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dani123
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Homework Statement
A 1200 kg car rounds a banked curve of radius 70 m. If the banked angle is 12° and the car is traveling at 90 km/h, show that friction is necessary in order for the car to safely make the turn.
Homework Equations
Force of gravity: Fg=m*g
to calculate the normal force: opposite side of angle: Fncos(12°)=mg
centripetal force: Fc=mv2/R
Fc=Ff
μ=Ff/Fn
The Attempt at a Solution
So these are the things we know:
m=1200kg
R=70m
θ=12°
v=90km/h=25m/s
car is traveling on a banked curve
With this information we can start off by calculating the Fg=mg=(1200kg)*(9.8m/s2)=11760N
The normal force can be calculated from:
Fncos(12°)=mg
Fn=11760N/cos(12°)= 12022.725N
From here, I calculated the Fc:
Fc=(1200kg)*(25m/s)/(70m)= 10714.286 N
I think we are to assume that Fc=Ff, because its a banked curve? I am not entirely sure on this reasoning... some clarification would be appreciated :)
so from here I calculated μ:
μ=Ff/Fn=10714.286N/12022.725N= 0.89
END OF MY FIRST ATTEMPT
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BEGINNING OF SECOND ATTEMPT (for this attempt I used another problem as a guide that I found online... not entirely sure of the explanation given at the end, I have a hard time visualizing it)
I calculated the Force trying to pull the car down the bank:
from the adjacent side: F=mgsin(12°)= (1200kg)*(9.8m/s2)sin(12)= 2445.041N
The needed centripetal force:
Fc=mv2/R= (1200kg)*(25m/s)2/(70m)= 10714.286N
The component of Fc that is parallel to the road surface is:
opposing side: Fc*cos(12°)=10480.153 N
The difference between component of centripetal force parallel to the roadway and the force due to gravity component parallel to the roadway is:
10480.153 N- 2445.041 N= 8035.111N
This force must be made up by friction, toward the center of the circle but parallel to the roadway surface.
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SO, if anyone could lend me a hand and let me know where I went wrong and which attempt better suites this type of problem that would be greatly appreciated! Thank you so much in advance!