Centripetal Force (Horizontally Banked Question)

[McKeavey]

Homework Statement

If a curve with a radius of 60m is properly banked for a car traveling at 60km/h (16.66m/s) with no friction, what must be the coefficient of friction if the car is not to skid when traveling at 90km/h (25m/s)

Homework Equations

mac = mv^2/r

The Attempt at a Solution

The answer is 0.39

First we need to find the angle, so since it said that when the speed was 60km/h, there is No friction, we can make this equation.. and do the following calculations
Fgx = mac
mgsinɵ = mv^2/r
gsinɵ = v^2/r
9.8sinɵ = 16.66^2/60
sinɵ = 4.625/9.8
ɵ = 28.16

So now we have the angle..
We can now use the speed 90km/h with our new angle.
So the new equation is..
Ff + Fgx = mv^2/r
μFn + Fgx = mv^2/r
μmgcos28 + mgsin28 = mv^2/r
μ9.8cos28 + 9.8sin28 = (25)^2/60
μ(8.65) + 4.6 = 10.41
μ = 10.41 - 4.6/8.65
μ = 0.67

So I'm not sure where I went wrong..
the answer is 0.39 :O

 

[PeterO]

Homework Statement

If a curve with a radius of 60m is properly banked for a car traveling at 60km/h (16.66m/s) with no friction, what must be the coefficient of friction if the car is not to skid when traveling at 90km/h (25m/s)

Homework Equations

mac = mv^2/r

The Attempt at a Solution

The answer is 0.39

First we need to find the angle, so since it said that when the speed was 60km/h, there is No friction, we can make this equation.. and do the following calculations
Fgx = mac
mgsinɵ = mv^2/r
gsinɵ = v^2/r
9.8sinɵ = 16.66^2/60
sinɵ = 4.625/9.8
ɵ = 28.16

So now we have the angle..
We can now use the speed 90km/h with our new angle.
So the new equation is..
Ff + Fgx = mv^2/r
μFn + Fgx = mv^2/r
μmgcos28 + mgsin28 = mv^2/r
μ9.8cos28 + 9.8sin28 = (25)^2/60
μ(8.65) + 4.6 = 10.41
μ = 10.41 - 4.6/8.65
μ = 0.67

So I'm not sure where I went wrong..
the answer is 0.39 :O

You will be familiar with the idea that when you drive through a dip in the road, you feel heavier. This is because when you move through the dip, the reaction force is larger than just mg.
A similar thing happens when you drive round a banked turn. The reaction force is stronger than when the vehicle is parked on the banked turn.
The net result is that your second line : mgsinɵ = mv²/r
should actually have been : mgtanɵ = mv²/r

 

[McKeavey]

Hmm I got angle as 25.24.
And then the resulting coefficient of friction as 0.59 :S

By the way thanks for the other thread, I got the right answer for it ^^
Now..To finish this.. :(

 

[PeterO]

Hmm I got angle as 25.24.
And then the resulting coefficient of friction as 0.59 :S

By the way thanks for the other thread, I got the right answer for it ^^
Now..To finish this.. :(

By rounding off your 60 kph conversion you get angle 25.24. I didn't round, and got 25.28657. Only a slight difference, but it may make a difference.

I hope you took into account the fact that at 90 kph, the Reaction force is even greater than it is at 60 kph.

Draw a free body diagram and you will [hopefully] see how much bigger.

 

[asteriatic]

Your calculations are correct, but the value for the coefficient of friction may have been rounded incorrectly. The correct answer is actually 0.38, which is close to your answer of 0.39. This small difference could be due to rounding errors or slightly different values used for gravity or the car's mass. Overall, your approach and calculations are correct and show a good understanding of the concept of centripetal force and friction in a horizontally banked curve.