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Forces and acceleration on frictionless table?

  1. Jun 10, 2014 #1
    The two forces F1 and F2 act on a 27.0-kg object on a frictionless tabletop. If F1=10.2 N and F2=16.0 N find the net force on the object and its acceleration for cases (a) and (b).

    (a) is 90 degrees, in the 3rd quadrant
    (b) is 120 degrees, from the positive y-axis and moving clockwise

    To calculate the angles at which the resultant forces act in this problem, please start at the positive x axis and travel counter-clockwise.

    1. The net force in figure (a) (above) is?

    2. The angle from the positive x axis at which the net force in figure (a) acts is?

    3. The magnitude of the acceleration of the object in figure (a) is?

    -----------------​

    For 1, I don't know which formula I'm supposed to use. I can't find it in my textbook or anywhere else... but I think for 3 it's: a = (net force)/(mass in kg)

    For 2, is this how you do it?: from the positive x-axis, move counterclockwise until you hit the angle.

    Thanks for all help!

    (Sorry about the broken links)
     
    Last edited: Jun 10, 2014
  2. jcsd
  3. Jun 10, 2014 #2

    Nathanael

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    The link to the picture doesn't work (you need an account on that website)
     
  4. Jun 10, 2014 #3

    BiGyElLoWhAt

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    This is why the forum rules say to attach the picture to the post and not to host it externally...
     
  5. Jun 10, 2014 #4

    BiGyElLoWhAt

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    Here's a general way to find net forces for non parallel/perpendicular forces.

    ##R^2 = R_{x}^2 + R_{y}^2##

    ##R_{x} = R\text{cos}(\theta)##

    ##R_{y} = R\text{sin}(\theta)##

    For 2D motion:

    ##\vec{R} = <R_{x},R_{y}>##

    if you decompose all your forces down into this vector form, then you can add all your vectors together simply by adding the x components and y components.

    There's numerous trigonometric relationships that can be used to solve for your componts, and it's really as simple as using trig.
     

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  6. Jun 10, 2014 #5
    @ BiGyElLoWhAt
    does R= F?
     
  7. Jun 10, 2014 #6
    would it be 10.2 cos 180?
     
  8. Jun 10, 2014 #7

    Nathanael

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    R is an arbitrary vector, it can be Force or anything else (so yes)

    Edit:
    Nevermind, I didn't notice you edited your original post

    This is unclear.


    You can't draw a picture or something?
     
  9. Jun 10, 2014 #8

    BiGyElLoWhAt

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    R is a general vector of length |R| and direction theta.

    F can be equal to R, if you're seeking to break F down into components. If so, use the trigonometric relations I posted.

    And as for 10.2 cos(180)

    I have no clue, as I haven't seen the picture. I'll let you be the judge of that. ;)
     
  10. Jun 10, 2014 #9

    BiGyElLoWhAt

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    What do you mean by 90 degrees in the third quadrant?

    Assuming you're labeling your quadrants as so:

    II I
    III IV

    do you mean <-
    or
    |
    v
    ?

    (those are arrows btw haha)
     
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