COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 4, Problem 73P

(a)

To determine

The tension in the rope.

(a)

Expert Solution
Check Mark

Answer to Problem 73P

The tension in the rope is T=88N

Explanation of Solution

Write the expression for the force for crate in terms of Newton’s second law along x direction.

  Fx=T+fkm1gsinθ=m1ax        (I)

Here, Fx is the frictional force, fk is the friction, T is the tension, m1 is the mass of the crate, ax is the acceleration along x direction, and g is the acceleration due to gravity.

Similarly write the expression for the force for the crate along y direction.

  Fy=Nm1gsinθ=0        (II)

Here, Fy is the frictional force, N is the normal force, m1 is the mass of the crate, and g is the acceleration due to gravity.

Write the expression for force for the box along  x direction.

  Fx=0        (III)

Similarly write the expression for the force for the box along y direction.

  Fy=Tm2g=m2ay

Here, Fy is the frictional force, m2 is the mass of the box, g is the acceleration due to gravity, ay is the acceleration along y direction and, T is the tension.

According to the problem ax=ay , and solve for ax from expression (I) and (II),

  m1ax=T+fkm1gsinθ=T+μkNm1gsinθ=T+μkm1gcosθm1gsinθax=Tm1+μkgcosθgsinθ        (V)

Solve for ay from the expression (IV)

  Tm2g=m2ayay=Tm2g        (VI)

Conclusion:

Eliminate for ax and ay  and solve for T , by equating the expression (V) and (VI),

  Tm1+μkgcosθgsinθ=Tm2gT(1m1+1m2)=g(1+sinθμkcosθ)T=m1m2m1+m2(1+sinθμkcosθ)        (VII)

 Substitute 15kg for m1 , 8.0kg for m2 , 9.8m/s2 for g , 60° for θ , and 0.30 for μk in expression (VII)

T=(15kg)(8.0kg)(15kg)+(8.0kg)(1+sin(60°)(0.30)cos(60°))=88N

The tension in the rope is T=88N

(b)

To determine

The time required by the crate to slide down.

(b)

Expert Solution
Check Mark

Answer to Problem 73P

The time required by the crate to slide down is 2s.

Explanation of Solution

Write the equation of motion, where v0=0

  Δx=12axt2        (I)

Here, Δx is the displacement, ax is the acceleration, and t is the time

Re-write the expression in terms of t, and substitute gTm2 from expression (VI) in sub part (a) as ax=ay

  t=2Δxaxt=2ΔxgTm2        (II)

Conclusion:

Substitute 2.00m for Δx , 88N for T, , 8.0kg for m2 , and 9.8m/s2 for g expression (II),

  t=2(2.00m)9.8m/s288N8.0kg=2s

The time required by the crate to slide down is 2s.

(c)

To determine

The force to push the crate.

(c)

Expert Solution
Check Mark

Answer to Problem 73P

The force to push the crate is 70N.

Explanation of Solution

Write the expression for the force for crate in terms of Newton’s second law along x direction.

  Fx=T+Pfkm1gsinθ=0P=fkm1gsinθT        (I)

Here, Fx is the frictional force, P is the push, fk is the friction, T is the tension, m1 is the mass of the crate, ax is the acceleration along x direction, and g is the acceleration due to gravity.

Similarly write the expression for the force for the box along y direction.

  Fy=Tm2g=0T=m2g        (II)

Here, Fy is the frictional force, m2 is the mass of the box, T is the tension, and g is the acceleration due to gravity.

Solve for P by substituting for T and fk from expression (I) and (II),

  P=fkm1gsinθm2g=g(m1(μkcosθ+sinθ)m2)        (III)

Conclusion:

Substitute 15kg for m1 , 8.0kg for m2 , 9.8m/s2 for g , 60° for θ , and 0.30 for μk in expression (VII)

  P=(9.8m/s2)(15kg(0.30cos(60°)+sin(60°))8.0kg)=70N

The force to push the crate is 70N.

(d)

To determine

The smallest mass to keep the crate from sliding down.

(d)

Expert Solution
Check Mark

Answer to Problem 73P

The smallest mass is m2=10kg

Explanation of Solution

Write the expression for the force for crate in terms of Newton’s second law along x direction.

