- #1
wallercp
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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?
http://www.freewebs.com/cwaller/Car.jpg
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
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
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
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