1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Magnitude of forces
  1. Magnitude of force (Replies: 9)

  2. Magnitude of force (Replies: 1)

  3. Magnitude of a Force (Replies: 4)

  4. Magnitude of Force (Replies: 2)

  5. Force and Magnitude (Replies: 4)