(II) How large must the coefficient of static friction be betw...

Question

(II) How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 85 m at a speed of 95 km/h?

Accepted Answer

Hey, everyone in this problem, we're asked, what is the static friction coefficient required between the wheels and the track for a vehicle to safely navigate a level turn with a radius of 80 m at a speed of 90 kilometers per hour. We're given four answer choices. Option A 0.8 option B 0.02 option C 3.33 and option D 5.68. So let's get started. And what we're gonna do is we're gonna draw a couple of different free body diagrams. So if we look at this kind of from a top down view and we have our vehicle, we're just gonna draw a circle and, and it's going around some turns, we can kind of mark that out in this dotted line. And as it's going around, the static friction force is actually gonna act towards the center of this circle that it's making. Ok. So as it turns kind of forming this center and that frictional force is going to act towards that center because it's opposing the motion and the motion. If we don't have enough static friction that this vehicle is gonna wanna make is to slide out off of this turn. OK. So, static friction acts towards the center and that is the direction of our centripetal force. OK. That centripetal force we call a to the middle, it's a center seeking force. And so we can think about that kind of a top down view. Um And then we can think of a regular free body diagram if we're looking at our car from the side or the front, and we're gonna have a normal force pointing upwards for our vehicle because it's on a surface and we have a force of gravity pointing downwards. OK. So what we're looking for is this static friction coefficient. So let's start with our forces in this centripetal direction. So the sum of the forces in the centripetal direction will be equal to the mass multiplied by this centripetal acceleration. I recall that standard equation. Now the sum of the forces in the centripetal direction, the only force we have acting is that force of static friction, it's acting in the positive direction. So we get FS on the left that's gonna be equal to the mass m multiplied by the centripetal acceleration. Now, we aren't given any information about the cental acceleration. We are, however, given information about the speed. So we're told that our vehicle is moving with a speed of 90 kilometers per hour. So instead of writing this in terms of our centripetal acceleration, what we can do is recall that the centripetal acceleration can be written as V squared that speed divided by R. OK. So the radius of this term, no, what that allows us to do is use the information we were given in the problem. We know V and we know R so now we know that entire OK. So we have the force of static friction on the left hand side, we want the static friction coefficient required. And so what we're gonna look at is the maximum static friction force. And if we have a maximum static friction force, we can recall that that is equivalent to the coefficient of static friction MS multiplied by the normal force. And that's gonna be equal to the mass multiplied by V squared divided by R. They were looking for us. Once again, we know V and we know R but now we need to figure out what this normal force N is. OK. And we can use that second free body diagram we drew in order to do that. So if we take a look at that second Freebo diagram, OK, are normal forces acting in the regular Y direction. The force of gravity is also acting in that direction and it's acting in a negative Y direction. So we can say that the sum of the forces in the Y direction, this is actually gonna be equal to zero. If this car is not moving up and down off of the road, it's staying on the road. It has no motion in the Y direction. So the sum of the forces is equal to zero. When we sum the forces, we get the normal force minus the force of gravity that's gonna be equal to zero, which tells us that the normal force is equal to the force of gravity. And recall that the force of gravity is given by the mass multiplied by the acceleration due to gravity. So we have N is equal to MG. OK. So we figured out a normal force, we can go back to our equation where we were looking at the centripetal force and we get new app multiplied by MG. Now substituting in what we found for that normal force is gonna be equal to M V squared divided by R the mass and we can divide on both sides, it's gonna divide out. OK. So the mass does not matter for our answer here and we can divide both sides by G as well. So we get that this coefficient of static friction us that we're looking for is going to be equal to V squared divided by G. Last thing we need to do is substitute in these values. On the right hand side, we know all of them. So we'll substitute them in. And so, and what we get is that this coefficient of static friction me us will be equal to 90 kilometers per hour. We have to be a little bit careful with that. One because we need our units to cancel up us has no units. It is dimensionless. So we need to make sure that our units match up. So we're gonna switch that speed V of 90 kilometers per hour into our standard unit of meters per second. So we're gonna multiply and we know that in a kilometer, we have 1000 m. So we multiply by 1000 m divided by one kilometer, the unit of kilometer will divide out and we multiply by one hour for 3600 seconds. Again, the unit of hour will divide out. Now we're left with meters per second. And when we work this out, we get that V is equal to 25 m per second. OK. So going back to our static friction coefficient calculation us is going to be 25 m per second all squared. OK. After converting that to our standard unit divided by the acceleration due to gravity 9.8 m per second squared, multiplied by the radius of this turn, which we're told is 80 m. And when we work this out, we get that us is equal to 0.7972. So that is that coefficient of static friction that is required to keep this car from slipping in this turn. If we compare this to the answer choices we were given and we round what we're gonna see is that the correct answer is option A this coefficient of static friction is about 0.8. Thanks everyone for watching. I hope this video helped see you in the next one.

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(II) How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius 85 m at a speed of 95 km/h?

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