  Fx=T+fkm1gsinθ=0        (I)

Here, Fx is the frictional force, fk is the friction, T is the tension, m1 is the mass of the crate, ax is the acceleration along x direction, and g is the acceleration due to gravity.

Similarly write the expression for the force for the crate along y direction.

  Fy=Nm1gsinθ=0        (II)

Here, Fy is the frictional force, N is the normal force, m1 is the mass of the crate, and g is the acceleration due to gravity.

Write the expression for force for the box along  x direction.

  Fx=0        (III)

Similarly write the expression for the force for the box along y direction.

  Fy=Tm2g=0        (IV)

Here, Fy is the frictional force, m2 is the mass of the box, g is the acceleration due to gravity, ay is the acceleration along y direction and, T is the tension.

Conclusion:

Thus, m2=Tg , and find for T

  T=m1gsinθfs=m1gsinθμsm1gcosθ        (V)

Thus, substitute the expression (V) in expression (IV) and re-write the expression in terms m2 

  m2g=m1gsinθμsm1gcosθm2=m1(sinθμscosθ)        (IV)

Substitute 15kg for m1 , 9.8m/s2 for g , 60° for θ , and 0.40 for μs in expression (IV)

  m2=15kg(sin(60°)0.40cos(60°))=10kg

The smallest mass is m2=10kg

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!

Chapter 4 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 4.4 - Prob. 4.8PPCh. 4.5 - Prob. 4.5ACPCh. 4.5 - Prob. 4.10PPCh. 4.5 - Prob. 4.5BCPCh. 4.6 - Prob. 4.11PPCh. 4.6 - Prob. 4.6CPCh. 4 - Prob. 1CQCh. 4 - Prob. 2CQCh. 4 - Prob. 3CQCh. 4 - Prob. 4CQCh. 4 - Prob. 5CQCh. 4 - Prob. 6CQCh. 4 - Prob. 7CQCh. 4 - Prob. 8CQCh. 4 - Prob. 9CQCh. 4 - Prob. 10CQCh. 4 - Prob. 11CQCh. 4 - Prob. 12CQCh. 4 - Prob. 13CQCh. 4 - Prob. 14CQCh. 4 - Prob. 15CQCh. 4 - Prob. 16CQCh. 4 - Prob. 17CQCh. 4 - Prob. 18CQCh. 4 - Prob. 19CQCh. 4 - Prob. 1MCQCh. 4 - Prob. 2MCQCh. 4 - Prob. 3MCQCh. 4 - Prob. 4MCQCh. 4 - Prob. 5MCQCh. 4 - Prob. 6MCQCh. 4 - Prob. 7MCQCh. 4 - Prob. 8MCQCh. 4 - Prob. 9MCQCh. 4 - Prob. 10MCQCh. 4 - Prob. 11MCQCh. 4 - Prob. 12MCQCh. 4 - Prob. 13MCQCh. 4 - Prob. 14MCQCh. 4 - Prob. 15MCQCh. 4 - Prob. 16MCQCh. 4 - Prob. 17MCQCh. 4 - Prob. 18MCQCh. 4 - Prob. 19MCQCh. 4 - Prob. 20MCQCh. 4 - Prob. 21MCQCh. 4 - Prob. 1PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69PCh. 4 - Prob. 70PCh. 4 - Prob. 71PCh. 4 - Prob. 72PCh. 4 - Prob. 73PCh. 4 - Prob. 74PCh. 4 - Prob. 75PCh. 4 - Prob. 76PCh. 4 - Prob. 77PCh. 4 - Prob. 78PCh. 4 - Prob. 79PCh. 4 - Prob. 80PCh. 4 - Prob. 81PCh. 4 - Prob. 82PCh. 4 - Prob. 83PCh. 4 - Prob. 84PCh. 4 - Prob. 85PCh. 4 - Prob. 86PCh. 4 - Prob. 87PCh. 4 - Prob. 88PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92PCh. 4 - Prob. 93PCh. 4 - Prob. 94PCh. 4 - Prob. 95PCh. 4 - Prob. 97PCh. 4 - Prob. 98PCh. 4 - Prob. 99PCh. 4 - Prob. 100PCh. 4 - Prob. 101PCh. 4 - Prob. 102PCh. 4 - Prob. 103PCh. 4 - Prob. 104PCh. 4 - Prob. 105P
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.
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
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