Banked Curve Algebra With Static Friction

[Extremist223]

Homework Statement

9. A 2000 kg car is racing around a banked circular path whose radius of curvature is 50.0m. The coefficient of static friction between the car and the road is 0.100. What is the maximum possible speed for the car to remain on the road if the angle of the curve is 15°?

a) 45.7 km/h
b) 49.0 km/h
c) 60.0 km/h
d) 130 km/h
e) 197 km/h

Homework Equations

Fsmax = Us FN
FN = cos15mg
Fc = mV^2/r

The Attempt at a Solution

Once again stuck on another problem. From what I see, there is a component of the normal force that points towards the center of the radius as well as a component of the static friction that points towards the center of the circle. These two values added should give me the centripetal force, if I am right. Assuming this is true this is what I have tried.

sin15(cos15)mg= X component normal force pointing towards center
Us(cos15mg)(cos15)= x component of FsMax that points towards center

sin15(cos15)mg + Us(cos15mg)(cos15) = Fc
Fc=mv^2/r
divide the equation by mass
sin15(cos15)g + Us(cos15g)(cos15) = v^2/r
multiply the left side by r and root for velocity
Root([(sin15)(cos15)g + Us(cos15g)(cos15)]r) = v
v = 12.96m/s

What am I missing here? I have a midterm tomorrow please answer as soon as you can thank you.

 

[kuruman]

The normal force is (mg)cosθ only if the acceleration is parallel to the incline. In general, the normal force is what is necessary to provide the observed acceleration. Redo your free body diagram to find what the normal force is in this case.

 

[kerimek]

I would first like to commend you on your attempt at solving this problem using the appropriate equations and understanding the components involved. However, I would like to point out a few things that may help you arrive at the correct answer.

Firstly, when calculating the normal force, it is important to remember that it is equal to the component of the car's weight perpendicular to the surface, which in this case is cos15mg. This means that the normal force will be equal to the weight of the car in this situation.

Secondly, when calculating the maximum static friction force, it is important to remember that it is equal to the coefficient of static friction (0.100) multiplied by the normal force. So in this case, it would be (0.100)(cos15mg).

Now, when adding these two components together to find the total centripetal force, it is important to remember that the normal force and the static friction force are acting in opposite directions, so they should be subtracted instead of added.

Therefore, the correct equation would be:

Fc = (mv^2)/r = (cos15mg)sin15 - (0.100)(cos15mg)cos15

Solving for v, we get v = 49.0 km/h, which is option b.

I hope this helps and good luck on your midterm! Remember to always double check your equations and make sure you are using the correct components for each force involved.