1. Sep 24, 2011

### farleyknight

1. The problem statement, all variables and given/known data

I've already answered the first part. The second part is what's giving me trouble.

2. Relevant equations

Here's how I solved #10:

Since the car is at rest, we're only worried about the force of friction and the parallel component of gravity:

F_f = F_g * sin(theta) = m * g * sin(theta)

Since F_f = mu_s * F_N = mu_s * F_g * cos(theta) = mu_s * m * g * cos(theta)

Then:

mu_s * m * g * cos(theta) = m * g * sin(theta)

Solving for theta, we get:

mu_s = tan(theta)

Or arctan(mu_s) = theta

3. The attempt at a solution

The second situation is different. We need to know the radius, and that is only involved when we have the centripetal force (F_C).

In this case, we are worried about having enough F_C to keep the car from veering off. Now the force of friction is equal, but going in the opposite direction (the same direction as the parallel component of gravity):

F_C * cos(theta) = F_f + F_g * sin(theta)

We already have F_f = mu_s * m * g * cos(theta), and F_g * sin(theta) = m * g * sin(theta). Also, since we're dealing with centripetal force, the proper equation for acceleration is:

m * a = m * (v^2 / R)

So that gives:

F_C * cos(theta) = m * (v^2 / R) = mu_s * m * g * cos(theta) + m * g * sin(theta)

Cancelling m and dividing by cos(theta):

(v^2 / R) = mu_s * g + g * tan(theta)

Since tan(theta) = mu_s:

(v^2 / R) = 2 * mu_s * g

This equation makes sense, to me, because previously we requires the parallel gravitation component to be equal (and opposite) to force of friction. Now we require that both of those forces act together to keep the car on the road, thus giving 2 * mu_s * g.

Solving for R:

R = v^2 / (2 * mu_s * g)

Which is dimensionally correct. Punching the values into my calculator gives:

v^2 / (2 * mu_s * g) = (50/3)^2 / (2 * 0.15 * 9.81) = 94.38592517

But this was rejected :(

Any idea where my mistake is at?

2. Sep 24, 2011

### PeterO

When a car is travelling around a banked track, the Normal reaction force is higher than when it is parked on the track [a similar thing happens when a car is driven through a dip rather than parked in a dip]. This increases the frictional force, an Fn is bigger.

3. Sep 24, 2011

### farleyknight

Yep, that's what I needed. Thanks! :)