A car (whose mass is 1500 kg) rounds a sharp curve (of...

A car (whose mass is 1500 kg) rounds a sharp curve (of radius 60 m) at a constant speed of 20 m/s. The coefficients of static and kinetic friction between the tires and the road are 0.70 and 0.50, respectively. (a) [5 pts.] Draw a free body diagram of the car. What force(s) give(s) it its centripetal acceleration? (b) [10 pts.] Can the car make it around the curve without skidding? (c) [5 pts.] Find the magnitude of the frictional force on the car. (d) [5 pts.] Suppose now that the car goes into the curve fast enough so that it does skid. What will be the direction of the frictional force on the car? Describe the path that the car will then follow. (e) [5 pts.] Suppose instead that the curve is banked as shown. Now what force(s) give(s) the car its centripetal acceleration?

Transcript

00:01 Hello students here the mass radius and velocity of a car is given and this is the coefficient of static friction and coefficient of kinetic friction.

00:13 For the first question they have asked which force causes it to move or which course gives the car a centripetal acceleration.

00:24 So first we find out the centipatel acceleration that is as equal to b squared.

00:34 By r so here v is 20 squared by r is 60 you get 6 .67 meter per second squared therefore here the frictional force acting between the tires and road use the centripetal acceleration and for the second part the car can make around the curve without skidding if it is possible only if force due to friction is greater than force due to centipatal force that is frictional force is greater than centripetal force then only the car can make around the curve without skidding so first calculate the centipatel force fc that is mv squared by r here m is 1 ,500 and v square by r is 6 .67 so this will give you force value as 10 ,000 5 newton and force due to friction is equal to force due to kinetic friction so that is 0 .70 into 2500 into 9 .8 this will give you 10 ,090 newton here force due to friction is greater than force due to the centipital force so therefore the car can make around the curve make around the curve without skidding for the third part of the question they have asked the magnitude of frictional force first magnitude of fictional force due to static friction that is given by new s m g that we have calculated as 10 ,290 newton in the first part and force due to kinetic friction this is given by new k into m g that is 0 .5 into 500 into 9 .8.

03:44 This will give you 7350 newton and for fourth part if the car moves fast if the car moves fast and it does it so in this care situation the direction of frictional force direction of frictional force is opposite to the direction of motion of the car opposite to the direction of motion of the car then in this case the car makes a straight line path car makes the straight line path and for fifth part of the question if the curve is banked then the component of normal force which gives centripetal acceleration which gives centripetal acceleration is perpendicular to the surface of the road perpendicular to surface of the road...

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