Coefficient of Static Friction on a Banked Curve

[wingman358]

Homework Statement

If a curve with a radius of 90 m is properly banked for a car traveling 68 $$\frac{km}{h}$$, what must be the coefficient of static friction for a car not to skid when traveling at 98 $$\frac{km}{h}$$?

Homework Equations

I know I need
$$\Sigma \vec{F} = m \vec{a}$$$$f_s <= \mu_s N$$

and

$$a_c = \frac{v^2}{r}$$

The Attempt at a Solution

I first solved for $$\theta$$, which is the angle of embankment at which a car traveling 68 $$\frac{km}{h}$$ will need no friction to travel along a curve of radius 90 m.

Given

$$\Sigma \vec{F} = m \vec{a}$$

with pertinent variables:

$$f_s = ma_c$$

substituting for $$a_c$$:

$$f_s = m\frac{v^2}{r}$$

substituting for $$f_s$$:

$$\mu_s N = m\frac{v^2}{r}$$

substituting for $$N$$:

$$\mu_s mg\tan{\theta} = m\frac{v^2}{r}$$

dividing by m:

$$\mu_s g\tan{\theta} = \frac{v^2}{r}$$

solving for $$\mu_s$$:

$$\mu_s = \frac{v^2}{gr\tan{\theta}}$$

Now substituting in for all known variables ($$v, g, r, and \theta$$), $$\mu_s$$ equals 2.077, which MasteringPhysics does not accept.

Now, I'm very confused on this problem, mainly because I'm unsure if my free body diagram is correct:

http://img227.imageshack.us/img227/5083/physfbd.th.png

Note: image is not to scale

Any help or hints would be appreciated!

 

[LowlyPion]

Strictly speaking, when the speed is balanced to the curve you really have

Cosθ *m* V²/R = Sinθ *m*g

To balance that equation however when the speed is greater than optimal you have a frictional force that is made up of

μ *m*(sinθ*V²/R + cosθ*g)

don't you?

 

[wingman358]

Strictly speaking, when the speed is balanced to the curve you really have

Cosθ *m* V²/R = Sinθ *m*g

To balance that equation however when the speed is greater than optimal you have a frictional force that is made up of

μ *m*(sinθ*V²/R + cosθ*g)

don't you?

LowlyPion, thank you for your quick response, but I'm not sure what you mean.

in the equation you gave me:

Cosθ *m* V²/R = Sinθ *m*g

where do the Cosθ and Sinθ come from?

 

[LowlyPion]

LowlyPion, thank you for your quick response, but I'm not sure what you mean.

in the equation you gave me:

Cosθ *m* V²/R = Sinθ *m*g

where do the Cosθ and Sinθ come from?

Those are the components of the horizontal force (centripetal) and the vertical force (gravity) that balance || along the surface of the incline.

You may be more familiar with seeing it as

V²/R = Sinθ/cosθ *m*g = tanθ *m*g

 

[wingman358]

Those are the components of the horizontal force (centripetal) and the vertical force (gravity) that balance || along the surface of the incline.

You may be more familiar with seeing it as

V²/R = Sinθ/cosθ *m*g = tanθ *m*g

I apologize for being so inept, but I'm still lost. Can you show/tell me step-by-step how to get to the correct answer?

 

[LowlyPion]

I apologize for being so inept, but I'm still lost. Can you show/tell me step-by-step how to get to the correct answer?

Well you originally used:

$$\mu_s N = m\frac{v^2}{r}$$
substituting for N:
$$\mu_s mg\tan{\theta} = m\frac{v^2}{r}$$

To determine the angle θ at the optimal speed for the curve, you are only balancing the downward force of gravity along the incline mg*sinθ against the upward force from the centripetal acceleration (m*v²/r)*cosθ. Or ...

mg*sinθ = (m*v²/r)*cosθ

This formula gives you the optimal angle for radius and velocity with no reliance on friction.

Now that you are entertaining higher speeds, you have another component that needs to be considered (it was previously 0 for the optimal angle at 68km/h). And that is the additional term I mentioned earlier derived from the normal components of gravity and centripetal acceleration and expressed through the coefficient of friction necessary to resist slipping at the higher speed.

μ *m*(sinθ*V²/R + cosθ*g)

 

[wingman358]

Well you originally used:To determine the angle θ at the optimal speed for the curve, you are only balancing the downward force of gravity along the incline mg*sinθ against the upward force from the centripetal acceleration (m*v²/r)*cosθ. Or ...

mg*sinθ = (m*v²/r)*cosθ

This formula gives you the optimal angle for radius and velocity with no reliance on friction.

Now that you are entertaining higher speeds, you have another component that needs to be considered (it was previously 0 for the optimal angle at 68km/h). And that is the additional term I mentioned earlier derived from the normal components of gravity and centripetal acceleration and expressed through the coefficient of friction necessary to resist slipping at the higher speed.

μ *m*(sinθ*V²/R + cosθ*g)

Okay so I understand that the equation
mg*sinθ = (m*v²/r)*cosθ
indicates the force acting down the slope is equivalent to the force acting up the slope (hence, no acceleration up or down the slope and therefore no friction needed).

Furthermore, if I understand you correctly,

$$f_s$$ = μ *m*(sinθ*V²/R + cosθ*g)

where f_s is force of static friction. Is that right?

 

[LowlyPion]

Yes. That's correct.

So what they want is what minimum value of μ to keep it from slipping at the increased speed.

 

[wingman358]

So what they want is what minimum value of μ to keep it from slipping at the increased speed.

I understand that, but I don't know how to do the set up an equation to solve. Could you draw a free body diagram of the car with the added force of static friction? I just can't visualize the forces...

 

[wingman358]

Okay I managed to figure it out with the help of this website: http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Circular Motion/banked_with_friction.htm

 

[LowlyPion]

Okay I managed to figure it out with the help of this website: http://www.batesville.k12.in.us/physics/PhyNet/Mechanics/Circular Motion/banked_with_friction.htm

(Those are better drawings than you could have expected from me I can say.)

You have figured then that the frictional force must account for the excess the centripetal force from the speed and can be added to the gravitational component of force down the incline. Knowing the other values then you can solve for μ.