Mechanical Engineering: Ch 11: Friction (46 of 47) Rolling Friction: Example 2

In this video I will calculate b=?, flatting, of a tire rolling down a 2% incline at constant speed.

Next video in this series can be seen at:

[swimcoach]

In this series I've been somewhat frustrated by the many uses of the 'close enough' method of simplifying equations. But it seems to work and if it didn't planes would be falling out of the sky and trains jumping the tracks. And actually calculus itself is based on the 'close enough' philosophy, without which we would have no calculus if we had insisted on absolute accuracy. This is one of the concepts that you can explore in this venue that a college instructor would not have the time to get into. I really enjoy exploring the various engineering disciplines with you that I didn't have time for in college. Never retire!

[malekstudies]

Awesome lesson as always!
I have a question. In one of the questions posted in the comments, by "burhan muhyddin", they asked if the reaction force =mgcos(theta) is different from the normal force. You said that the normal force is a reaction force.

If thats the case, then why on a flat surface is the reaction force not equal to the normal force, but equal to sqrt(Frictional Force^2 + Normal Force^2)

[malekstudies]

I feel like The reactionary force is simply equal to the weight of the object... correct me if im wrong

My proof:
If we imagine the previous example ( flat, non-inclined surface) . There were two important forces, the weight of the ball and the force the ball is being pushed to overcome the friction. The reactionary force will be the resultant of the opposite pairs of the two forces (Newtons 3rd law, normal force and the frictional force).
Taking that into consideration, if we look at the ball on an incline, the two "main forces" in this case would be the parallel and perpendicular components of the weight. Then the reactionary force will be the resultant force of the opposite pairs of the two forces (newtons 3rd law, the normal force and the frictional force).

taking that to consideration, and as stated in the problem that the ball will be moving at a constant speed, the frictional force must be equal to the parallel component of the weight, mgsin(theta). and the normal force = mgcos(theta)
Those two make a right angle, with the hypotenuse being the reactionary force. So naturally, that'd mean that the reactionary force = mg = weight of the object.
it also makes sense cause if you draw the reactionary force, it will act in the exact opposite direction of the weight...

Sorry to trouble you with all of this sir, thank you as always for these amazing videos

[TheNaaaaag]

Hi professor, can we use force friction formula to find b

[exodia]

why don't you use that sir?
T=I*a/r, t=Ffr*r, i=0.5mr^2 then Ffr=0.5*m*a
Fnet=m*a where Fnet=(m*g*sintheta ) - (0.5*m*a) then g*sintheta = 3/2 * a So a=0.130m/sec^2
Ff*r=Wcos(phi) * B where Ff=0.5 m*a, cos(phi)=1 and W=mg so B=(0.5*a*r)/g SO
@Michel van Biezen

[PSM]

Hi Michael, I couldn’t relate R = W cos(phi) as shown on the diagram…. or cos(phi) = R/W….as both R and W are acting through the centre?

[hareesvarman]

Is rolling friction just a torque caused by the shift in normal contact force or is it a force acting down the incline. I'm so confused 😕

[camurgo]

the two angles should be of the same value, no?

[burhanmuhyiddin]

Hi, thank you very much for your effort and lessons. However, in this video I have a question. You replaced 'R'(reaction force) with m*g*cos(fi). Isn't m*g*cos(fi) is 'N'(normal foce)? Aren't they different? Thank you :)

[umarpatel]

Why am I getting chicken wing ads for a mechanical engineering video???? lmao