If the coefficient of static friction between a car’s tires an...

Question

If the coefficient of static friction between a car’s tires and the pavement is 0.65, calculate the minimum torque that must be applied to the 66-cm-diameter tire of a 1250-kg automobile in order to “lay rubber” (make the wheels spin, slipping as the car accelerates). Assume each wheel supports an equal share of the weight.

Accepted Answer

Welcome back. Everyone in this problem for a 951 kg vehicle with 34 centimeter diameter tires and a static friction coefficient of 0.66 between the tires and the road, calculate the minimum torque needed to cause the wheel to spin and slip under equal weight distribution across all tires. For our answer choices. A says it's 1.2 multiplied by 10 square meter newtons. B 2.6 multiplied by 10 square meter, Newtons. C 4.3 multiplied by 10 square meter, Newtons and D 5.2 multiplied by 10 squared meter newtons. Now first, let's make sure we understand what's going on here. So we're talking about tires uh on the road. Ok. For a 951 kg vehicle. Ok. I know we know that the weight of the vehicle is supported by four tires equally. Ok. So that means that the normal force on each tire is going to be a quarter of the width of the vehicle. In other words, our normal force is going to be a quarter of mg. We already know here that each tire has a diameter of 34 centimeters So that means its radius is a half of 34 or 17 centimeters. Finally, we also know that there's torque applied here. Ok. So we have our applied torque and for our surface here, our friction is going in the direction of motion because our wheel is not slipping. And now we want to figure out the minimum torque needed to cause the wheel to spin and slip under that equal weight distribution. Now, for the tires to be on the verge of slipping, there must be an applied torque which is equal to the torque supplied by the static friction. So to satisfy that condition then OK, we're here, we have our friction. Then at that point, our applied torque applied talk must be equal to our to due to static friction. OK. Now what do we know about both of those that can help us? Well, first of all, let's start with our applied torque recall. OK. That our torque is the rotational force of our object over a distance defined as a radius multiplied by our force. OK. And we also know that our frictional force, frictional force is equal to our coefficient of static friction multiplied by our normal force. So for our applied torque here, since on the verge of slipping, it's equal to the torques applied by static friction. That force is going to be the force due to friction. OK. In other words, or applied torque, this tells us then that our applied torque is going to be equal to the radius multiplied by our frictional force. Remember, our frictional force equals mu multiplied by our normal force. So that's R multiplied by Mu FN. And we know that our normal force is a quarter of mg. So that means our applied torque then is going to be equal to a quarter of MG multiplied by mu R. Nor do we have all this information here. Well, yes, we know the mass of our, the mass of our vehicle. OK. We know our coefficient of static friction is 0.66 and we know the radius of the tire is 17 centimeters and we can take G or acceleration due to gravity to be 9.8 m per second squared. So let's substitute all these values to sulfur or applied torque because no, that means it's going to be a quarter of 951 kg multiplied by 9.8 m per second squared or acceleration due to gravity multiplied by 0.66 or static coefficient multiplied by 0.17 m. OK? Or the radius of our wheel. Ok. So now we can go ahead and solve for our applied force or applied torque rather. Now when we do that, we get our applied torque to be equal to 2.6 sorry 2.6. OK. 2.6 multiplied by 10 square meter newtons. Thus our applied torque is 2.6 multiplied by 10 squared meter newtons. That means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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