Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 5, Problem 90P

(a)

To determine

To Find:The force of static friction exerted on the car by the runway surface.

The minimum coefficient of static friction necessary for the car to sustain this speed.

(a)

Expert Solution
Check Mark

Answer to Problem 90P

  500N ; 0.0255

Explanation of Solution

Given Information:

Mass of the car, m=2000kg

Air drag on the car, FD=500N

Speed of the car, v=100km/h

Formula Used:

Newton’s second law:

  Fnet=ma

Frictional force

  fs=μsN

Calculation:

The following is the free body diagram of the car as the car travels on the runway.

  Physics for Scientists and Engineers, Chapter 5, Problem 90P , additional homework tip  1

Assume the mass of the car as M . In the above figure, normal force on the car is N , the frictional force between the car and the road is, fs and the drag force is FD .

Convert the units of the speed of the car from km/h to m/s

Speed of the car, v=100km/h

  =(100 kmh)( 10 3 m 1km)( 1 h 3600s)=27.8 m/s

Since, the car is moving with constant velocity, the acceleration of the car is zero.

Net force acting on the car along z axis,

  Fnet,x=MaxfsFD=0...(1)fs=FD

Substitute 500 N for FD in equation (1) to solve for the static frictional force exerted by the runway on the car, fs=500N

Thus, the required static frictional force on the car is 500N .

Net force acting on the car along y axis:

  Fnet,y=MayNMg=0N=Mg

Rewrite equation (1)

  fs=FDμsN=FD       [Since, N=Mg]μsMg=FDμs=FDMg

Here μs is the coefficient of static frictional force

Substitute 500 N for FD and 2000 kg for M in this equation

  μs=500 N( 2000kg)( 9.81 m/s 2 )=0.0255

Conclusion:

Thus, the required coefficient of static frictional force is 0.0255 .

(b)

To determine

To Find: The minimum coefficient of static friction necessary for the car to hold the last radius of curvature without skidding.

(b)

Expert Solution
Check Mark

Answer to Problem 90P

The radius of the curve at θ=45ois 3.09 km .

The radius of the curve at θ=88ois 108 m .

The required coefficient of friction for the car without skidding is 0.73 .

Explanation of Solution

Given Information:

Mass of the car, m=2000kg

Air drag on the car, FD=500N

Speed of the car, v=100km/h

Formula Used:

Newton’s second law:

  Fnet=ma

Frictional force

  fs=μsN

Calculation:

The following is a free body diagram of the car on a circular path.

  Physics for Scientists and Engineers, Chapter 5, Problem 90P , additional homework tip  2

In the above figure, r represents the radius of the curve, fs represents the static frictional force exerted by the road on the tires of the car and θ is the angle made by the frictional force with the speed of the car.

Apply Newton’s second law to the motion of the car and calculate the net force acting on the car

The radial force acting on the car along x-axis.

  Fx=Macfssinθ=Mv2r....(2)

Here, ac is the centripetal acceleration and v is the speed of the car.

The tangential force acting on the car along x-axis.

  F tan=Ma1fscosθFD=0....(3)fscosθ=FD

Divide equation (2) by equation (3) .

  tanθ=Mv2rFD

Rearrange this equation for r .

  r=Mv2tanθFD..........(4)

Substitute 500N for FD,2000 kg for M,45o for θ and 27.8m/s for v in this equation,

  r=Mv2tanθFD=( 2000kg) ( 27.8m/s )2tan 450( 500N)=(3.09× 103m)( 1km 10 3 m)=3.09km

Thus, the radius of the curve at θ=45o is 3.09 km .

Similarly, use equation (3) to find the radius of the curve at θ=88o .

  r=Mv2tanθFD

Substitute 500N for FD,2000 kg for M,88o for θ and 27.8m/s for v in this equation.

  r=Mv2tanθFD=( 2000kg) ( 27.8m/s )2tan 880( 500N)=107.9m=108m

Thus, the radius of the curve at θ=88o is 108m .

Calculate the minimum coefficient of friction required for the car to hold at radius 108 m without skidding by making use of Newton’s second law of motion.

  F=Mafs=Mv2rμsMg=Mv2rμs=v2rg

Substitute 27.8 m/s for v and 108 m for r in this equation,

  μs=v2rg= ( 27.8m/s )2( 108m)( 9.81 m/s 2 )=0.73

Conclusion:

Thus, the required coefficient of friction for the car without skidding is 0.73 .

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Physics for Scientists and Engineers

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