Hi everyone,(adsbygoogle = window.adsbygoogle || []).push({});

I'm having a bit of trouble with this homework problem. Here it is:

A car is traveling at a constant speed along a hilly road, as the drawing shows. Point A is at the bottom of the dip, while point B is at the top of the hill. The radius of the curvature is 35.0 m at both points. The apparent weight of the car at point A exceeds the apparent weight at point B by a factor of four. What is the speed of the car?

http://www.freewebs.com/cwaller/Car.jpg [Broken]

Fc=Centripetal force

Fg=Force of gravity

Fn=Normal force

v=Velocity

Fc = mv^2/r

Fg = mg OR GMm/R (M = mass of larger mass, m = mass of smaller mass)

At point A

4Fn-Fg=mv^2/R

4Fn-mg=mv^2/R

At point B

Fg-Fn=mv^2/R

mg-Fn=mv^2/R

Simultaneous Equations...

4Fn-mg=mv^2/R

mg-Fn=mv^2/R

3Fn=2mv^2/35

105Fn=2mv^2

52.5Fn=mv^2

I still have three unknown variables and I'm really stumped. My teacher never said if we need a to get a number answer, I can easily solve this for v in terms of Fn and m. The only other possibility is that at point B the car is going at critical speed, but the wording of the problem doesn't lead me to believe that this is true. Any help is appreciated.

Edit: I made some progress with this problem.

Solve the equation at point B for Fn.

mg-Fn=mv^2/R

Fn=mg-(mv^2/r)

Fn=m(g-(v^2/r))

Substitute back into simultaneous equation...

3Fn=2mv^2/r

(3/2)Fn=mv^2/r

(3/2)(m(g-(v^2/r)))=mv^2/r

(3/2)g-((3v^2)/(2R))=mv^2/r

(3/2)g=(v^2/R) + (3v^2/2R)

(3/2)g=(5v^2)/(2R)

14.7=(5v^2)/70

205.8=v^2

v=14.34 m/s

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: A car travelling over a hill.

**Physics Forums | Science Articles, Homework Help, Discussion**