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Units of variable force?

  1. Nov 20, 2008 #1

    A particle experiences a variable force Fv = Fo * sin (cx) where x is displacement in meters. What units are the constants c and Fo in.

    2. Relevant equations

    3. The attempt at a solution
    Um ok....how the heck do i figure out what units they're in
    so i know my final units need to be kg*m/s^2 , and i already got my meters from the displacement x inside the sin function
    but the rest i can't figure out...
    my best guess is...one cycle of the sin function is 2pi radians...so delta x * c = 2pi rads
    delta x is just velocity so m/s*c = rad... does that make c in rad * s / m? cause thats not aunit ive heard of...
    if i do use that though, then Fo is just in the regular force units kg*m/s^2 ?
  2. jcsd
  3. Nov 20, 2008 #2
    I think that anything going into a (non inverted) trig function has to have units of degree's or radians. That is you can't have the unit meter inside that sin(). How would you get rid of the unit meter.

    Also F0 must have units of Kgms^-2, since it is a force. What does that tell you about the units inside the sin function?
  4. Nov 20, 2008 #3
    i guess the units of c are then 1/m to get rid of the meters inside the sin function ... so c would i guess be the frequency of the oscillation of the force per every meter travelled by the particle?
  5. Nov 20, 2008 #4
    Exactly right! :smile:
  6. Nov 20, 2008 #5
    Thank you very much!!! :) So I am workin on the rest of the question now....
    it says "at what position x does the force first reach its maximum value...your answer will be in terms of the constants Fo, c, and possibly some other numerical constants"....I don't get this part because it reaches its max value when x = Pi / 2c if im not mistaken...which doesn't use Fo anywhere...

    next, i gotta do "what is the particles velocity as it reaches the first peak of force in terms of Fo, Vo, m, and c

    the only thing i can think to do is consider the initial kinetic energy 0.5mvi^2...the velocity will be higher at the peak force, so kinetic energy will be higher too, and the change of kinetic energy is equal to the work done....and work done is area under force graph...so am i supposed to integrate sin(cx) from x = 0 to x = PI / 2c and it'll give me the additional work?
    so 0.5mvi^2 = 0.5 mvf^2 - Fo * cos (cx) / c from x=0 to x= pi/2c?
  7. Nov 20, 2008 #6
    This must be a vibrating/oscillating system, since for half a period the mass is being pushed to the right (say) and for the other half it's being pushed to the left. Since our force graph is sinusoidal the velocity graph must be cosine, so when the force is maximum I would have said the velocity is zero.

    I'm really sorry but my mechanics is a bit rusty/rubbish. Maybe one of the other people here can help?
  8. Nov 20, 2008 #7
    Your answer is correct. The question doesn't say that all the constants should be present in your answer :wink:

    You can also check your answer this way: You know that the position x where the force reaches its maximum should have units of distance, i.e. m. If your final expression involves Fo, then you have mass(kg) and time(s) units in the expression, which cannot be gotten rid of using only other numerical constants. (Note: This is only for quick verification, you should always solve the problem!)

    Well done. You have the right approach, but there is a sign error in your final expression. The change in kinetic energy = {Final energy} - {Initial energy}, and you equate this to the area under the force graph.


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