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A highway curve with radius 900.0 $\mathrm{ft}$ is to be banked so that a car traveling 55.0 mph will not skid sideways even in the absence of friction. (a) Make a free-body diagram of this car. (b) At what angle should the curve be banked?

$12.7^{\circ}$

Physics 101 Mechanics

Chapter 6

Circular Motion and Gravitatio

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

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the road is making an angle beata with the horizontal. So that's the angle beta that's the road is making from the horizontal. So if the car is around here, let's say that's our car. So if the cars around here, that means the normal force acting on the car is perpendicular to the ah road. So that's where the normal forces acting. And if we actually draw a preventable line from the horizontal, will get something like thiss and we can prove. And we and that's our Y axis, by the way, because if we consider this as our exact is and the perpendicular from exactly this will be the Y axis, so we can prove that this will be angle better as well. On the way, we prove that is for me. So we have the banking and ah, then we can actually pushed this line towards the center and then we'll have the same angle because the line which is here is parallel to this line. So that means this will form angle better as well. And if we are drawing a perpendicular from here, so this is basically 90. Mine is better and again if you're drawing a preventable from this line. That means this is 90 degree, and if this part is 19 miners better, so let's actually write that down. This is 19 minus better. So this angle right here will be 90 minus this part, which is 90 minus 90 minus pita, which gives us a beer. So yeah, so this will be seen as the one that's making in Horace of that making an angle from the horizontal? No, since where we need to draw the free body diagram. We see that this force can be can have toe components one in the white direction one the extraction. So let's actually draw that. So that's the component in extra Y direction and that there's a component in the extraction and let's call it and why and an ex. And if we know if they used a trick here, we can actually write and why equals two and co sign beta and in X equals two and sign better. And on top of it, since the car is making an arts, that means they're the radial acceleration towards this way, and we'll call it a radio and ah, again the car will have the gravitational pull. We must not forget that. So that means there's Ah, I'm so sorry about trying. There's ah, gravitation and pull acting directly downwards its will be m times c. So let's actually jor clean diagrams so that we can understand what's going on. So first off Ah, let's draw the X and Y axis. So this is our Y axis. This one is our exact is Now we have an ex. We have the normal force, which is making an angular beta from the vertical. Then we have the component off the normal force, which is and why? And then we have the component off the Norman force on extraction, which is end ex On top of it, we have the gravitational pull that's acting downwards. That's md. So that's the complete freeway diagram off the car. And also we shouldn't forget that we have a radio acting this way. All right. Ah, now, since we have sorted out that now we need to find out beta or the angle that the road needs to be so that the car doesn't slip. So let's first check the Y direction forces. So we have f y equals and y So that's a subway on DH. Then we have f sub X equals and sub X. And from there, if we first look at the extraction. So in extraction we have the radial acceleration. So from there we can get the force which we have mass times the rate of acceleration and then also will have the component of the normal force on extraction. So that will be and sign beta. It is equal to I am re squared over r because X is b squared over r in our case and we are live already mentioned that an ex is Ensign Vita. Yeah, and ah from here What we can do is wait can set why as well So anyways, in cause better which will be ableto mg so and course I in beta is equal to mg are from here we can write n equals two mg over co sign beta and we can use this term and plug it here and producer will get mg course I in beta time. Signed later equals toe em. Reece squared over r and now we can get rid of the EMS on. We need beta, so we know that sign made over cost. Vader gives us 10 better and we can take G on the right on will get me squared over Z times are so from here. If you want to solve for beta, that's gonna be 10. Inverse off. He's Greg over gr right now we're given the velocity is 55 miles per hour. We can actually convert that to ah, foot per second. And if we do so, we'll have 18 0.7 foot per second on DA Our gravity will be 32 put per second squared and our are given as 900 feet. So So this is the velocity, right? So now we know that 10 beta is equal to Visegrad over gr So for these great, it's going to be 80.7 foot per second squared, divided by G is 32 with per second squared. Then we have our which is 900 put. And from here we if we solved for us, we get 0.226 And from here to solve or better get an inverse Sarah point total six cities 12 0.73. Thank you

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