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Lawn mower force physics help

  1. Jul 7, 2008 #1
    A person pushes a 16-kg with constant speed with a force of 80N directed along the handle. Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds (assuming the same retarding force).

    k well so far i got the acceleration, which is 0.6m/s/s. I also got FN which is FN= F(y)+mg, but i dont know if thats important. My retarding force is 56.6N does that mean my F(push)= Fretarding?
     
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  3. Jul 7, 2008 #2

    G01

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    Re: Forces

    How did you find all these values? Can you show your work?
     
  4. Jul 7, 2008 #3

    dynamicsolo

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    Re: Forces

    Let's go back to the first part. When you are pushing the lawn mower at constant speed, what is the net force on it? Why would the speed be constant if you are applying an 80-N force to it? What does this tell you about the retarding force? (We will use this same retarding force for the main question.)
     
  5. Jul 7, 2008 #4
    Re: Forces

    Wouldn't the Net force = to 0 since acceleration = 0?, and the speed would be constant if you are applying 80 - N force because the retarding force = 80N i think >.<
     
  6. Jul 7, 2008 #5

    dynamicsolo

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    Re: Forces

    All right! Now you have found the acceleration for the lawn mower for the main question. What must the net force on it be this time? If the retarding force is unchanged, what force must you apply to it to obtain that net force?
     
  7. Jul 7, 2008 #6
    Re: Forces

    Well, my accerlation rate is 0.6m/s/s so i went Fnet=ma and got (16)(0.6)= 9.6N +56.6N
    which is the retarding force

    so basically my equation looks like this : Fnet = ma + F(retarding)
     
  8. Jul 7, 2008 #7

    dynamicsolo

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    Re: Forces

    I thought you found that the retarding force is 80 N?

    Consider it this way. Your applied force is being treated as if it is entirely in the forward direction. The retarding force points backwards. So the net force will be

    F_net = F_applied - F_retarding = ma .

    Your result for the net force is correct, so F_applied = ?
     
  9. Jul 7, 2008 #8
    Re: Forces

    Hmm, so this is what i would get

    9.6 = F_applied - 56.6 = ma?

    whats confusing me is the "=ma" part
     
  10. Jul 8, 2008 #9

    dynamicsolo

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    Re: Forces

    Back in post #6, you yourself calculated the net force on the mower as F_net = ma = (16 kg)·(0.6 m/(sec^2)) = 9.6 N. Then you said in post #4 that the retarding force is 80 N. (Where are you getting 56.6 N from?)
     
  11. Jul 8, 2008 #10
    Re: Forces

    56.6N came from my horizontal component (Cos 45)(80)
     
  12. Jul 8, 2008 #11

    dynamicsolo

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    Re: Forces

    Oh, I was wondering about that "force along the handle"... There's no mention in your original post of the 45º angle of the mover handle. (That's kind of an important detail. This is why the helpers on this Forum ask you to show your calculation up front...)

    All right, so the horizontal component of the applied force is 56.6 N forward, so the horizontal retarding force is 56.6 N backward. The net force is F_net = ma = 9.6 N forward, so the equation you had in post #8 is correct:

    9.6 N = F_applied - 56.6 N .
     
  13. Jul 8, 2008 #12
    Re: Forces

    K well i hope this is correct, cause the book saids its 94 N.
     
  14. Jul 8, 2008 #13

    dynamicsolo

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    Re: Forces

    OK, right, now that we have the 45º involved in this, the horizontal component of the applied force is 9.6 + 56.6 N = 66.2 N.

    So this now means that

    (F_applied)·(cos 45º) = 66.2 N .

    That should do it.

    (Please be careful in future not to leave out details of a problem. Sometimes they matter a lot.)
     
  15. Jul 8, 2008 #14
    Re: Forces

    Alright, I finally reached the answer, using pythag thnx alot for the help, this question has been puzzling me since 6pm today >.< I finally feel relived. Once again thnx a bunch
     
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