay. You should check your signs in that equation.
    For equilibrium (no acceleration), ΣF = 0. Another way to think about it (which I prefer) is:
    Σ(upward forces) = Σ(downward forces). That is what I originally thought you were doing. You have to be consistent in your signs and directions. Typically, the positive direction will be a positive value, and the negative direction will be a negative value.

    It may very well be obvious to you, but in case not, I am going to explain it. How do you know whether or not there is acceleration in the y direction? The y component of the applied force has to exceed the weight of the object. If it is less than the weight, the object will not accelerate in the y direction and the y component forces will sum to zero. The reason I'm telling you that is to try to help you understand the concept, as opposed to trying to find a formula that fits.

    Except for that one sign error, I think the remainder of the problem was worked correctly.
     
  19. Feb 25, 2017 #18

    haruspex

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    Frequently a bad move. You need to check that your set-up complies with the preconditions for the equation. If they're not specified, don't trust it. Even if it does match, it might not be a trustworthy source.
    As TomHart wrote, the equation you can rely on is ΣF=ma, where the sum is over (the components in some chosen direction of) all forces acting on the rigid body and a is its resulting acceleration in that direction. For a statics question, as here, a=0, of course. The online equation you found only considered two forces with vertical components; you have three.
     
  20. Feb 25, 2017 #19
    FinalPic.jpg

    I changed the signs and I think I got it this time. The problem I made is I wasn't accounting the fact that Fg has a negative value so Fapp-y should be subtracted from Fg (I think) since Fapp-y is positive and moving in the opposite direction of Fg.

    Ps. Ignore #3 on my picture, it was for a different part of the question. Also, sorry once again for the terrible quality.
     
  21. Feb 25, 2017 #20

    kuruman

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    Your answer agrees with mine.
     
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