A car of mass M = 900 kg traveling at 65.0 km/hour enters a banked turn covered

Question

A car of mass M = 900 kg traveling at 65.0 km/hour enters a banked turn covered with ice. The road is banked at an angle , and there is no friction between the road and the car's tires as shown in Figure 1 Use g = 9.80 m/s2 throughout this problem.

What is the radius r of the turn if = 20.0 (assuming the car continues in uniform circular motion around the turn)

I can't seem to get the right answer. My process so far has been

Fx=ma

Fgsin(theta)=ma

Fgsin(theta)=m(v^2/r)

gsin(theta)=v^2/r

r=v^2/(gsin(theta)

r=96.7

But this is incorrect. I think you have to account for the Fy portion but I can't figure out how to include it. Thanks

Explanation / Answer

Mass, M = 900 kg
speed, u = 65 km/hr = 18.0556 m/s
banking anlgle = theta
friciton = 0
g = 9.8 m/s/s
radius of turn = r
theta = 20

From Force balance in vertical direction
Ncos(theta) = mg [ where N is the normal reaction]
From horizontal force balance
Nsin(theta) = mv^2/r
by dividing both
tan(theta) = v^2/rg
tan(20) = (18.0556)^2/r*9.80
r = 91.4071 m,

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