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Homework Help: Find the Net Force and Acceleration Question

  1. Jun 20, 2010 #1
    1. The problem statement, all variables and given/known data
    The two forces F1 and F2 (shown in the figure) (i attached the figure hopefully I did it correctly) act on a 27.0 kg object on a frictionless tabletop. If F1=10.2N and F2= 16.0N, find the net force on the object and its acceleration for a and b



    2. Relevant equations
    F=ma


    3. The attempt at a solution
    If I'm approaching this correctly I believe its a simple vector addition problem?

    for a I have:
    x= -10.2
    y= -16.0
    this vector is = SQRT(-10.22 + -16.02) = 18.97N

    F=ma
    19.97 = 27.0kg (a)
    a= 0.70 m/s2

    For b I have:
    F2: x=0 y=16
    F1: x=cos30 (10.2) = 8.8 y= sin30 (10.2) = -5.1

    adding the x's together = 8.8
    adding the y's together = 10.9

    finding the vector= SQRT(8.82+ 10.92) = 14.0N
    F=ma
    14.0 = 27.0 kg (a)
    a= 0.52 m/s2

    Does this look correct?? Or did I even approach this correctly??

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 21, 2010 #2

    kuruman

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    It looks correct, but I think the problem really wants you to find the x and y components of the acceleration, not just its magnitude. Or, if you prefer to calculate magnitudes, then you must also specify a direction to define a vector unambiguously.
     
  4. Jun 21, 2010 #3
    Hi Kuruman,
    Thanks for your help on this problem. So if I decide to find the magnitude of a vector I must also specify the direction? If I'm not mistaken is this done by taking the inverse tangent of opposite/adjacent?

    for the first case tan-1(-16/-10.2) = 57.5o

    and for the second case tan-1(10.9/8.8) = 51o


    So for the part of the question that it asks for the net force is it safe to say that it is:

    case 1 = 18.97N, 57.5o

    case 2 = 14.0N, 51o

    thanks.
     
  5. Jun 22, 2010 #4

    kuruman

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    You still need to be a little careful and define the axis and direction with respect to which the angle is measured. The most common convention is "clockwise with respect to the positive x-axis". Personally and unless the problem states otherwise, I prefer to define vectors in terms of their components not magnitudes and angles.
     
  6. Jun 22, 2010 #5
    So if I'm following you correctly:

    The components of the first situation is:
    x= -10.2
    y= -16.0
    Theata = 51o (when doing this I'm not entirely sure from what axis it is, I did tan-1(-16/-10.2)

    The components of the second situation:
    x = 8.8
    y = 10.9
    theta = 51o (again I dont know from which axis this is from, I did tan-1(10.9/8.8)

    Thanks for helping me with this!
     
  7. Jun 22, 2010 #6

    kuruman

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    Part (a)
    To find the angle you need a right triangle. It is the right triangle such that

    tanθ = opposite/adjacent.

    Draw yourself a right triangle such that one right side is 10.2 units to the left and the other right side is 16.0 unit down. To do this go to the tip of F1 and draw a line segment of 16.0 units straight down. Connect the tip of that segment to the origin. There is your right triangle. Now answer the following questions

    1. What is the angle that the hypotenuse makes with respect to the negative x-axis?
    2. What is the angle that the hypotenuse makes with respect to the positive x-axis?
    3. What does the 51o that you found have to do with the above two angles?

    If you can do this, then proceed the same way with part (b). In part (b) use the components that you found to draw another right triangle.
     
  8. Jun 22, 2010 #7

    Ok, so I first drew a line with magnitude 10.2 going to the right (east), then I drew a line down (south) from the tip of east line. From there I drew a line from the tail of the east line to the head of the south line (tail to tip method). I have 10.2 as my x and 16.0 as my y. my hypotenuse has a magnitude of 18.97N pointing south east. my angle of the triangle is tan-1= (16/10.2) = 57.5o

    1. The hypotenuse makes a 57.5o with respect to the negative x axis.
    2. the hypotenuse makes a 237.5o with respect to the positive x axis.
    3. The 51o I initially got was probably a calculator error on my end, the angle of the line should be what is above in 1 and 2..

    hopefully I'm now getting this, sorry to drag you through this with me, thanks again for your help!

    If this is correct I know how to do it for the second case.
     
  9. Jun 23, 2010 #8

    kuruman

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    All of the above is correct. I think you understand what is going on. Proceed with part (b). By the way, in my earlier posting I was incorrect. The convention is "anti-clockwise with respect to the positive x-axis", which is what you calculated anyway. Sorry about the confusion.
     
  10. Jun 23, 2010 #9
    Awesome! I think this thread has helped me understand more than just the problem but on how to calculate the direction of a vector within a plane. Thank you so much for all your help, its greatly appreciated!
     
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