This one is almost impossible to do without the FBD in front of you. In this problem the mg force is down the normal is purpendicular to the track and friction is parallel to the track.
Equation in y:
Fn cosθ + Ff sinθ = mg
Equation in x (causing circ motion):
Fn sinθ - Ff cosθ = mv^2/r
Third eq you need mu Fn = Ff
Now you have three eqs and three unknowns.The math is less than delightful.
1300 kg car rounds a banked curve of radius= 50m. If car traveling at 75 km/h and the coefficient of static friction between the tire and road is 0.45 ...
1.) what is the minimum angle of the bank needed to keep the car in the turn?
2.) If the car changes tires, which increases the Coef. of Friction to 0.65.
3.) What will be the max. Velocity for the car to be able to turn (assuming the same banked angle)?
3 answers
Sorry, that should be
y direction:
Fn cosθ = mg + Ffsinθ
x direction:
Fn sinθ + Ff cosθ = mv^2/r
And:
mu Fn = Ff
We're trying to not fly off the track.
I get a value of about 17.1o which seems reasonable.
y direction:
Fn cosθ = mg + Ffsinθ
x direction:
Fn sinθ + Ff cosθ = mv^2/r
And:
mu Fn = Ff
We're trying to not fly off the track.
I get a value of about 17.1o which seems reasonable.
How did you get 17.1
0 answers