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Force from car

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    a car of mass 1500kg and decelerating 15ms-2. what is the total force of car on road from all tires?

    2. Relevant equations

    W=mg
    F=ma

    3. The attempt at a solution
    I'm not sure if the force is simply mg or if it is ma in this case. There is also a frictional force but I don't know if it applies here.
    mg=15000?
    ma=22500?
     
  2. jcsd
  3. Oct 31, 2011 #2

    Redbelly98

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    Try drawing a free body diagram of the car. What are the forces that are present, and in what direction are they?

    Also, in what direction (horizontal or vertical) does the car decelerate? Which force or forces are acting in that direction?
     
  4. Oct 31, 2011 #3
    well theres the weight of the car downwards, the normal force from ground probably equal to that since car isn't moving upwards. then in the horizontal there is each wheel friction in direction of car moving
     
  5. Oct 31, 2011 #4

    Redbelly98

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    Yes: gravity downwards, normal force from ground upwards, and friction force horizontally. And two of those forces are exerted by the ground, so the car exerts equal-but-opposite forces on the ground.
     
  6. Oct 31, 2011 #5
    so would that be mg+ma because opposite of gravity would be mg and the force opposite friction seems to be ma or is there a separate y force and x force
     
  7. Oct 31, 2011 #6

    cmb

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    Difficult to tell. I expect this is not the answer the question is looking for, but it means you are in the process of having an accident so your tyres might not be generating any force at all!

    I do not believe your tyres can generate 1.5g (I think the limit for road cars is around 1.1g with latest tech), unless you are in a race car with aerodynamic downforce, but then the car wouldn't weight more than 605kg. (due to Formula 1 construction rules 605kg or Formula 3 540kg.)
     
  8. Oct 31, 2011 #7

    Redbelly98

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    The two forces act in different directions, x or y, so we can't simply add them. But yes, they turn out to be equivalent to mg and ma -- I guess this is what you were saying in your first post. But technically you need to say in what direction those forces are acting. So for the 15,000 N force you calculated from mg, in what direction does it act? Likewise for the 22,500 N force you calculated from ma.

    While you bring up a valid point, since this is intro physics it's probably best to keep things simple; neglect air resistance, and use the numbers given. It's likely that the question author was not thinking through about the implications of a 1.5g acceleration due to friction -- or the 15 m/s2 was a simple typo.
     
    Last edited: Oct 31, 2011
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