Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 5, Problem 33P
(a)
To determine
To Find: The magnitude of the
(b)
To determine
To Find:The magnitude of the force of friction acting on the block.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A box of bananas weighing 40.0 N rests on a horizontal surface.The coefficient of static friction between the box and the surfaceis 0.40, and the coefficient of kinetic friction is 0.20. If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it?
A box is pulled, at constant speed, across a horizontal floor by a cable. The 15 N tension in the cable is applied at an angle of 48 degrees above the horizontal. Determine the force of friction acting on the box.
A block of mass 12.0 kg rests on a horizontal plane. The coefficient of a static friction between surfaces is 0.450. What is the maximum possible static frictional force that could act on the block of a force is applied to it?
Chapter 5 Solutions
Physics for Scientists and Engineers
Ch. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10P
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
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
- The x and y coordinates of a 4.00-kg particle moving in the xy plane under the influence of a net force F are given by x = t4 6t and y = 4t2 + 1, with x and y in meters and t in seconds. What is the magnitude of the force F at t = 4.00 s?arrow_forwardA box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of 18.0 N, what is the magnitude of the friction force and what is the box’s acceleration?arrow_forwardA box of bananas weighing 40.0 N rests on a horizontal surface.The coefficient of static friction between the box and the surfaceis 0.40, and the coefficient of kinetic friction is 0.20. If the monkey applies a horizontal force of 18.0 N, what is the magnitude of the friction force and what is the box’s acceleration?arrow_forward
- A man exerts a horizontal force of 121 N on a box with a mass of 34.7 kg. (a)If the box doesn't move, what's the magnitude of the static friction force (in N)? (b)What is the minimum possible value of the coefficient of static friction between the box and the floor? (Assume the box remains stationary.)arrow_forwardThe coefficients of static and kinetic friction between the 100-kg block and the inclined plane are 0.30 and 0.20, respectively. Determine (a) the friction force Facting on the block when P is applied with a magnitude of 200 N to the block at rest, (b) the force P required to initiate motion up the incline from rest, and (c) the friction force F acting on the block if P = 600 N. %3D P 100 kg 20° Hs = 0.30 Hk = 0.20 15° 5.arrow_forward(a) A flatbed truck moving at 28 m/s carries a steel girder that rests on its wooden floor. The girder is not strapped down, in violation with USDOT regulations. If the coefficient of static friction between steel and wood is 0.52, what is the minimum distance over which the truck can come to a stop without the girder sliding toward the cab of the truck? (answer: 77 m) (b) What is the minimum time over which the truck can accelerate forward from 0 m/s to 28 m/s with a constant acceleration without the girder sliding off the back? (answer: 5.5 s)arrow_forward
- (a) A flatbed truck moving at 28 m/s carries a steel girder that rests on its wooden floor. The girder is not strapped down, in violation with USDOT regulations. If the coefficient of static friction between steel and wood is 0.52, what is the minimum distance over which the truck can come to a stop without the girder sliding toward the cab of the truck? (answer: 77 m) (b) What is the minimum time over which the truck can accelerate forward from 0 m/s to 28 m/s with a constant acceleration without the girder sliding off the back? (answer: 5.5 s) FNET = ma fs,max = UsN W =mg v² = vo² + 2aAx %3D %3D V = Vo+ at g = 9.81 m/s?arrow_forward(a) A flatbed truck moving at 28 m/s carries a steel girder that rests on its wooden floor. The girder is not strapped down, in violation with USDOT regulations. If the coefficient of static friction between steel and wood is 0.52, what is the minimum distance over which the truck can come to a stop without the girder sliding toward the cab of the truck? (answer: 77 m) (b) What is the minimum time over which the truck can accelerate forward from 0 m/s to 28 m/s with a constant acceleration without the girder sliding off the back? (answer: 5.5 s) FNET = ma fs.max = μsn W = mg v² = v₁² + 2aAx V = Vo+ at g=9.81 m/s²arrow_forwardSuppose a 25.0 kg block rests on a horizontal plane, and the static friction coefficient between the surfaces is 0.220. (a) What is the maximum allowable static frictional force that could act on the block? (b) What is the actual static frictional force acting on the block if a 25.0 N external force acts horizontally? Assume g = 9.80 m/s2.arrow_forward
- A 0.85 kg block of wood is stationary on an incline that is15◦above horizontal. A force of 3.0 N is being applied toit 75◦below horizontal, directly into the incline (the force isperpendicular to the incline). What is the minimum coefficientof static friction required to keep the block stationary?arrow_forwardA 3.5 kg block is pushed along a horizontal floor by a force of magnitude 15 N at an angle 40 with the horizontal . The coefficient of kinetic friction between the block and the floor is 0.25. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.arrow_forwardA toy chest and its contents have a combined weight of180 N.The coefficient of static friction between toy chest and flooris 0.42.The child attempts to move the chest across thefloor by pulling on an attached rope. (a) If u is 42, what is the magnitudeof the force that the child must exert on the rope to putthe chest on the verge of moving? (b) Write an expression for themagnitude F required to put the chest on the verge of moving as afunction of the angle u. Determine (c) the value of u for which F isa minimum and (d) that minimum magnitude.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Newton's Second Law of Motion: F = ma; Author: Professor Dave explains;https://www.youtube.com/watch?v=xzA6IBWUEDE;License: Standard YouTube License, CC-BY