$4.14^\circ$

### Solution video

# OpenStax College Physics, Chapter 6, Problem 25 (Problems & Exercises)

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*theta*and we can use normal force multiplied by sine of

*theta*to find this component in the x direction or radial component. We know this is

*theta*because you can imagine this line here parallel to the ramp and then you can imagine another line here which is horizontal and this would be

*theta*because these would be interior opposite angles. With that being

*theta*and this being 90 because this is a normal force which is perpendicular to the surface of the ramp, this dot here plus

*theta*has to add up to 90. But this dot here plus whatever this angle is here, let's assume we don't know it's

*theta*yet, also has to make 90 since this

*F n y*is pointing straight up and that makes a 90 degree angle there. So if this dot plus

*theta*makes 90 and this dot plus the angle in here also makes 90, that means this must equal that and so it is

*theta*in there. All right. So the y component of the normal force is the normal force multiplied by cosine

*theta*because the y component is the adjacent leg of this triangle. We know that that has to balance gravity. So that's vertical, this

*F n y*is vertically upwards and so it has to balance the

*m g*force of gravity downwards. That means

*F n*cos

*theta*is

*m g*. We can solve for

*F n*by dividing both sides by cos

*theta*. That gives

*F n*is

*m g*over cos

*theta*and the reason that's useful is we can substitute that back into this formula replacing

*F n*in the

*F n x*formula in order to well, eventually solve for

*theta*. But let's take it one step at a time. So re-writing

*F n x*and noticing that it is the centripetal force, in which case it must be

*m v*squared over

*r*because that's the formula for centripetal force. Now we can say that

*m v*squared over

*r*is this expression we had before for

*F n x*which is

*F n*sine

*theta*. But now we'll replace

*F n*with what we figured out in this part here in green,

*m g*over cos

*theta*times sine

*theta*

*. Now sine**theta*over cos*theta*can be written as tan*theta*and we'll divide both sides by*m*as well. We have*v*squared over*r*equals*g*tan*theta*. Then divide both sides by*g*and you get tan*theta*is*v*squared over*r g*. That means*theta*is the inverse tangent of*v*squared over*r g*. So that's the inverse tangent of 105 kilometers per hour converted into meters per second and then square that, and divide by 1.2 kilometers converted into meters, times 9.8 meters per second squared. This all gives 4.14 degrees is the ideal banking angle for this particular speed and this radius of curvature. This assumes there is no friction by the way.## Comments

Are the answers in all problems shown with the correct number of significant figures??

Hi rileyhubsmith, thank you for the question. Yes, the final answers are rounded to the correct number of significant figures.

All the best,

Shaun

Why isn't friction taken into account? If there were no friction the car would slide off the road, correct?

Hi e_than, thanks for the question. From here in the textbook: *In an “ideally banked curve,” the angle ? is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road.*

I agree that it seems counter-intuitive that a car could negotiate a turn with **zero** friction. Consider what friction force would provide if it existed in this scenario? It would give a force perpendicular to the direction of motion that would change the direction of motion - make the car move along a curve. This curve is imagined to be part of a circle with the car velocity tangent to the circle and the force perpendicular to the velocity. The force, being perpendicular to velocity is directed toward the center of the circle (we know this because the tangent/radius angle is ninety degrees. The force is along the circle's radius in otherwords). Since the force is toward the center it is called **centripetal**.

The horizontal component of the normal force provided by the banked curve could be described the same way as a hypothetical friction force. The horizontal component of the normal force is directed perpendicular to the car's velocity, causing it to go along an arc of a circle (causing it to go around the turn, in other words). If the horizontal component of the normal force is large enough to provide make the car follow the turn, then friction isn't necessary. In the real world, friction is normally needed to supplement the horizontal component of the normal force.

Hope that wasn't too wordy and helps,

Shaun