How Does Static Friction Affect a Car on a Curve?

[.NoStyle]

Homework Statement

A car drives around a curve with radius 410 m at a speed of 32m/s. The road is not banked. The mass of the car is 1400kg.

A. What is the frictional force on the car?

B. Does the frictional force necessarily have magnitude μsN? Explain.

Homework Equations

I'm not quite sure, I assume that:

fs,max = μsN

though I don't know how to find the coefficient of static friction since I don't know fs,max.

The Attempt at a Solution

I don't even know where to begin to find the coefficient of static friction... any hints would really help. Thanks

 

[dynamicsolo]

You won't need to know the coefficient of friction for this problem. What you do need to know is that static friction is what is holding the car on its turn (after all, what horizontal forces are there?).

Keep in mind that centripetal force is not a physical force, but the net force that causes an object to follow uniform circular motion. What is/are the physical force(s) providing the centripetal force on the car?

For part (b), you also don't need to know mu_s; you simply are being asked if it is necessary to be at the maximum static frictional force in order to make a turn.

 

[ace123]

I believe the centripital force will equal the force of friction since nothing else is acting the horizontal direction and you can find Fn and solve for the static coeffecient

 

[ace123]

O yea you don't need the coeffecient for this. My mistake

 

[.NoStyle]

You won't need to know the coefficient of friction for this problem. What you do need to know is that static friction is what is holding the car on its turn (after all, what horizontal forces are there?).

Keep in mind that centripetal force is not a physical force, but the net force that causes an object to follow uniform circular motion. What is/are the physical force(s) providing the centripetal force on the car?

For part (b), you also don't need to know mu_s; you simply are being asked if it is necessary to be at the maximum static frictional force in order to make a turn.

thanks dynamicsolo,

so if I'm understanding correctly, part A is asking for the physical forces that are providing centripetal force on the car? So this would be, N + mg + fs ??

And for B, you would not need to be at maximum static frictional force to make a turn because that is just the safety limit. Anything above would be safer than the fs,max. Is that correct?

 

[.NoStyle]

O yea you don't need the coeffecient for this. My mistake

thanks ace123 for the help.

 

[dynamicsolo]

so if I'm understanding correctly, part A is asking for the physical forces that are providing centripetal force on the car? So this would be, N + mg + fs ??

Which of these (N, mg, f_s) are horizontal?

And for B, you would not need to be at maximum static frictional force to make a turn because that is just the safety limit. Anything above would be safer than the fs,max. Is that correct?

Let's look at it this way: once we sort out the issue of what is providing the centripetal force, we can figure out what the coefficient of static friction would be for which the car would just hold the turn. If the actual coefficient between tires and pavement is larger than that, then the surfaces will not be exerting maximum static friction while the car is turning. The car would hold the turn without being on the verge of skidding. If the actual coefficient were less that the value we find, the maximum static friction would be exceeded and the car could not avoid skidding out of the turn.

 

[ace123]

Fn and mg are acting in the vertical. They are not providing the centripital force.

Edit: Posted one second before me dynamic

 

[.NoStyle]

ok thanks guys, so fs is the frictional force I'm looking for. I understand now!dynamicsolo, so when it asks, "Does the frictional force necessarily have magnitude μsN? Explain."

so fs,max = μsN

sorry if I'm being redundant, but it seems like you're confirming that you don't need to be at maximum static friction to make a turn. Anything above will allow you to safely make the turn without losing grip with the road. Sorry if I'm being confusing.

thanks for your help

 

[dynamicsolo]

sorry if I'm being redundant, but it seems like you're confirming that you don't need to be at maximum static friction to make a turn. Anything above will allow you to safely make the turn without losing grip with the road.

I think it's the phrase "anything above" that might be confusing. I'm presuming that you mean any speed that would require a higher centripetal force than maximum static friction can provide will lead to a skid, which would be correct. (EDIT: On reading your statement again, another way of saying this would be that if the coefficient of static friction is larger than the critical value worked out below, then the amount of static friction required to hold the car in the turn is less than the maximum friction, so the car won't skid.)

If we look at the situation where this turn would take place at the static frictional limit, we have

F_cent = m(v^2)/R = F_s_max = (mu_s) · mg ,

giving us

mu_s = (v^2) / gR .

For the values in your problem, we find mu_s = 0.255 . For present-day tires on dry highway pavement, the coefficient is probably more like 0.6 or so , so this car isn't very close to reaching the limit of static friction. (If the pavement were icy, the coefficient would be rather lower and the car could be close to skidding on this turn at highway speeds.)

Notice, by the way, that the coefficient we found has no dependence on the mass of the vehicle. Any vehicle having the same coefficient of friction with the road will skid on the same turn at the same speed.

 

[.NoStyle]

thanks dynamicsolo!