1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pushing a car & rolling resistance of tires

  1. Oct 12, 2012 #1
    I just want to know if I am understanding this correctly.

    I am wondering what effect the rolling resistance has on the motion of a car when pushed, or when rolling to a stop. The mass of the car is 1,000kg.

    I looked up on a table to find the rolling resistance coefficient of a normal car tire on asphalt and got .03 so that is the value I am using.

    Frr = .03 * 1,000kg * 9.8m/s² = 294 N

    a = 294 N / 1,000 kg = .294m/s²

    Question 1 :

    Does this mean that if a 1,000kg car is at rest in nuetral, and I want to get that car rolling, I would need to apply a force greater than 294 N before the car would even begin to roll?

    Question 2 :

    If this car was already in motion at say, 3m/s, would it come to a stop in about 10.2 seconds? (t = 3m/s / .294m/s² = 10.2s)

    I am ignoring any other sources of friction(ball bearings, etc), and the asphalt is perfectly level.

    Any help would be greatly appreciated. Thank you.
     
  2. jcsd
  3. Oct 12, 2012 #2
    That is a reasonable number for rolling resistance, which relates to the coefficient of dynamic friction. It would take more to get the car started.

    Think of a locomotive, which could never start a train of cars at the same time because the coefficient of static friction is too high. But because it only has to start one car at a time, it has no trouble keeping them moving.
     
    Last edited: Oct 12, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pushing a car & rolling resistance of tires
  1. Rolling Resistance (Replies: 11)

  2. Car roll over (Replies: 7)

Loading...