### Re: What friction is experienced by a car when its brakes are off?

Date: Mon Jan 8 16:14:29 2007
Posted By: Joel Chapman, Undergraduate, Mechanical Engineering, NC State
Area of science: Physics
ID: 1167608809.Ph
Message:
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Reference:
Hibbeler, RC "Engineering Mechanics:  Statics", Tenth Edition p424-427

For a vehicle which has the brakes disengaged and the clutch out, the
things which will resist motion are friction within the axles and any parts
which the axles touch, air resistance, and rolling resistance.  I think
that since you're traveling at relatively low speeds on this conveyor belt,
you can neglect air resistance for the most part.  Air resistance is a
function of the speed of the object on the conveyor, and it would also act
to slow the conveyor belt itself if the plane is not slipping or rolling,
and these are complex things to deal with which are not really causing that
much of a change to begin with, so just neglect air resistance.

So, I think you should focus on the friction acting on the axles, and the
rolling resistance.  Friction within the axles is denoted as "mu" (looks
like a u), and for your case with wheels, the moment (or torque) required
to overcome this friction is R*r*u, where R is the radius of your tire, r
is the radius of the axle, and u is "mu", the friction on the axle.  Once
you exceed this torque, however, the frictional force does not increase, so
running the conveyor belt faster will not increase your speed.  However, if
you do not exceed this torque, the plane will not roll at all (assuming
that the tires do not slip on the conveyor).

Rolling resistance also comes into play.  Since the tires deform, there are
forces which are required to deform the tires (force of deformation), and
also a force which is caused by the tire expanding back into shape (force
of restoration).  However, the force of deformation is always greater (yes,
always) than the force of restoration, so this causes a resistance to
motion.  The force acts horizontally (that is, tangent to the surface and
tangent to the bottom of the tire), and the force required to overcome it
is approximately W*a/r  where W is the weight of the plane, a is the
horizontal distance between where the tire first deforms and the center of
the tire, and r is the natural outer radius of the tire.  Once this
required force is exceeded, however, the resistance does not increase.  It
acts just like friction in that it will match any force up to the required
force to overcome it, but will not increase past that required force.  Do
not forget that for an object with multiple tires/wheels, you must treat
each wheel individually due to the fact that there are different weights
supported by those tires/wheels...

However, since these frictions are constant, try to get a general idea of
how big these forces are for the plane you are trying to build.  The one
that will be the determining factor is the larger one.  If the rolling
resistance is really high, even if the axles are frictionless, the wheels
still will not roll.  If there is no rolling resistance, but the friction
in the axles is really high, the wheels will not roll.  Find the larger of
the two frictions, find (or approximate) a set value, and then use that as
a constant number within the program.  If the force caused by the conveyor
belt is less than this number, there will be no rolling of the plane's
wheels.  If the force is greater than this number, the force put out by
these resistances will still not exceed that number.

Note that F=m*a and T=f*d.  So, when you're trying to figure out how fast
the plane will accelerate, you must take the force (resistance) and the
mass of the plane into account.  If you have a rolling resistance, the
required TORQUE is f*d (that's the force times the radius of the wheel, in
this case).  Torque is not simply force, so do not forget that!

Also, technically, the friction the axles would be higher when the wheels
are not rolling than when they are rolling.  So, up until the wheels roll,
you have a high resistance, but once the wheels roll, the friction force
actually drops, and the force output by that friction is actually lower.
You  may want to take this into account in your program!

I hope this helps you out!

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