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72_Cutlass_S
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Hello, I'm towards the end of my first semester of Physics 1 (cal based) and I've had this homework problem that has been making want to pull my hair out. I know that the coeffeicent of friction is between 0 and 1 and I have worked out the problem like my professor did but keep getting the wrong answer. Thank you for any help.
Michael
A banked circular highway curve is designed for traffic moving at 70 km/h. The radius of the curve is 215 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?
f-mg*sin(x)=ma*cos(x)
N-mg*cos(x)=ma*sin(x)
tan(x)=v^2/Rg
where:
m= mass = cancels out
g= gravity = 9.8 m/s^2
a= acceleration = v^2/R = 11.11^2/215 = .574 m/s^2
x= angle
v= velocity = 40 km/h = 11.11 m/s
R= radius = 215m
coefficent of friction = (g*sin(x)-a*cos(x))/(g*cos(x)+a*sin(x))
x= tan^-1(11.11^2/(215*9.8))
x= 3.353
f = (9.8*sin(3.353)-.574*cos(3.353))
(9.8*cos(3.353)-.574*sin(3.353))
f = -1.01*10^-15
Michael
Homework Statement
A banked circular highway curve is designed for traffic moving at 70 km/h. The radius of the curve is 215 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?
Homework Equations
f-mg*sin(x)=ma*cos(x)
N-mg*cos(x)=ma*sin(x)
tan(x)=v^2/Rg
where:
m= mass = cancels out
g= gravity = 9.8 m/s^2
a= acceleration = v^2/R = 11.11^2/215 = .574 m/s^2
x= angle
v= velocity = 40 km/h = 11.11 m/s
R= radius = 215m
The Attempt at a Solution
coefficent of friction = (g*sin(x)-a*cos(x))/(g*cos(x)+a*sin(x))
x= tan^-1(11.11^2/(215*9.8))
x= 3.353
f = (9.8*sin(3.353)-.574*cos(3.353))
(9.8*cos(3.353)-.574*sin(3.353))
f = -1.01*10^-15