Pushing a car & rolling resistance of tires

AI Thread Summary
Rolling resistance significantly impacts a car's motion when pushed or rolling to a stop, with a coefficient of 0.03 for normal tires on asphalt resulting in a force of 294 N needed to initiate movement for a 1,000 kg car. To get the car rolling, a force greater than 294 N must be applied. If the car is already in motion at 3 m/s, it would take approximately 10.2 seconds to come to a stop due to this rolling resistance. The discussion highlights the difference between static and dynamic friction, emphasizing that higher static friction makes it harder to start movement compared to maintaining it. Understanding these forces is crucial for accurately predicting vehicle motion.
Hope Lansing
Messages
7
Reaction score
0
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.
 
Physics news on Phys.org
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:
This is from Griffiths' Electrodynamics, 3rd edition, page 352. I am trying to calculate the divergence of the Maxwell stress tensor. The tensor is given as ##T_{ij} =\epsilon_0 (E_iE_j-\frac 1 2 \delta_{ij} E^2)+\frac 1 {\mu_0}(B_iB_j-\frac 1 2 \delta_{ij} B^2)##. To make things easier, I just want to focus on the part with the electrical field, i.e. I want to find the divergence of ##E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2##. In matrix form, this tensor should look like this...
Thread 'Applying the Gauss (1835) formula for force between 2 parallel DC currents'
Please can anyone either:- (1) point me to a derivation of the perpendicular force (Fy) between two very long parallel wires carrying steady currents utilising the formula of Gauss for the force F along the line r between 2 charges? Or alternatively (2) point out where I have gone wrong in my method? I am having problems with calculating the direction and magnitude of the force as expected from modern (Biot-Savart-Maxwell-Lorentz) formula. Here is my method and results so far:- This...
Back
Top