Max. velocity with given coefficient of static friction (μ_s)

[Hendrick]

Homework Statement

A car goes around a curved stretch of flat roadway of radius R = 104.5 m. The magnitudes of the horizontal and vertical components of force the car exerts on a securely seated passenger are, respectively, X = 236.0 N and Y = 572.0 N.

This stretch of highway is a notorious hazard during the winter months when it can be quite slippery. Accordingly the DMV decides to bank it at an angle φ = 23.5 ° to the horizontal.

(1) What is the maximum speed at which the car could safely negotiate this banked curve during a wintry night when the coefficient of static friction between the road and tyres is μs = 0.24 ?

Homework Equations

F_s = (m*v^2)/r
F_s = μ_s * n
n = m*g
where n is the normal force and μ_s is the coefficient static friction.

The Attempt at a Solution

v = ((F_s*r)/m)^1/2
v = ((μ_s*n*r)/m)^1/2
v = ((μ_s*m*g*r)/m)^1/2
v = (μ_s*g*r)^1/2

v = (0.24*9.81*104.5)^1/2 * 3.6 (to convert ms^-1 to kmh^-1)
= 56.5 kmh^-1 (3sf)

Actual answer is 100 kmh^-1 (to 3sf)

 

[hage567]

For one thing, you didn't take into account that the curve is now banked at an angle of 23.5 degrees. That's important.

 

[MathematicalPhysicist]

the equations should be:
$$fcos(\phi)-\frac{mv^2sin(\phi)}{R}-Nsin(\phi)=0$$
$$mg-Ncos(\phi)-fsin(\phi)=0$$
with the appropiate sketch it's also understood why.
and ofcourse f=mu*N.

 

[Hendrick]

the equations should be:
$$fcos(\phi)-\frac{mv^2sin(\phi)}{R}-Nsin(\phi)=0$$
$$mg-Ncos(\phi)-fsin(\phi)=0$$
with the appropiate sketch it's also understood why.
and ofcourse f=mu*N.

I still don't quite understand why I can't use the v = (μ_s*g*r)^1/2 formula.
For this question I used it:

At what speed could the car now negotiate this curve without needing to rely on any frictional force to prevent it slipping upwards or downwards on the banked surface?

v = (μ_s*g*r)^1/2
v = (tanφ*g*r)^1/2
v = (tan23.5*9.81*104.5)^1/2 * 3.6 (to convert ms^-1 to kmh^-1)
= 76.0 kmh^-1 (3sf)

 

[MathematicalPhysicist]

you can't use this formula, cause the force diagram of yours is incorrect.
i reackon you have a sketch of the problem, look at it and then draw the forces.
this question is a question of statics, the maximum speed is reached when there it neither slips to the sideways or inwards of the road, this we get when the net force is zero.

 

[Hendrick]

you can't use this formula, cause the force diagram of yours is incorrect.
i reackon you have a sketch of the problem, look at it and then draw the forces.
this question is a question of statics, the maximum speed is reached when there it neither slips to the sideways or inwards of the road, this we get when the net force is zero.

I did draw a diagram originally to get part a of the question and it was right. The only thing that has changed is that they gave me a coefficient of static friction. It should be the same diagram still since the angle of 23.5 degrees still applies...

I got two questions from it (the x and y components):

1) Sigma F_x = m.g.sinφ - F_s = m.a_x = 0
2) Sigma F_y = n - m.g.cosφ = m.a_y = 0

I have tried using your equations in every way but I get no where near the answer, would you be so kind and provide me with a helpful hint to steer me in the right direction? I'm really going around in circles at the moment :(...

---Thanks