# Homework Help: Uniform Circular Motion (free-body diagram)

1. May 16, 2012

### Joshuarr

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

Curved portions of highways are always banked (tilted) to prevent cars from sliding off the highway. When a highway is dry, the frictional force between the tires and the road surface may be enough to prevent sliding. When the highway is wet, however, the frictional force may be negligible, and banking is then essential. Figure 6-13a represents a car of mass m as it moves at a constant speed v of 20 m/s around a banked circular track of radius R = 190 m. If the frictional force from the track is negligible, what bank angle θ prevents sliding?

2. Relevant equations

Uniform circular motion: a = v^2/R

Magnitude of centripetal force: F = m*a = m*v^2/R

I used this identity in solving the problem: sin(2θ) = 2*sin(θ)*cos(θ)

3. The attempt at a solution

So, I managed to solve this problem, but the answer I got seems to have a conceptual difference than the book. So I'm wondering how I can correct my thinking.

I started by drawing a similar diagram in 6-13b, EXCEPT, I assumed that Fg has the largest magnitude, and when decomposing the vectors, I chose it as the hypotenuse.

This lead me to the relationships: F_N = mg*cos(θ) and F_Nr = mg*cos(θ)*sin(θ) = mg*sin(2θ)/2

I labeled F_Nr as the centripetal force, so: F_Nr = mg*sin(2θ)/2 = m*v^2/R,

which gives: θ = Acos( 2*v^2/(R*g))/2 = Acos(2*(20)^2/(190*9.8))/2 = 12.722°

The book got: Atan(v^2/(g*R)) = Atan(20^2/(9.8*190)) = 12.124°

They are similar, and normally I'd just ignore the difference, but there seems to be a difference in the formulas themselves [for example, they get F_n = mg*(1/cos(θ)) as opposed to my F_n = mg*cos(θ)], which leads me to believe that my thinking was wrong, and I just got a very close answer by serendipity.

I don't understand why F_N is larger than F_g. It seems to me the force that "causes" all the other forces is the force of gravity, so it should be the largest.

How do I know that that isn't the case?

The diagram the book gives, including their free-body diagram, is attached.

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2. May 16, 2012

### azizlwl

Actually only 2 forces exerted on the car(no friction).
1. Mg
2. Normal force(tangent to the bank)

If you take vertical direction, the net force is zero
Only horizontal net force equal to centripetal force.

Thus Ncosθ=mg.
and Nsinθ=mv2/r

You can also replace(resolve) mg to mgcosθ and mgsinθ.
Now you have 3 forces which makes the calculation harder since both mgsinθ and mgcosθ has component to centripetal force.

Last edited: May 16, 2012
3. May 16, 2012

### Joshuarr

Ahh, I think I see what you mean.

So I first start by drawing the forces (in the free-body diagram) and then decompose it into its components and then create relationships.. This sounds obvious now. lol.

I was trying to derive all of the relationships from just F_g. Thanks this helped a lot.

So, I think I understand it mathematically now - but why is F_N bigger than F_g though?

At first, I considered F_N simply a component of F_g, so I just projected F_n onto F_g, but apparently - that's not correct.

I guess the moral of the story is just not to presume I know what F_N is.

4. May 16, 2012

### azizlwl

Yes that what i did wrong previously.
If you try to solve a problem of a block on an inclined plane and the incline is accelerating, you will see that normal force is more than mgcosθ so the net have the component to accelerate the block forward.

Fn has to "produce" 2 forces.
1. To hold the block from falling down(into the ground)
2. To accelerate the car to the center(to change the direction of the car).

Mg has no component to centripetal force.

Last edited: May 16, 2012
5. May 16, 2012

### Joshuarr

I understand. Thanks. This makes much more sense now.

I really appreciate your response. :)