A curve of radius 76 m is banked for a design speed of 100 km/h

  • #1
rperez1
7
0

Homework Statement


If the coefficient of static friction is 0.38 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]

Homework Equations



I am trying to figure out the Vmax

3. The Attempt at a Solution


I was able to find the Vmin in doing:

ΣFx = ma = mv^2/r
Fnsinθ = mv^2/r (1)

Designed for 100km/h ~ 27.78m/s so there's no friction involved for now. Fn is the normal force.
Fnsinθ/Fncosθ = mv^2/mgr
tanθ = v^2/gr
θ = arctan(v^2/gr) = arctan((27.78)^2/(9.8*76))
= 46.02°

When you go very slow, the car's tendency is to slide down, so the friction acts up the ramp:

ΣFx = mv^2/r
Fnsinθ - fcosθ = mv^2/r (1)

ΣFy = 0
Fncosθ + fsinθ - mg = 0
Fncosθ + fsinθ = mg (2)

To eliminate Fn , f and m, i used the definition of friction:

f = μ*Fn
Fn = f/μ

And replaced Fn by f/μ in both equations:

fsinθ/μ - fcosθ = mv^2/r (1)*
fcosθ/μ + fsinθ = mg (2)*

I divided (1)* by (2)* to eliminate some values:

(fsinθ/μ - fcosθ)/(fcosθ/μ + fsinθ) = mv^2/mgr

(Factor f out and cancel it)
(sinθ/μ - cosθ)/(cosθ/μ + sinθ) = v^2/gr

Plugging in θ = 46.02° I found the min speed:

v = 18.73m/s = 67.43km/h

I have to find the max speed now but this time friction acts down the ramp to prevent car from shooting up. I just can't figure out how?
 
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  • #2
rperez1 said:

Homework Statement


If the coefficient of static friction is 0.38 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]

Homework Equations



I am trying to figure out the Vmax

3. The Attempt at a Solution


I was able to find the Vmin in doing:

ΣFx = ma = mv^2/r
Fnsinθ = mv^2/r (1)

Designed for 100km/h ~ 27.78m/s so there's no friction involved for now. Fn is the normal force.
Fnsinθ/Fncosθ = mv^2/mgr
tanθ = v^2/gr
θ = arctan(v^2/gr) = arctan((27.78)^2/(9.8*76))
= 46.02°

When you go very slow, the car's tendency is to slide down, so the friction acts up the ramp:

ΣFx = mv^2/r
Fnsinθ - fcosθ = mv^2/r (1)

ΣFy = 0
Fncosθ + fsinθ - mg = 0
Fncosθ + fsinθ = mg (2)

To eliminate Fn , f and m, i used the definition of friction:

f = μ*Fn
Fn = f/μ

And replaced Fn by f/μ in both equations:

fsinθ/μ - fcosθ = mv^2/r (1)*
fcosθ/μ + fsinθ = mg (2)*

I divided (1)* by (2)* to eliminate some values:

(fsinθ/μ - fcosθ)/(fcosθ/μ + fsinθ) = mv^2/mgr

(Factor f out and cancel it)
(sinθ/μ - cosθ)/(cosθ/μ + sinθ) = v^2/gr

Plugging in θ = 46.02° I found the min speed:

v = 18.73m/s = 67.43km/h

I have to find the max speed now but this time friction acts down the ramp to prevent car from shooting up. I just can't figure out how?
(I see you're new here, but ...) Please state your problem in the body of the Original Post of your thread. Don't post information in the title without including it in the thread, if it's important information.

I assume the "100 k" refers to 100 km/hr. Right ?
 
  • #3
rperez1 said:
I have to find the max speed now but this time friction acts down the ramp to prevent car from shooting up. I just can't figure out how?
Wouldn't that simply change the sign you use for the force of friction ?
 
  • #4
SammyS said:
(I see you're new here, but ...) Please state your problem in the body of the Original Post of your thread. Don't post information in the title without including it in the thread, if it's important information.

I assume the "100 k" refers to 100 km/hr. Right ?
yes it refers to 100 km/hr
 

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