EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 5, Problem 97P
To determine

To Find: The safe range of speeds

Expert Solution & Answer
Check Mark

Answer to Problem 97P

  20.1km/hr to 56.1 km/hr

Explanation of Solution

Given Information:

Radius of the curve is 30m .

Mass of the car, m=950kg car

Speed of the car is 40.0km/h

The coefficient of static friction between the road and the tires is 0.300 .

Formula Used:

Newton’s second law of motion:

  F=ma

Where, m is mass and a is acceleration.

Calculation:

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 5, Problem 97P

From the Free body diagram is shown above. When the coefficient of static friction is zero along x-axis,

  Fnsinθ=mvmin2r...(1)

And along y-axis

  Fncosθmg=0Fncosθ=mg...(2)

Along x-axis

  Fnsinθfscosθ=mvmin2r...(3)

And along y-axis

  Fncosθ+fssinθmg=0...(4)

Now the force of static friction is given as

  fs=fs,max=μsFn...(5)

Substituting (5) in (3) and (4)

  Fn(μscosθsinθ)=mv min3r...(6)andFn(cosθ+μssinθ)=mg...(7)

Now dividing (6) and (7)

  Fn( μ s cosθsinθ)Fn( cosθ+ μ s sinθ)=m( v min 2 r )mg( μ s cosθsinθ)( cosθ+ μ s sinθ)=( v min 2 rg)vmin= rg( sinθ+ μ s cosθ ) ( cosθ+ μ s sinθ )vmin= ( 30 m )( 9 .81 m/s 2 )( sin 22.8 o 0.3cos 22.8 o ) ( cos 22.8 o +0.3sin 22.8 o )vmin=5.59m/svmin=20.1km/h

Now when the car is traveling at maximum speed:

Along x-axis

  Fnsinθ+fscosθ=mvmax2r...(8)

And along y-axis

  Fncosθfssinθmg=0...(9)

Now the force of static friction is given as

  fs=fs,max=μsFn...(10)

Substituting (10) in (8) and (9)

  Fn(μscosθ+sinθ)=mv max2r...(11)and Fn(cosθμssinθ)=mg...(12)

Now dividing (11) and (12)

  Fn( μ s cosθsinθ)Fn( cosθ+ μ s sinθ)=m( v max 2 r )mg( μ s cosθ+sinθ)( cosθ μ s sinθ)=( v max 2 rg)vmax= rg( μ s cosθ+sinθ ) ( cosθ μ s sinθ )vmax= ( 30 m )( 9 .81 m/s 2 )( 0.3cos 22.8 o +sin 22.8 o ) ( cos 22.8 o 0.3sin 22.8 o )vmax=15.6m/svmax=56.1km/h

Conclusion:

The safe range of speed is 20.1km/hr to 56.1 km/hr .

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Chapter 5 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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