PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 5, Problem 94P

(a)

To determine

To Find:The normal force.

(a)

Expert Solution
Check Mark

Answer to Problem 94P

  8.24kN

Explanation of Solution

Given Information:

Radius of the curve is 150m

Banking angle is 10o .

Mass of car is 800kg .

Speed of the car is 85 km/h .

Formula Used:

The acceleration of the car along x- axis is the centripetal acceleration, that is,

  a=v2r

Here, v is the speed of the car and r is the radius of the curve.

Calculation:

The free-body diagram for the car on the curved road is as follows

  PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS, Chapter 5, Problem 94P

In the free body diagram shown below, the normal force and the static friction force, both contribute to the centripetal force according to the situation described in this problem.

The speed of the car is

  v=85km/h( 5 18 m/s km/h)=23.61 m/s

The net force on the x -axis is

  Fnsinθ+fscosθ=ma

Substitute v2r for a in above equation.

  Fnsinθ+fscosθ=m(v2r)...(1)

Multiply the equation (1) with sinθ .

  fssinθcosθ+Fnsin2θ=m(v2r)sinθ...(2)

The net force on the y-axis is,

  Fncosθfssinθ=mg...(3)

Multiply the above equation with cosθ .

  Fncos2θfssinθcosθ=mgcosθ...(4)

Add the equations (2) and (4)

  Fnmgcosθ=m(v2r)sinθ

Rearrange the above equation for the normal force Fn .

  Fn=mgcosθ+m(v2r)sinθ

Substitute 800 kg for m,9.81m/s2 for g,10o for θ,23.61 m/s for v and 150 m for r in above equation.

Conclusion:

The normal force exerts on the car is 8.24 kN .

(b)

To determine

To Find:The frictional force exerted by the pavement on the tires.

(b)

Expert Solution
Check Mark

Answer to Problem 94P

The frictional force exerted by the pavement on the tires is 1.56 kN

Explanation of Solution

Given Information:

Radius of the curve is 150m

Banking angle is 10o .

Mass of car is 800kg .

Speed of the car is 85 km/h .

Formula Used:

The net force along the y-axis is

  Fncosθfssinθ=mg

Calculation:

Rearrange the above equation:

  fs=Fncosθmgsinθ

Substitute 8244.8N for Fn,10ofor θ,800kg for m and 9.81m/s2 for g in above equation.

  fs=( 8244.8N)cos 10o( 800kg)( 9.81 m/s 2 )sin 10o=( 8244.8N)( 0.9848)-( 800 kg)( 9.81 m/s 2 )( 0.1736)=1563.8N=1563.8 N( 1kN 10 3 N)=1.56kN

Conclusion:

The frictional force exerted by the pavement on the tires is 1.56kN .

(c)

To determine

To Find:The coefficient of the static frictional force.

(c)

Expert Solution
Check Mark

Answer to Problem 94P

  0.193

Explanation of Solution

Given Information:

Radius of the curve is 150m

Banking angle is 10o .

Mass of car is 800kg .

Speed of the car is 85 km/h .

Formula Used:

The equation for the frictional force is

  fs=μs,minFn

Here, Fn is the normal force.

Calculation:

Rearrange the above equation for μs,min .

  μs,min=fsFn

Substitute 1.56kN for fs, and 8.24kN for Fn in the above equation.

  μs,min=1.56kN8.24kN=0.193

Conclusion:

The coefficient of the minimum static frictional force is 0.193 .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
CH 57. A 190-g block is launched by compressing a spring of constant k = = 200 N/m by 15 cm. The spring is mounted horizontally, and the surface directly under it is frictionless. But beyond the equilibrium position of the spring end, the surface has frictional coefficient μ = 0.27. This frictional surface extends 85 cm, fol- lowed by a frictionless curved rise, as shown in Fig. 7.21. After it's launched, where does the block finally come to rest? Measure from the left end of the frictional zone. Frictionless μ = 0.27 Frictionless FIGURE 7.21 Problem 57
3. (a) Show that the CM of a uniform thin rod of length L and mass M is at its center (b) Determine the CM of the rod assuming its linear mass density 1 (its mass per unit length) varies linearly from λ = λ at the left end to double that 0 value, λ = 2λ, at the right end. y 0 ·x- dx dm=λdx x +
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 161 cm , but its circumference is decreasing at a constant rate of 15.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf E induced in the loop after exactly time 9.00 s has passed since the circumference of the loop started to decrease. please show all steps

Chapter 5 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

Ch. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Prob. 55PCh. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - Prob. 58PCh. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 65PCh. 5 - Prob. 67PCh. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Prob. 71PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Prob. 74PCh. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 84PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Prob. 87PCh. 5 - Prob. 88PCh. 5 - Prob. 89PCh. 5 - Prob. 90PCh. 5 - Prob. 91PCh. 5 - Prob. 92PCh. 5 - Prob. 93PCh. 5 - Prob. 94PCh. 5 - Prob. 95PCh. 5 - Prob. 96PCh. 5 - Prob. 97PCh. 5 - Prob. 101PCh. 5 - Prob. 102PCh. 5 - Prob. 103PCh. 5 - Prob. 104PCh. 5 - Prob. 105PCh. 5 - Prob. 106PCh. 5 - Prob. 107PCh. 5 - Prob. 108PCh. 5 - Prob. 109PCh. 5 - Prob. 110PCh. 5 - Prob. 111PCh. 5 - Prob. 112PCh. 5 - Prob. 113PCh. 5 - Prob. 114PCh. 5 - Prob. 115PCh. 5 - Prob. 116PCh. 5 - Prob. 117PCh. 5 - Prob. 118PCh. 5 - Prob. 119PCh. 5 - Prob. 120PCh. 5 - Prob. 121PCh. 5 - Prob. 122PCh. 5 - Prob. 123PCh. 5 - Prob. 124PCh. 5 - Prob. 125PCh. 5 - Prob. 126PCh. 5 - Prob. 127PCh. 5 - Prob. 128PCh. 5 - Prob. 129PCh. 5 - Prob. 130PCh. 5 - Prob. 131PCh. 5 - Prob. 132PCh. 5 - Prob. 133PCh. 5 - Prob. 134PCh. 5 - Prob. 135PCh. 5 - Prob. 136PCh. 5 - Prob. 137PCh. 5 - Prob. 138PCh. 5 - Prob. 139P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
What Is Circular Motion? | Physics in Motion; Author: GPB Education;https://www.youtube.com/watch?v=1cL6pHmbQ2c;License: Standard YouTube License, CC-BY