Maximizing Banked Turn Angle for Runner on Track

[mvs1c]

Homework Statement

A runner is running on the banked turn of a track. The radius of the turn is 30 meters. The speed of the runner is 10 m/s. The coefficient of static friction between the runner's shoes and the track is .71. What is the maximum banking angle that will still allow the runner to run through the turn?

µf : .71
speed of runner : 10 m/s
radius of the turn: 30 m

Choices of answers:
33.7 degrees
23.8
54.1
46.8

Homework Equations

F = m(V^2/R)
Fs-max = µs*Fn
Fn= normal force

The Attempt at a Solution

I haven't had much luck. My professor has covered a few example problems that were similar, but they had negligible friction or a different angle set up.

I did try taking µs*Fn*sinθ = m(-v^2/R) and dividing it by Fncosθ = mg
resulting in tan θ = -V^2*µs/(g*R), which is equivalent to tan θ = -100^2*.71/(30*9.81)
but that resulted in an angle of 13.56 degrees, which isn't even on the answer choices.
I would appreciate any help you can give =)

 

[willem2]

I did try taking µs*Fn*sinθ = m(-v^2/R) and dividing it by Fncosθ = mg

the first equation isn't valid because you forgot the horizontal component of the normal force.

the second equation isn't valid because you forgot the vertical component of the friction force.

 

[mvs1c]

in one of the solved problems, they approached it with a radial calculation, which was the first I used, and a vertical calculation, which was the second.

Could you possibly clarify a bit more so I understand how I can fix it?

 

[willem2]

Both the normal force and the friction force have horizontal and vertical components.

The vertical components of the friction force and the normal force have to balance gravity.
The horizontal components of these forces have to produce the acceleration.
The maximum angle is the angle that the runner nearly slips down. The friction force will be pointed along the slope and upwards.
What you tried to do was valid except for forgetting those components. You'll get a much more complicated expression for tan(theta). It will be really easy to make a sign error here.