What Speed Range Keeps a Car Safe on a 65m Banked Curve?

[demente182]

Homework Statement

A curve of radius 65m is banked for a design speed of 105 km/h. If the coefficient of static friction is 0.34 (wet pavement), at what range of speeds can a car safely make the curve?

I am not sure how can I get range of velocities at which the car can travel without skidding

Homework Equations

F = mA
F = ukFN
A = (v^2) / R

The Attempt at a Solution

uk x FN = mA
uk x mg =m(v^2)/r
v = sqrt [uk x r x g]
v = sqrt[ 0.34 x 65 x 9.8] = 14.72 m/s -> 53 km/hr

 

[lightgrav]

the curve is "banked" at some angle above the horizontal;
draw your car with velocity INTO the page, so all the Force vectors ,
and the acceleration vector, are diplayed ON the paper with angles showing.
at the "design speed", mg + F_N = ma_c , even with no friction.

if a car goes too slow, friction is needed ... also if a car goes too fast.

 

[demente182]

This is what I found out after doing the diagram:
*theta = angle at which the curve is banked.

F_N = -ma_c / cos ("theta" +90)
so I replaced into:

mg + sin ("theta" +90) x F_N = ma_c

mg + sin ("theta" +90) x ( (-ma_c) / (cos("theta" +90) ) = ma_c

(-9.8) + tan ("theta"+90) x ( v^2 / r ) = v^2 / r

Theta = 29.76181434

Is this correct? if so, I understand this is the angle at which the car can travel at v = 29.166 m/s without friction needed, but I still don't know how to find the range of speeds.

 

[lightgrav]

mg (down) + F_N(diagonal) = ma_c (horizontal) ... right triangle .
F_N is the same angle from vertical , that the road is from horizontal .
(your v^2/r is 13m/s^2 , right? ma_c this is bigger than gravity!)

If a car is trying to creep around this curve, it will slide down the steep slope.
what direction does friction push this car? ... how hard?

 

[demente182]

mg (down) + F_N(diagonal) = ma_c (horizontal) ... right triangle .
F_N is the same angle from vertical , that the road is from horizontal .
(your v^2/r is 13m/s^2 , right? ma_c this is bigger than gravity!)

If a car is trying to creep around this curve, it will slide down the steep slope.
what direction does friction push this car? ... how hard?

Yes, my v^2/r is 13.08 m/s^2.
If the car slides down the slope friction will act in the opposite direction the car is sliding, so friction acts up the slope.
Force of friction = F_N x uk.
Am I right?

 

[demente182]

I tried again starting from zero, and this time I set the sum of the forces in the Y direction to zero, but now I get a negative angle.
So:
F_N = -ma_c / cos (90 + "theta")
Replaced into :
sin (90 + "theta") * F_N = mg
and I got
"theta" = -126.825869

 

[Ms. Marsh]

Hello demente182!

I watch these forums to discourage students from cheating. Please solve the problem on your own.

Ms. Marsh