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Determining unknown Force

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A 3.7 kg mass has a constant acceleration from an initial velocity of 2.33 ihat m/s to a final velocity of (18.9 ihat - 10.7 jhat) m/s in 2.71 s. Two forces act on the mass: gravity and an unknown applied force. Determine the acceleration and unknown force.


    2. Relevant equations


    3. The attempt at a solution
    For the acceleration i managed to get 16.25 m/s, and for the force of gravity it is -36.26 m/s.


     
  2. jcsd
  3. Feb 16, 2015 #2

    BvU

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    Hello Jose, welcome to PF :smile:

    Can you show your attempt ? The numerical value of the answer isn't as important as how you found it.
    And what equation would you like to use to determine the applied force ? I assume that's where you run into problems ? Where precisely ?
     
  4. Feb 16, 2015 #3
    I believe the a I got the first time is wrong.
    a = (V-Vi)/t, so a = (6.1ihat - 3.95 jhat) m/s

    The problem is on on incline starting at 0,0 and having an angle below the x-line.

    So the force on the x axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N

    22.6 +14.6 = 37.2 N

    37.2N - 36.26 N = 0.94 N
    0.94 N = unknown force
     
  5. Feb 16, 2015 #4

    lightgrav

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    gravity around here, all by itself, causes acceleration 9.8 m/s2 ... that's 32 ft/s2 , if anybody still uses feet.
    So the Force by gravity is 36.26 Newtons ... watch your units.
     
  6. Feb 16, 2015 #5

    jbriggs444

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    I agree with the corrected numbers. Thank you for showing your work.

    22.6 N in one direction plus 14.6 N in the direction at right angles equals what?

    Same problem. 37.2 N in what direction? Minus 36.26 N in what direction?
     
  7. Feb 16, 2015 #6
    What stumps me is the how can i do this if there is no angle.
     
  8. Feb 16, 2015 #7

    jbriggs444

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    You have referred to an "incline", an "x-axis" and a point at "0,0". Possibly there is a diagram that has not been shared? A reasonable assumption would be that gravity acts in the -y direction. You do not need an angle if you consider gravity and the y component of the unknown force together and the x component of the unknown force separately.
     
  9. Feb 16, 2015 #8

    haruspex

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    The problem statement is still unclear to me.
    If it is 'on an incline', there would be a normal force from the surface. Is that the unknown force, or considered a consequence of the gravity?
    Please state the entire question word for word.
     
  10. Feb 16, 2015 #9
    The problem doesn't mention an incline i just assumed it was on an incline, because of its final velocity.
     
  11. Feb 16, 2015 #10

    BvU

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    I subscribe to the a = (6.1ihat - 3.95 jhat) m/s, and also to force on the x axis is 3.7 kg times 6.1 = 22.6 N, and the y force 3.7kg time 3.95 = 14.6 N
    but from there you start losing me:

    If you go one step to your right (##\hat\imath##) and one step forward (##\hat\jmath##), the result is that your displacement is √2 steps at 45 degrees.
    It's called vector addition. The two forces have to be added as vectors -- if you want to add them at all. But in this exercise you don't want that. They give velocities as ##\ ( a\; \hat\imath + b\; \hat\jmath\ )\ ##, so an answer in the same terms should be OK.​


    The gravity force that works in the ##\hat\jmath## direction is ##\ m\;\vec g\ = 3.7 * -9.81\; \hat\jmath\ ##. If the net force is ##\ -14.6\; \hat\jmath\ ## N (I hope you didn't forget the minus sign ?) then the applied force in the y-direction must be quite positive.

    This whole escapade can take place on an inclined plane or in thin air -- if the exercise tekst says "Two forces act on the mass: gravity and an unknown applied force" then that is enough to calculate that one unknown force. Which is all the exercise wants you to determine.
     
  12. Feb 16, 2015 #11
    ok, so an answer for this can be given as vectors? (22.60ihat N -14.6 jhat N) ?
     
  13. Feb 16, 2015 #12

    BvU

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    What you describe here is the net force. That isn't the answer for this exercise. I think the exercise asks for the second force.
     
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