A car travelling over a hill.

1. Nov 21, 2006

wallercp

Hi everyone,

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?

[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

Last edited by a moderator: Apr 22, 2017 at 2:10 PM
2. Nov 21, 2006

turdferguson

Looks good. You didnt need to go to substitution if you had multiplied the point B equation by 4 before eliminating. Thats why it was wierd

3. Nov 21, 2006

G01

If you divide everything by m, the mass, you get:

A
$$4a_n - g = \frac{v^2}{R}$$

B
$$g - a_n = \frac{v^2}{R}$$

where $$a_n$$ is the acceleration caused by the normal force. Get $$a_n$$ to cancel and you can solve for v. Try that it makes sense in my head anyway! Good Luck. (If I made a mistake, someone please correct me)

(Actually, all u gotta do is get Fn to add out in the equations, then m cancels and you can solve for v)

Last edited: Nov 21, 2006