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Learning Goal: To set up and evaluate the equations of motion in a normal-tangential coordinate system. A car of weight 11.5 kN is traveling around a curve of constant curvature ρ. A - The car is traveling at a speed of 25.5 m/s, which is increasing at a rate of 1.75 m/s^2, and the curvature of the road is ρ = 190 m. What is the magnitude of the net frictional force that the road exerts on the tires? Express your answer to three significant figures and include the appropriate units. B - Suppose that the tires are capable of exerting a maximum friction force of 4630 N. If the car is traveling at 17.5 m/s and the curvature of the road is ρ = 150 m, what is the maximum tangential acceleration that the car can have without sliding? Express your answer to three significant figures and include the appropriate units. C - Suppose that the tires are capable of exerting a maximum net friction force of 3390 N. If the car is traveling at 22.5 m/s, what is the minimum curvature of the road that will allow the car to accelerate at 1.65 m/s^2 without sliding? Express your answer to three significant figures and include the appropriate units.

          Learning Goal:
To set up and evaluate the equations of motion in a normal-tangential coordinate system.
A car of weight 11.5 kN is traveling around a curve of constant curvature ρ.
A - The car is traveling at a speed of 25.5 m/s, which is increasing at a rate of 1.75 m/s^2, and the curvature of the road is ρ = 190 m. What is the magnitude of the net frictional force that the road exerts on the tires? Express your answer to three significant figures and include the appropriate units.
B - Suppose that the tires are capable of exerting a maximum friction force of 4630 N. If the car is traveling at 17.5 m/s and the curvature of the road is ρ = 150 m, what is the maximum tangential acceleration that the car can have without sliding? Express your answer to three significant figures and include the appropriate units.
C - Suppose that the tires are capable of exerting a maximum net friction force of 3390 N. If the car is traveling at 22.5 m/s, what is the minimum curvature of the road that will allow the car to accelerate at 1.65 m/s^2 without sliding? Express your answer to three significant figures and include the appropriate units.

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Added by Jennifer D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vishal Gupta

05/25/2022

Step 1

The centripetal force is given by Fc = m*v^2/r, where m is the mass of the car, v is the speed of the car, and r is the radius of curvature of the road. The tangential force is given by Ft = m*a, where a is the tangential acceleration of the car. The net Show more…

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Learning Goal: To set up and evaluate the equations of motion in a normal-tangential coordinate system. A car of weight 11.5 kN is traveling around a curve of constant curvature ρ. A - The car is traveling at a speed of 25.5 m/s, which is increasing at a rate of 1.75 m/s^2, and the curvature of the road is ρ = 190 m. What is the magnitude of the net frictional force that the road exerts on the tires? Express your answer to three significant figures and include the appropriate units. B - Suppose that the tires are capable of exerting a maximum friction force of 4630 N. If the car is traveling at 17.5 m/s and the curvature of the road is ρ = 150 m, what is the maximum tangential acceleration that the car can have without sliding? Express your answer to three significant figures and include the appropriate units. C - Suppose that the tires are capable of exerting a maximum net friction force of 3390 N. If the car is traveling at 22.5 m/s, what is the minimum curvature of the road that will allow the car to accelerate at 1.65 m/s^2 without sliding? Express your answer to three significant figures and include the appropriate units.

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Transcript

00:01 Hi, here this problem is based upon the motion of an object over a circular turn.

00:08 This is the circular turn, the horizontal circular turn it may be.

00:13 And a car moving over it like this.

00:19 The car is having an accelerated motion.

00:23 So it will be having a tangential force as well the centrifugal force acting on it.

00:35 Radially outward f c so the resultant force will be like this so the force of friction it should be equal and opposite to this resultant force f f weight of the car that is given as 11 .5 kilo newton or we can say this is 1150 and then using an expression for the weight we get the value of the mass of the car w by g means this is 11 5500 newton divided by g 9 .8 meter per second square and finally it comes out to be equal to mass of the car is equal to 1173 .5 kilogram okay now in the first part of the problem, as we have seen, force of friction is equal to the net force and that will be found using vector addition.

01:50 The forces are having an angle of 90 degree between them.

01:53 So we will use pythagoras theorem and that is f c square plus f t square.

02:00 Then expression for centrifugal force that is m into v square by r whole square plus f m .a using newton's second of motion m .a.

02:14 Now plugging in all the known values here for the mass of the car that is 1173 .5 into square of speed and in this case in the first case the speed of car is given as 25 .5 square aperture of this path that is 190 meter and having a a square of it plus again mass 117 .3 .5 multiplied by acceleration and tangential acceleration is 1 .75 meters per second is square.

02:56 Having a whole square of it.

02:58 So finally it comes out to be equal to the force of friction which should be acting here that will be given by square root of the huge number here, the big number, 161 ,28.

03:14 614 .03 plus 4217 -37 .64 .64 and then adding them and then taking a square root.

03:27 Finally we get 4 ,510 .6 newton is the force of friction and then to convert it into three significant figures only as required, it may be written as 4 .5 .5.

03:43 1 kilo.

03:45 Answer for the first part of the problem...

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