(II) A 1250-kg car rounds a curve of radius 68 m banked at an ...

Question

(II) A 1250-kg car rounds a curve of radius 68 m banked at an angle of 14°. If the car is traveling at 85 km/h, will a friction force be required? If so, how much and in what direction?

Accepted Answer

Welcome back everyone. A 1251 kg vehicle takes a 73 m radius. Turn on a truck inclined at 15 degrees. If the vehicle moves at 86 kilometers per hour, will it require a frictional force if necessary? What will be the magnitude and direction of this force? A says the force is 2.2 multiplied by 10 to the third newtons. B 4.8 multiplied by 10 to the third newtons. C 6.3 multiplied by 10 to the third newtons and D 8.5 multiplied by 10 to the third newtons. No, so far, we know that our vehicle has a mass of 1251 kg. A radius of 73 m or a track is inclined at an angle theater of 15 degrees. And our vehicle moves at 86 kilometers per hour, which if we convert that to si units, meters per second, it's going to be equivalent to 23.89 m per second. Ok, I know from this information. We want to figure out two things first, we'd like to know if it requires a frictional force. Ok. And second, if it does, we want to figure out what the magnitude and direction of this force will be. So, let's answer that first question. Will it require a frictional force? Let's draw a diagram to help us better understand what's going on here. So we're gonna draw our incline. Ok. And we're gonna have our vehicle where a normal force is going to be exerted from the surface onto the vehicle while the vehicle's weight will be acting downwards as MG, our vehicle is inclined at the degrees and we also know that our normal force is sorry, our vertical force is at theta degrees to the normal force. Ok. Now, if we want to figure out if it requires a frictional force, first, let's try to figure all the horizontal force without friction because if the horizontal force without friction is greater than the centripetal force, then we know that it will not require a frictional force because the frictional force is what would be needed for the vehicle to roam the curve. So what do we know about both the horizontal and centripetal forces? Well, here from Newton's second law, we know that the horizontal forces, the sum of forces in the horizontal direction is going to be equal to the mass multiplied by the acceleration in the horizontal direction. And from our diagram, we know that there's going to be a horizontal component of the normal force acting here which is going to be the normal force multiplied by the sine of theta. Now, we're on our way to finding the horizontal force, but we don't know what the normal force is. Do we know anything else that can help us? Well, let's consider the vertical component. We know that according to Newton's second law, the sum of the vertical forces are going to be equal to zero because there is no acceleration in the vertical direction. And if we resolve those components notice that normal force has a vertical component and MG is also a vertical component acting downwards. So the sum of the forces in the Y direction will be FN cas theta minus MG being equal to zero. From this equation, we can solve for our normal force because no, if we add MG to both sides and divide by cost, the the normal force equals MG divided by cost data. And now we can substitute that into our formula for our horizontal force because now this tells us then that based on what we have here, our horizontal force is going to be equal to MG divided by cost theta multiplied by sine theta time the divided by cost, theta is tan theta. And now we know our mass is 1251 kg. Let's take G to be 9.8 m per second squared. And we know our angle is 15 degrees. Thus, our horizontal force is going to be 3.285 multiplied by 10 to the third newtons. So we know what our horizontal force is. What about our central pal force? Well, recall that the centripetal force equals MV squared divided by R and we're told that our mass is 1251 kg. Our speed is uh earlier, we said it was 23.89 m per second. So all of that is gonna be squared and then we divide that by our radius of 73 m to give us a centripetal force of 9.7797 multiplied by 10 to the third new task. So what does our result tell us our centripetal force is greater than our horizontal force. The horizontal force is not enough. The vehicle requires more force that can be supplied by the normal force to roam the curve to provide that extra centripetal force, that extra force must have a horizontal component to the right of our diagram and the frictional force points downwards on the plane. Therefore, the vehicle cannot make the curve without friction. So now that we know that a frictional force is necessary, we need to figure out what it is and we can do that by essentially doing the same thing, but no accounting for our frictional force. So let's take our diagram. It's gonna be the same diagram basically. But no, we're going to account for our frictional force on our diagram that we can represent here. OK. As FFR, no, adding a frictional force. Well, let me write this in red, adding a friction of where again, let's try to resolve our components in the X and Y direction. Let's start in the X direction. We know our centripetal force equals MV squared divided by R. And that centripetal force should be the horizontal force. And if we think about our forces in the horizontal direction, it will be the X component of the normal force. So let me, let me just make a note here. So this is our forces in the X direction, we should put that in red. So that's going to be the X component of our normal force. F and sine theta plus the X component of our fictional force F class theta. And that is gonna be equal to MV squared divided by art. What about our forces in the Y direction or for the Y axis? Well, the same holds true. The forces in the Y direction will be equal to zero. So that means the Y component of the normal force FN plus theta minus or weight MG minus the vertical component of the frictional force, which is F sine theta is going to be equal to zero. Like our previous problem. Now, we can solve our normal force and substitute it into our equation for the horizontal forces to solve for that frictional force. Because no, this tells us, this tells us that our normal force is going to be equal to MG plus or frictional force multiplied by sine theta all divided by the cosine of theta. We add it to both sides and then we divide the entire creation by class. The know that we have an equation for our normal force. OK. Let's substitute it into our equation for our horizontal forces because now this implies then, yeah, that FN sine theater is now going to become MG plus F sine theta divided by the cosine of theta multiplied by sine theta plus F class theater being equal to MV squared divided by R. Now, here's the thing. Let's multiply through our entire equation by cost the HM Now when we do that, our equation now becomes MG plus F sign theater. OK. Multiplied by sound theater. Yeah, plus our frictional force equals MV squared divided by our cost theta if we expand sine theta here, OK. That means MG sine theta plus F sine squared theta. OK. My apologies here in our previous equation, we should have had cost the multiplied by cost theta which is cost squared, theta, my apologies. OK. Plus our frictional force multiplied by that expression being equal to MV squared divided by R cost data. No, no, this here we can isolate for a fictional force. So it's gonna be sin square, theta plus cos square theta equals MV square divided by R KS theta minus MG sine theta. We're going a little bit deep into the math here and no recall from our Trigon trick, identity sine squared, theta plus co square theta equals one. So that means our frictional force is really MV squared divided by R class theta minus MG sine theta. Now we can go ahead and solve for our frictional force because if we factor out to our mass and substitute the rest of the values, we know that's going to be 1251 kg multiplied by 23.89 m per second or squared. That was our, that was our or speed view divided by a radius of 73 m multiplied by the cosine of 15 degrees. OK. Minus G which is 9.8 m per second squared multiplied by the sine of 15 degrees. Now, we can go ahead and solve for fictional Force. And when we do that to two significant figures, we get our fictional Force to be 6.3 approximately 6.3 multiplied by 10 to the third newtons where the fictional Force acts horizontally toward the center of the term. Therefore, if we go back to our answer choices, that means see is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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(II) A 1250-kg car rounds a curve of radius 68 m banked at an angle of 14°. If the car is traveling at 85 km/h, will a friction force be required? If so, how much and in what direction?

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