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Magnitude of forces

  1. Jan 4, 2006 #1
    A 2.7 kg mass accelerates at 7 m/s^2 in a
    direction 28 degrees north of east. One of the two
    forces acting on the mass has a magnitude of
    12.6 N and is directed north.
    Determine the magnitude of the second
    force. Answer in units of N.

    I did F2= sqr rt (ma^2+ F1^2)

    = 22.715 N

    This is apparently incorrect. This seems too easy and yet I cannot figure it out. I'm assuming F2 is in the positive "x" direction w/o a "y" component.

    Any help appreciated.
  2. jcsd
  3. Jan 4, 2006 #2
    I think your formula should be [itex]F_2 = \sqrt{(ma)^2 - F_1^2}[/itex] (remember [tex]c^2 = a^2 + b^2[/tex]) to get 14.09 N.
  4. Jan 9, 2006 #3
    This is not the answer... tried it and got it wrong :(

    I also tried splitting up the the acceleration into x and y components and trying to solve it F1x+F2x = MAx; F1y+F2y=MAy F1x=0 F1y=12.6N

    I solved it and then did sqr rt(F2x^2+F2y^2)

    = 16.2662N

    I only have 1 try left so it would be great if someone can help me out, please.
  5. Jan 9, 2006 #4
    How come the force acting north has a magnitude of 12.6N? It should be [tex]masin(\theta)=F_N[/tex] which is 8.87N. Was 12.6 given?
  6. Jan 9, 2006 #5
    The force you're trying to find must also have an y-component, otherwise cscotts formula would've worked.
  7. Jan 9, 2006 #6
    The question states (as found in the original post):

    "One of the two forces acting on the mass has a magnitude of 12.6 N and is directed north."

    So assume the sole direction of F1 is North, or in the positive y direction.

    Any ideas as to an equation or problem-solving technique to solve this?
  8. Jan 9, 2006 #7
    To make sure I have this right:

    -The net force (magnitude) of the two components is the [tex]7 \frac{m}{s^2} * 2.7kg = 18.9N[/tex]

    -This magnitude has an angle of 28º north of east

    -The north acting component of the net force is 12.6N

    Unless physics isn't allowed to use vectors anymore, this is wrong.
  9. Jan 9, 2006 #8
    hmm... i don't know. I cut and pasted the question -- maybe the question is wrong?
  10. Jan 9, 2006 #9


    User Avatar
    Homework Helper

    ma cos28 = (F2) cos@
    ma sin28 = (F1) + (F2) sin@

    @ is the angle F2 makes north of east

    eliminate @ to get F2

  11. Jan 9, 2006 #10

    ma cos28 / cos@ = ma sin28 - F1 / sin@

    tan@ = ma sin28 - F1/ ma cos28

    @ = tan -1 (ma sin28 - F1/ ma cos28)

    F2x = 17.09883283
    F2y = -17.09883283

    F2 = sqr rt ( 17.09883283^2 * 2)

    = 24.18140129 N?

    This comes up as wrong. I dont know -- maybe I made a math mistake.

    Thanks for the help though.
  12. Jan 9, 2006 #11


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    Homework Helper

    solve in this way

    F2 cos@ = macos28 = 16.688N
    F2 sin@ = masin28 - F1 = -3.727 N

    squaring and adding F2 = 17.1 N (please check the calculations)

    Who told you that the two forces are perpendicular or F2 is towards east?

  13. Jan 9, 2006 #12

    I think that's it. Thanks for the help!
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