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Two forces act on an object

  1. Nov 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Two forces act on a 55 kg object. One has a magnitude of 65 N directed 59 degrees clockwise from the pos. x axis. The other has magnitude 35 N at 32 o clockwise from the pos. y axis.What is the acceleration of the object?

    The answer is given as 1.1 m/ s squared,
    (I got a similar answer 1.0 m/s squared.) , but not confident that I approached the solution correctly.

    Thanks in advance for any help :-)

    2. Relevant equations


    3. The attempt at a solution
    -- Is the correct approach to find the sum of: all x components, and y components of the forces, (65N, 35N, and weight)?
     
    Last edited by a moderator: Nov 4, 2016
  2. jcsd
  3. Nov 4, 2016 #2
    I get 1.0577 m/s^2
     
  4. Nov 4, 2016 #3
    Thank you! :-)
     
  5. Nov 4, 2016 #4

    haruspex

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    I confirm that.
    @SDTK, if that is not what you got, please post your detailed working. Most likely you rounded some intermediate answer, keeping insufficient precision.
     
  6. Nov 4, 2016 #5
    thank you
     
  7. Nov 4, 2016 #6
    I also looked at doing vector addition, but it doesn't give a right triangle.

    upload_2016-11-4_20-39-37.png
     

    Attached Files:

  8. Nov 5, 2016 #7

    I also looked at vector addition, but I would not get a right triangle. I'm thinking that vector addition would not be appropriate--- Is that correct?

    upload_2016-11-4_20-39-37-png.108478.png
     
  9. Nov 5, 2016 #8
    SDTK, you wrote:
    ΣFx = sin(32)(35) + cos(59)(65) = max
    620.19 = max

    How did you get from the first line to the second line?

    Edit: It looks like you may have multiplied those 2 terms, rather than add them.
     
    Last edited: Nov 5, 2016
  10. Nov 5, 2016 #9

    gneill

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    The problem states that two forces act on the object. Gravity was not one of them.

    Cartesian vector components add in quadrature: square root of the sum of the squares.
     
  11. Nov 5, 2016 #10
    Not sure what I did, thank you! for pointing it out :-)
     
  12. Nov 5, 2016 #11
    gneill
    Thank you!
    Is this what you are telling me? (work shown below)
    (I'm concerned because I get 0.99 m/s^2, which is different from the answers of @haruspex 1.0577m/s^2, @ObjectivelyRational 1.0577m/s^2, and the answer given with the problem 1.1m/s^2.)
    Thanks

    upload_2016-11-5_18-30-46.png
     
  13. Nov 5, 2016 #12
    Find x component of the net force, then find it in Y direction. THESE are of course 90 degrees wrt each other. Calculate the magnitude of the total net force. then calculate a from F and m.
     
  14. Nov 5, 2016 #13

    gneill

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    No. The given forces won't in general form a right angle triangle (and they don't in this case). You need to sum their components.

    What I was getting at was, once you've summed the individual x and y components of the vectors to form the resultant vector's components, then the magnitude of that resultant is found by summing its x and y components. You still need to find the components of the resultant vector using the method that you used in post #6 (only forget the gravitational force since it plays no role here).
     
  15. Nov 5, 2016 #14

    haruspex

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    Isn't it more direct just to draw the vector triangle and apply the cosine rule?
     
  16. Nov 5, 2016 #15

    haruspex

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    In the triangle of forces you drew, what is the angle between the 65N and the 35N? Just a bit of elementary geometry needed.
     
  17. Nov 5, 2016 #16

    gneill

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    Yes, if you have the angle between the forces. In general though I expect a student to be faced with summing multiple forces to find a resultant. Doing them one at a time via the cosine rule could be tedious.
     
  18. Nov 5, 2016 #17
    Thank you!
    I found
    f net x to be 67.92N sin(32)35N + cos(59)65N
    f net y to be -26.03 N cos(32)35N - sin(59)65N

    If I draw a vector, magnitude of 68N on the positive x axis, and a vector magnitude 26N on the negative y axis,... then draw a vector between the two that meets the segment representing the vector sum, is that a vector representing the total net force?

    upload_2016-11-5_19-57-22.png




     
  19. Nov 5, 2016 #18
    63 degrees
     
  20. Nov 5, 2016 #19

    gneill

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    Cartesian components of a vector form a rectangle with the sides paralleling the coordinate axes and the vector itself forming the diagonal:

    upload_2016-11-5_20-19-40.png

    Note that the components ##f_x## and ##f_y## of vector ##f## are at right angles to each other and are parallel to the axes.

    The magnitude of f is ##|f| = \sqrt{f_x^2 + f_y^2}##

    I don't agree with the value of the x_component that you calculated in your post #17. Can you give more details of that calculation? The y_component looks okay.
     
  21. Nov 5, 2016 #20

    haruspex

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    Ok, so you have a choice of approaches. For a simple resultant of two vectors, you can use the angle between them (when drawn in the nose-to-tail format of the force triangle) and apply the cosine rule, or for multiple applied forces you can reduce each to its X and Y components and find the sum along each axis.
     
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