Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 4, Problem 19P

A box weighing 77.0 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is hung from the other end (Fig. 4-45). Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs (a) 30.0 N, (b) 60.0 N, and (c) 90.0 N.

Chapter 4, Problem 19P, A box weighing 77.0 N rests on a table. A rope tied to the box runs vertically upward over a pulley

(a)

Expert Solution
Check Mark
To determine

The force exerted on the box by the table when 30 N weight is hanging on the other side.

Answer to Problem 19P

Solution:

For exerted on the box by the table is 47N .

Explanation of Solution

Given:

Weight of the box at the table, WT=77 N

Weight of the box hanging on the other side, Wh=30 N

System is at rest as hanging weight is less than weight on the table, a=0 m/s2

To find:

(I) Tension along the rope, T=?

(II) Force exerted on box by the table, FT=?

Formula:

From Newton’s second law of motion:

  F=ma

Where, m is mass and a is acceleration.

Calculations:

(I) Fortension in the rope.

  Physics: Principles with Applications, Chapter 4, Problem 19P , additional homework tip  1

  F=maWHT=ma30T=m(0)T=30N

(II) For force that table exerts on the block.

  Physics: Principles with Applications, Chapter 4, Problem 19P , additional homework tip  2

  F=maFTWT+T=maFT+ 30 77=m(0)FT=47N

Conclusion: Using Newton’s law, the force on the box bythe table is equal to 47N .

(b)

Expert Solution
Check Mark
To determine

The force exerted on the box by the table when 60 N weight is hanging on other side.

Answer to Problem 19P

Solution:

Force exerted by the table on the box is 17 N.

Explanation of Solution

Given:

Weight of the box at the table, WT=77 N

Weight of the box hanging on the other side, Wh=60 N

System is at rest as hanging weight is less than weight on the table, a=0 m/s2

>

To find:

(I) Tension along the rope, T=?

(II) Force exerted on the box by the table, FT=?

Formula:

From Newton’s second law of motion:

  F=ma

Where, m is the mass and a is the acceleration.

Calculations:

(I) For hanging weight.

  Physics: Principles with Applications, Chapter 4, Problem 19P , additional homework tip  3

  F=maWhT=ma60T=m(0) T=60 N

(II) For a mass on the table.

  Physics: Principles with Applications, Chapter 4, Problem 19P , additional homework tip  4

  F=maFT+TWT=maFT+ 6077 = m(0)FT=17N

Conclusion: Newton’s second law of motion helps to calculate the force exerted by the table on the box equal to 17N .

(c)

Expert Solution
Check Mark
To determine

The force exerted on the box by the table when 90 N weightis hanging on another side.

Answer to Problem 19P

Solution:

Force exerted on the box equal to 0 N.

Explanation of Solution

Given:

Weight of the box at the table, WT=77 N

Weight of the box hanging on the other side, Wh=90 N

System is at rest as hanging weight is less than weight on the table, a=0 m/s2

To find:

(I) Tension along the rope, T=?

(II) Force exerted on box by the table, FT=?

  • Hanging weight is more than weight on the table.
  • This arrangement makes the hanging box move downward and box on the table move upward.
  • Box moving upward no longerremains in contact with the table.
  • The normal force FT is contact force.
  • Hence, in absence of contact, contact force FT ceases to zero instantly.

Conclusion: In absence of contact, table exerts no force on the box.

08:18

Chapter 4 Solutions

Physics: Principles with Applications

Ch. 4 - Prob. 11QCh. 4 - Prob. 12QCh. 4 - Prob. 13QCh. 4 - Prob. 14QCh. 4 - Prob. 15QCh. 4 - Prob. 16QCh. 4 - Prob. 17QCh. 4 - Prob. 18QCh. 4 - A block is given a brief push so that it slides up...Ch. 4 - Prob. 20QCh. 4 - Prob. 21QCh. 4 - What force is needed to accelerate a sled (mass =...Ch. 4 - Prob. 2PCh. 4 - How much tension must a rope withstand if it is...Ch. 4 - According to a simplified model of a mammalian...Ch. 4 - Superman must stop a 120-km/h train in 150 m to...Ch. 4 - A person has a reasonable chance of surviving an...Ch. 4 - What average force is required to stop a 950-kg...Ch. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - A box weighing 77.0 N rests on a table. A rope...Ch. 4 - Figure 4-46 Problem 21. 21. (I) Draw the free-body...Ch. 4 - Prob. 21PCh. 4 - Arlene is to walk across a “high wire" strung...Ch. 4 - A window washer pulls herself upward using the...Ch. 4 - One 3.2-kg paint bucket is hanging by a massless...Ch. 4 - Prob. 25PCh. 4 - A train locomotive is pulling two cars of the same...Ch. 4 - Prob. 27PCh. 4 - A 27-kg chandelier hangs from a ceiling on a...Ch. 4 - Prob. 29PCh. 4 - Figure 4-53 [shows a block (mass mA) on a smooth...Ch. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - 35. (Ill) Suppose the pulley in Fig. 4-55 is...Ch. 4 - Prob. 34PCh. 4 - A force of 35.0 N is required to start a 6.0-kg...Ch. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - A box is given a push so that it slides across the...Ch. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - 46. (II) For the system of Fig. 4-32 (Example...Ch. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - A person pushes a 14.0-kg lawn mower at constant...Ch. 4 - Prob. 49PCh. 4 - (a) A box sits at rest on a rough 33° inclined...Ch. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - A 25.0-kg box is released on a 27° incline and...Ch. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - The crate shown in Fig. 4-60 lies on a plane...Ch. 4 - A crate is given an initial speed of 3.0 m/s up...Ch. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - The coefficient of kinetic friction for a 22-kg...Ch. 4 - On an icy day, you worry about parking your car in...Ch. 4 - Two masses mA= 2.0 kg and mB= 5.0 kg are on...Ch. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 66GPCh. 4 - Prob. 67GPCh. 4 - Prob. 68GPCh. 4 - Prob. 69GPCh. 4 - Prob. 70GPCh. 4 - Prob. 71GPCh. 4 - Prob. 72GPCh. 4 - Prob. 73GPCh. 4 - Prob. 74GPCh. 4 - Prob. 75GPCh. 4 - Prob. 76GPCh. 4 - Prob. 77GPCh. 4 - Prob. 78GPCh. 4 - Prob. 79GPCh. 4 - Prob. 80GPCh. 4 - Prob. 81GPCh. 4 - Prob. 82GPCh. 4 - Prob. 83GPCh. 4 - Prob. 84GPCh. 4 - Prob. 85GPCh. 4 - Prob. 86GPCh. 4 - Prob. 87GPCh. 4 - Prob. 88GPCh. 4 - Prob. 89GP
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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