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Net force of object indicated in following diagram

  1. May 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the net force acting on the object indicated in the following diagrams. Show your work.

    1)
    upload_2016-5-4_15-20-43.png

    2)
    upload_2016-5-4_15-21-18.png


    2. Relevant equations
    upload_2016-5-4_15-22-33.png

    upload_2016-5-4_15-22-57.png

    3. The attempt at a solution

    1)
    8.0 N (north) + 10.0 N (south)
    - 8.0 N (south) + 10.0 N (south)
    = 2.0 N (south)

    c= [ (2.0 N) 2 + (17.0 N) 2 - 2 (2.0 N) x (17.0 N) cos 45° ] 1/2
    c= 15.6 N

    sinA/2.0N = sin45°/15.6N

    A= 5.2°
    A= 45° - 5.2° = 39.8°

    Fnet = 15.6 N [N 39.8° W]

    2)
    c= [ (12 N) 2 + (15 N)2 - 2 (12 N) x (15 N) cos124°]
    c= 23.9 N

    sinA/15N = sin124°/ 23.9 N

    A = 31°

    Fnet = 23.9 N [ N 31° E]

    is this correct?
     
  2. jcsd
  3. May 4, 2016 #2

    haruspex

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    I agree with both magnitudes.

    In 1, you do not show a diagram for the triangle of forces, so I'm not sure what your angle A is. You seem to have used it inconsistently, leading to the wrong answer for direction of force.

    In 2, the angle you found is not the one you want. It seems to be the angle between the resultant and the 12N force.

    Can you post your force triangle diagrams?
     
  4. May 6, 2016 #3
    1)
    Here is the diagram for 1) :

    upload_2016-5-6_20-13-21.png
    I corrected my answer for 1) :

    8.0 N (north) + 10.0 N (south)

    -8.0 N (south) + 10.0 N (south)

    = 2.0 N (south)

    c = [(2.0 N)2 + (17.0 N)2- 2 (2.0 N) x (17.0 N) cos 45°]1/2

    c = 15.6 N

    sinA/a = sinB/b = sinC/c

    sinA/2.0 N = sin 45°/15.6 N

    A = 5.2°

    45° - 5.2° = 39.8°

    Fnet = 16 N [ W 40° N]

    is this better?
     
    Last edited: May 6, 2016
  5. May 6, 2016 #4
    2)
    Here is the diagram for 2):
    upload_2016-5-6_20-26-14.png

    I also corrected my answer for 2) :

    c2 = a2 + b2 - 2abcosC

    c = (a2 + b2 - 2abcosC)1/2

    c = [(12 N)2 + (15 N)2- 2 (12 N) x (15 N) cos 124°]1/2

    c = 23.9 N

    sinA/a = sinB/b = sinC/c

    sinA/15 N = sin 124°/ 23.9 N

    A = 31°

    32° - 31° = 1°

    90° - 1° = 89°

    Fnet = 24.0 N [N 89° E] or 24.0 N [E 1° N]

    better?
     
    Last edited: May 6, 2016
  6. May 6, 2016 #5

    haruspex

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    Looks right.

    Edit: the angle might be a bit inaccurate. Retry it keeping more significant digits at each step.
    Doing it a slightly different way (using arctan) I got about 42 degrees.
     
    Last edited: May 6, 2016
  7. May 6, 2016 #6

    haruspex

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    yes, but again the angle could be a touch more accurate.
     
  8. May 6, 2016 #7
    Ok, I will keep a few more significant digits to make the angle more accurate. Anyway, thanks for the help :)
     
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