#1

Here's the question. Its for an online homework assignment: A 700 kg car travels at 90 km/h around a curve with a radius of 150 m. What should the banking angle of the curve be so that the only force between the pavement and tires of the car is the normal reaction force?

I already tried breaking the components of Normal Force into Fn*Sinx = m*V^2/r and Fn*Cosx=m*g. When you combine the equations you get Tanx = V^2/(r*g) or x = InverseTan( V^2 / (R*g) where V is 90 R is 150 and g is 9.81.

The answer I got was about 79.71 degrees, which is a hell of a bank it seems, plus my homework program said it was wrong... Any idea what I'm doing wrong?

I already tried breaking the components of Normal Force into Fn*Sinx = m*V^2/r and Fn*Cosx=m*g. When you combine the equations you get Tanx = V^2/(r*g) or x = InverseTan( V^2 / (R*g) where V is 90 R is 150 and g is 9.81.

The answer I got was about 79.71 degrees, which is a hell of a bank it seems, plus my homework program said it was wrong... Any idea what I'm doing wrong?

#2

I'm taking physics, but we havn't quite gotten there yet... sorry I'm not any help, some aspects of that sound familiar, but unfortunately not enough of it.

Good luck.

Good luck.

#3

Asking the pit is what you're doing wrong.

#4

That's a pretty wide turn for such a low speed. No idea how to solve, but maybe you got the equations all right but you got the inverse. In other words, maybe a bank of 11.29 degrees?

#5

think of it as a centripetal motion problem and a banking curve problem. as the curve goes up, the more friction it puts against the curve. so say you figure out the normal force and the acceleration down the slope. the acceleration down should be equal or greater than the centripetal ("centrifugal inertia") force,

#6

It's obviously 111 miles per day on an average of six.

#7

We just finished that section, the question's a bitch. Do you guys use the Giancoli book?

#8

I got the angle to be 23 degrees. The only problem I see is that you used 90 km/h instead of 25 m/s.

#9

I got 25.16 degrees.

Centripetal force is equal to mgsin(theta)

mv^2/r = mgsin(theta)

solve for theta.

In this case the centripetal force is supplied completely by gravity and thus no friction is needed. So between the tire and the road there is only a normal force.

I think...

Centripetal force is equal to mgsin(theta)

mv^2/r = mgsin(theta)

solve for theta.

In this case the centripetal force is supplied completely by gravity and thus no friction is needed. So between the tire and the road there is only a normal force.

I think...

*Last edited by AznElliot518 at Feb 2, 2009,*

#10

It crashes so it's irrelevant.

#11

I got 25.16 degrees.

Centripetal force is equal to mgsin(theta)

mv^2/r = mgsin(theta)

solve for theta.

In this case the centripetal force is supplied completely by gravity and thus no friction is needed. So between the tire and the road there is only a normal force.

I think...

If I remember this chapter correctly, the force supplying the radial acceleration would be Fnsin(theta), since the centripetal force acts horizontally, not down the incline.

#12

Hmm you may be right. I sorta BSed that chapter :S

#13

I got the angle to be 23 degrees. The only problem I see is that you used 90 km/h instead of 25 m/s.

That was it. Good work

#14

That was it. Good work

Physics is the only reason I care to go to school anymore xD

#15

2009, cuz that's the year it is.