What Angle Should a Curve Be Banked for Frictionless Cornering?

[paolo_c10]

[SOLVED] just a problem that I'm stuck on..

Homework Statement

A car travels around a curve with a radius 55.0 m at a speed of 20.0 m/s. At what angle must the curve be banked so that the car does not have to rely on friction to remain on the road?

Homework Equations

tan[theta] = FC/FG
tan[theta] = v^2/rg

The Attempt at a Solution

there is no solution..I'm confused..
I tried rotating the equation on a different but rather same question..and I can't figure it out..
when got the answer, i tried transposing tan to the other side..it got me more confused..

 

[paolo_c10]

Force that car exerts on the ground must be equal to centripetal force. Since the curve iz banked, weight has two components. The component on the ground is centripetal force. You know the centripetal force and the weight of the car. You can get the angle with a little help of trigonometry.

but the weight of the car was not given, and the equation states that plugging in the numbers will give me the angle theta. I tried plugging in the numbers and I can't seem to get the right answer, which means I must be doing something wrong, or I'm missing on something..

 

[Dr. Jekyll]

Force that car exerts on the ground must be equal to centripetal force. Since the curve iz banked, weight has two components. The component on the ground is centripetal force. You know the centripetal force and the weight of the car. You can get the angle with a little help of trigonometry.

You can also replace the forces with accelerations (\vec{g}=\vec{a}_{cp}+\vec{a}).

 

[paolo_c10]

It doesn't make sense, i have all the formulas. I never encountered any problems when I solved the almost same question(but asking for velocity, instead of angle). And i plugged in the right numbers. Now with this question I just tweaked the formula, and i searched other sites for other formulas, I get the same one..i just can't seem to get the answer i need, w/c means I am definitely missing on something...

And isn't the centripetal force the middle line on my free-body-diagram?and not the component on the ground..

 

[Dr. Jekyll]

And isn't the centripetal force the middle line on my free-body-diagram?and not the component on the ground..

Yes, it is. My mistake.

 

[Kurret]

the force being equal to the centripetal force, is the component of the gravitation along the radius of the curve. You don't need the mass since you can cancel it out in the equations, or you can see it as the centripetal acceleration should be equal to the composant of the gravitational acceleration.

 

[paolo_c10]

yes, mass is not needed..so how do I correct this problem?how can i find the banking angle?

 

[paolo_c10]

got it..!it was just an easy mistake with my calculator... =S

 

[Eros]

How did u do it please...is the angle tan 0.72?

 

[paolo_c10]

no..the angle is about 36.55, but you're on the right track of using the formula. You only forgot to transpose the tan to the right side..so tan^-1(inverse function of your calculator) and 0.74 on your calculator gives you the banking angle.. =)

 

[Eros]

oh thanks ^^