Question

Alright, big test tomorrow. My teacher always asks about with a given coefficient of friction, what is the range of speed a car can travel on a banked curve. I just dont understand how to put it into the equtions.

Answer 1

Firstly, you must consider the forces invovled in the system. Let define the system to be the curved road, and car. Say the bank is at angle (\theta), the car weights (m) and the coefficient of friction is (\mu). Right so the first force that comes to mind is gravity.
\[\vec{F_g}=-Mg\hat{z}\]Since the force of gravity is straight down.
Then I think of the normal force next.
\[\vec{F_n}=\vec{F_\perp}\]We say the normal force is "F perp", the contact force that the car exerts on bank.
\[\vec{F_f}=-\mu \vec{F_\perp}=-\mu \vec{F_n}\]The friction is defined by this formula.
So, now we know everything about the forces in this problem. STOP if you think there are more forces and think about it. Radial acceleration is not a force. We actually have everything we need already!
Now we simply have to use math and newtons 2nd law.
\[-----\]
Use the math to arrange for "F perp" and thus the friction force. Now if we down from above of this system we see a road with a car travelling (v) on it. Up to this point we havn't said anything about the way the road turns. That's because we have left it general. You should now have all the forces, and you've used math to figure out what force of friction is. You could also change the coordinate system of the car right? You could break the friction into x,y components "up and down and left and right" or you could make them x',y' "up and down the slope and perpendicular to the slope". Infact, "F perp" is in the y' direction right?
Now, we stop being general and turn our attention to a problem.
We know,
\[\sum F=ma\]And remember we are looking from above down.
\[---\]If the car travels in a straight line and constant speed then \[\sum F=m\cdot0=0\] then we could plug in our forces and see that the car can travel as fast as it wants. For if the car travels straight, the forces are the same as if it was sitting still (a=0) and if the friction can support the car's weight then it dosen't matter how fast it travels.
\[---\]If the car travels in a circular motion then\[\sum F=ma_r=m(v^2/r)\]I presume this is the question you are after. If you could answer the previous question when the car travels in a straight line then you can answer this one. Remember, you can break the forces into x,y components instead of x',y'.

Good luck