Physics - Further Centripetal Force/Friction Question with Example

[sciencebooks]

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A 1000 kg car wants to round a curve on a flat road of radius 100 m at a speed of 72 km/hr (~20 m/s) without slowing down. What will happen to the car if the pavement is dry and the coefficient of static friction is 0.60?
A.The car will make the turn only if it speeds up.
B.The car will make the turn if it stays the same speed.
C.The car will make the turn only if it slows down.
D.There is not enough information to determine what will happen to the car.

Answer/Explanation:
C. The net force acting on the car is also known as Fc (centripetal force with acceleration directed towards the center of the circle) provided by friction. We must compare the net force required for the car to make the turn compared to the maximum frictional force that the pavement can provide. Fnet = Fc = mv2 / r = (1000)(20)2 / 100 = 4000 N. Maximum static friction = μsFN = μsmg = (0.6)(1000)(10) = 6000 N. The maximum force of static friction is strictly greater than the force needed for the centripetal force. As such, the car will make the turn.

I feel like the explanation goes against the answer?

Also, in general, for a turn to occur in such a scenario, we want static friction to be greater to or equal to the centripetal force required, correct?

Thanks!

 

[MissionStanford]

Fc = Ff = uN

Centripetal force equals the force of friction, which equals the coefficient of static friction times the normal force.

The normal force equals the weight since we're not accelerating up or down.

W = mg = (1,000kg)(10m/s^2) = 10,000N

Ff = uN = (0.6)(10,000N) = 6,000N. This is the centripetal force provided.

We also know another formula for centripetal force.

Fc = mv^2 / r = (1,000kg)(20m/s)^2 / 100m = 4,000N

Since the centripetal force required is only 4,000N, but since we have a centripetal force of 6,000N, we're providing too much force, so I believe we would need to decelerate (slow down) to decrease the amount of force provided to only 4,000N in order to make the turn.

Thus, I believe C is the correct answer.

 

[Pre-Med Village]

Friction must deliver 4000N of centripetal force to keep the car on the circular path at the speed given (mv^2/r). The threshold of static friction is 6000N (0.6 X Normal Force) so the car should be able to make the turn. It looks to me like the answer is B.

However, in your post, you don't give the Answer/Explanation of the question writer. Is the answer I reached consistent with their explanation?

 

[MissionStanford]

Friction must deliver 4000N of centripetal force to keep the car on the circular path at the speed given (mv^2/r). The threshold of static friction is 6000N (0.6 X Normal Force) so the car should be able to make the turn. It looks to me like the answer is B.

However, in your post, you don't give the Answer/Explanation of the question writer. Is the answer I reached consistent with their explanation?

Whoops. That makes sense. B sounds right, not C. We're giving 4,000N, but we can deliver up to 6,000N. Since we're not exceeding the static friction, the car should be able to travel in a circle.

 

[sciencebooks]

Friction must deliver 4000N of centripetal force to keep the car on the circular path at the speed given (mv^2/r). The threshold of static friction is 6000N (0.6 X Normal Force) so the car should be able to make the turn. It looks to me like the answer is B.

However, in your post, you don't give the Answer/Explanation of the question writer. Is the answer I reached consistent with their explanation?

It's in white! Sorry, I should make a note for people to highlight it but I put it in white so people could try it on their own first, 🙂.

But yes, I agree with you and their explanation seem to too... Mistype on their part?

 

[MissionStanford]

It's in white! Sorry, I should make a note for people to highlight it but I put it in white so people could try it on their own first, 🙂.

But yes, I agree with you and their explanation seem to too... Mistype on their part?

Looks like it was a mistype on the book's part. The last part of the answer seems to indicate that choice B is correct.

 

[Mornhavon]

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[MangoPlant]

Are you sure it's a mistype?

The question mentions the co-efficient of static friction, not kinetic, which would be lower and therefore closer to the Fc of the car...meaning you'd have to slow down...

But the tires of a car use the coefficient of static friction because the point of contact between the tire and the road does not move if the tire does not slide.

 

[Mornhavon]

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