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Frictional force by tires

  1. Jul 21, 2011 #1
    [1. The problem statement, all variables and given/known data

    An SUV pulls a trailer carrying a boat. The mass of the SUV is 2460 kg and the mass of the boat and trailer is 490 kg. The SUV starts from rest (velocity = 0) and accelerates for 10 seconds to final velocity 12 m/s; i.e., a = 1.2 m/s2.

    What is the frictional force exerted by the road on the tires of the SUV?


    2. Relevant equations

    F=ma


    3. The attempt at a solution

    so I used
    .5mv^2=[itex]\mu[/itex]mgd

    d=.5at^2
    d=60m

    thus [itex]\mu[/itex]=0.1223

    so then Ff on the road by the tires =M_suv*0.1223=300.92 N, but this is not correct, where am I going wrong this seems so simple
     
  2. jcsd
  3. Jul 21, 2011 #2
    Why are you equating the kinetic energy and energy dissipated by friction? No kinetic energy is being dissipated by friction here. Use Newton's second law.
     
  4. Jul 21, 2011 #3
    ok yeah I thought that was wrong, so then the net force on the SUV is 2950N, but I am confused as to where to go from here.
     
  5. Jul 21, 2011 #4
    Is this all the information you have? You can express the force on the SUV in terms of the frictional force and a forward force.
     
  6. Jul 21, 2011 #5
    This is all the info I have yes. So could you nudge me in the right direction at least, because at the surface this seems simple so I think I'm over looking something.
     
  7. Jul 21, 2011 #6
    When the SUV obtains it's final velocity, all forces on it must be balanced. So you need to find the forces on the SUV.
     
  8. Jul 21, 2011 #7
    so the forces needed here are just the forward force and frictional, so F=ma however how do you find the Forces with the info given, I would only think the net force can be found
     
  9. Jul 21, 2011 #8
    True, you need the forward force. I might be missing something here...let me think.
     
  10. Jul 21, 2011 #9
    Do you have to find the frictional force or the coefficient of friction?

    EDIT: It seems your original method is correct. From conservation of energy-

    0=(Kinetic Energy)+(Work due to friction)

    I think you forgot a g when calculating the frictional force. Instead of taking that long route you could just have divided the kinetic energy by the distance. F=ma isn't really applicable since you lack knowledge of all forces.
     
    Last edited: Jul 21, 2011
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