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

  • Thread starter Thread starter Maiia
  • Start date Start date
  • Tags Tags
    Car Friction
AI Thread Summary
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.
Maiia
Messages
78
Reaction score
0

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?
 
Physics news on Phys.org
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! :)
 
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top