What Is the Correct Banked Curve Angle for a Car Traveling at 64 km/hr?

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To determine the correct banked curve angle θ for a car traveling at 64 km/hr on a curve with a radius of 50 m, it's essential to convert the speed to meters per second. The relevant formula is tan(θ) = v²/(r*g), where g is the acceleration due to gravity (9.8 m/s²). The initial calculations mistakenly suggested an angle of approximately 90 degrees, but the error stemmed from not converting the speed units. After correcting the units and applying the formula, the correct angle can be calculated. Proper unit conversion is crucial for accurate results in physics problems.
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Homework Statement


A curve of radius r is banked at angle θ so that a car traveling with uniform speed v can
round the curve without relying on friction to keep it from slipping to its left or right.
The acceleration of gravity is 9.8 m/s2. If r = 50 m and v = 64 km/hr, what is θ?
Answer in units of ◦.

For this one, I drew a FBD. In my picture, Fny and Fg canceled each out, so the unbalanced force was Fnx. I used N2nd Fnet=ma and used components:
Fnx=max
Fnsintheta=max
Fny= mg
Fncostheta=mg
then i divided them:
Fsintheta max
--------- = --------
Fcostheta mg
The masses canceled out on the right and on the left I had tantheta.
on the right, ax/g= v^2/rg
When I plugged in numbers, I got theta= approx 90 degrees. However, when I submit this answer, it is wrong..could someone help me find out where I messed up?
 
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It's hard to follow your work but it looks like your resultant equation is correct: tan(x) = v^2/rg. Make sure to convert the km/hr to m/s before doing the calculation.
 
Maiia said:

Homework Statement


A curve of radius r is banked at angle θ so that a car traveling with uniform speed v can
round the curve without relying on friction to keep it from slipping to its left or right.
The acceleration of gravity is 9.8 m/s2. If r = 50 m and v = 64 km/hr, what is θ?
Answer in units of ◦.

For this one, I drew a FBD. In my picture, Fny and Fg canceled each out, so the unbalanced force was Fnx. I used N2nd Fnet=ma and used components:
Fnx=max
Fnsintheta=max
Fny= mg
Fncostheta=mg
then i divided them:
Fsintheta max
--------- = --------
Fcostheta mg
The masses canceled out on the right and on the left I had tantheta.
on the right, ax/g= v^2/rg
When I plugged in numbers, I got theta= approx 90 degrees. However, when I submit this answer, it is wrong..could someone help me find out where I messed up?

The force down the bank will be m*g*sinθ and the force from centrifugal acceleration is your m*v²*cosθ/r

That means that you have v²/(r*g) = tanθ

So to find θ that would be the tan-1( v²/(r*g)) = θ
 
oh that would be the problem, not converting the units >< thanks to both of you for your help! :)
 
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