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totallyclone
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Homework Statement
A curve of radius 60 m is banked for a design speed of 100km/h. If the coefficient of static friction is 0.30, at what range of speeds can a car safely make the turn.
Homework Equations
Fun=ma
Ff=μsFn
Fc=mV2/r
The Attempt at a Solution
So, when it is "designed" for 100km/h that means there is no friction at 100km/h or 27.8m/s so this is my fbd for this speed. I've tried working it out and I manage to get okay looking answers but I'm not sure if I'm doing it right... :P
So, there is no acceleration in the Y-component:
ƩFy=ma
Fny-Fg=0
Fny=Fg
Fncosθ=mg 1st eqn.
For the X-component, there is acceleration which is the centripetal acceleration:
ƩFx=ma
Fnx=mV2/r
Fnsinθ=mV2/r 2nd eqn.
Now, to find the angle of the bank at that "design", so I isolated for θ:
2nd eqn./1st eqn.
Fnsinθ/Fncosθ=(mV2/r)(1/mg)
tanθ=V2/rg
θ=tan-1(27.82/60x9.8)
θ=52.7°
Now that I found the angle it is for 100km/h to make the turn in the bank without the need for friction, I'm trying to find the maximum speed for it in order not to slide up the bank so now I'll throw in friction and this is my fbd.
So I'll use the same concept with the Fun=ma just that I have Ff now:
ƩFy=0
mg=μsFnsinθ-Fncosθ eqn.1
ƩFx=ma
Ffx-Fnx=mv2/r
μsFncosθ-Fnsinθ=mv2/r eqn.2
eqn.2/eqn.1
(mv2/r)(1/mg)=(μsFncosθ-Fnsinθ)/(μsFnsinθ-Fncosθ)
V2/rg=(μscosθ-sinθ)/(μssinθ-cosθ)
V=√(rg(μscosθ-sinθ))/(μssinθ-cosθ)
V=31.3m/s
V=113km/h
So, that is the maximum speed for it in order not to slip up the bank. Now, I want to find the minimum.
I just flip some signs around in my fbd so:
ƩFy=0
Fg-Fny=0
mg=Fncosθ eqn.1
ƩFx=ma
Fnx-Ffx=mv2/r
Fnsinθ-μsFncosθ=mv2/r eqn.2
eqn.2/eqn.1
(mv2/r)(1/mg)=(Fnsinθ-μsFncosθ)/Fncosθ
V2/rg=(sinθ-μscosθ)/cosθ
V=24.4m/s
V=87.8km/s
∴Range of speed in order to make the turn safely is from 87.8km/h to 113km/h.
I appreciate if there's any corrections or pointers that I missed or anything! :)