6th Edition

ISBN: 9780130606204

Author: Douglas C. Giancoli

Publisher: Prentice Hall

Concept explainers

If a very smooth ball is rolled down an inclined surface with zero friction, the ball will continue to roll indefinitely for an infinite time, in an attempt to reach to the same height on the opposite hill!

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Question

Chapter 4, Problem 32P

To determine

a)

Free body diagram of each block.

Expert Solution

Answer to Problem 32P

Solution:

Physics: Principles with Applications, Chapter 4, Problem 32P , additional homework tip 1

Explanation of Solution

On each block, normal force by the surface acts in upward direction and force of gravity act in a downward direction. the contact force acts between the blocks in the equal and opposite direction

To determine

b)

The acceleration of the system.

Expert Solution

Answer to Problem 32P

Solution:

$a = \frac{F}{m_{A} + m_{B} + m_{C}}$

Explanation of Solution

Consider the three blocks as one by combining the masses.

Given:

Mass of block A $= m_{A}$

Mass of block B $= m_{B}$

Mass of block C $= m_{c}$

Total Sum $= M$

The net force on the combination of block $= F$

Acceleration $= a$

Formula Used:

$F = m a$

Calculation:

The total mass of the blocks is given as

$M = m_{A} + m_{B} + m_{C}$

Acceleration is given as

$\begin{array}{l} a = \frac{F}{M} \\ a = \frac{F}{m_{A} + m_{B} + m_{C}} \end{array}$

To determine

c)

The net force on each block.

Expert Solution

Answer to Problem 32P

Solution:

$F_{A} = \frac{m_{A} F}{m_{A} + m_{B} + m_{C}}$

$F_{B} = \frac{m_{B} F}{m_{A} + m_{B} + m_{C}}$

$F_{C} = \frac{m_{C} F}{m_{A} + m_{B} + m_{C}}$

Explanation of Solution

Given:

Mass of block A $= m_{A}$

Mass of block B $= m_{B}$

Mass of block C $= m_{c}$

Acceleration $= a = \frac{F}{m_{A} + m_{B} + m_{C}}$

The net force on block A $= F_{A}$

The net force on block B $= F_{B}$

The net force on block C $= F_{C}$

Formula Used:

The net force is given as

$F = m a$

Calculation:

The net force on block A is given as

$\begin{array}{l} F_{A} = m_{A} a \\ F_{A} = \frac{m_{A} F}{m_{A} + m_{B} + m_{C}} \end{array}$

The net force on block B is given as

$\begin{array}{l} F_{B} = m_{B} a \\ F_{B} = \frac{m_{B} F}{m_{A} + m_{B} + m_{C}} \end{array}$

The net force on block C is given as

$\begin{array}{l} F_{C} = m_{C} a \\ F_{C} = \frac{m_{C} F}{m_{A} + m_{B} + m_{C}} \end{array}$ .

To determine

(d) To Determine:

The contact force on each block.

Expert Solution

Answer to Problem 32P

Solution:

$F_{A B} = F_{B A} = \frac{(m_{B} + m_{C}) F}{m_{A} + m_{B} + m_{C}}$

$F_{B C} = F_{C B} = \frac{m_{C} F}{m_{A} + m_{B} + m_{C}}$

Explanation of Solution

The contact forces between the two blocks are equal in magnitude and opposite in direction as per Newton’s third law.

Hence

$\begin{array}{l} F_{A B} = F_{B A} \\ F_{B C} = F_{C B} \end{array}$

Physics: Principles with Applications, Chapter 4, Problem 32P , additional homework tip 2

Given:

Mass of block A $= m_{A}$

Mass of block B $= m_{B}$

Mass of block C $= m_{c}$

Acceleration $= a = \frac{F}{m_{A} + m_{B} + m_{C}}$

The net force on block A $= F_{A} = \frac{m_{A} F}{m_{A} + m_{B} + m_{C}}$

The net force on block B $= F_{B} = \frac{m_{B} F}{m_{A} + m_{B} + m_{C}}$

The net force on block C $= F_{C} = \frac{m_{C} F}{m_{A} + m_{B} + m_{C}}$

Formula Used:

The net force is given as

$F = m a$

Calculation:

From the force diagram, force equation for block A along the horizontal direction is given as

The net force on block A is given as

$\begin{array}{l} F - F_{A B} = m_{A} a \\ F - F_{A B} = \frac{m_{A} F}{m_{A} + m_{B} + m_{C}} \\ F_{A B} = \frac{(m_{B} + m_{C}) F}{m_{A} + m_{B} + m_{C}} \end{array}$

Also

$\begin{array}{l} F_{B A} = F_{A B} \\ F_{B A} = \frac{(m_{B} + m_{C}) F}{m_{A} + m_{B} + m_{C}} \end{array}$

From the force diagram, force equation for block C along the horizontal direction is given as

$\begin{array}{l} F_{C B} = m_{C} a \\ F_{C B} = \frac{m_{C} F}{m_{A} + m_{B} + m_{C}} \end{array}$

Also

$\begin{array}{l} F_{B C} = F_{C B} \\ F_{B C} = \frac{m_{C} F}{m_{A} + m_{B} + m_{C}} \end{array}$

To determine

e)

Numerical answers to part b), c)and d)

Expert Solution

Answer to Problem 32P

Solution:

$\begin{array}{l} a = 3.2 m / s^{2} \\ F_{A} = F_{B} = F_{C} = 32 N \\ F_{A B} = F_{B A} = 64 N \\ F_{B C} = F_{C B} = 32 N \end{array}$

Explanation of Solution

Given:

$F = 96 N$

Mass of block A $= m_{A} = 10 k g$

Mass of block B $= m_{B} = 10 k g$

Mass of block C $= m_{c} = 10 k g$

Acceleration $= a = \frac{F}{m_{A} + m_{B} + m_{C}}$

The net force on block A $= F_{A}$

The net force on block B $= F_{B}$

The net force on block C $= F_{C}$

Formula Used:

$a = \frac{F}{m_{A} + m_{B} + m_{C}}$

The net force is given as

$F = m a$

Calculation:

Using the equation

$a = \frac{F}{m_{A} + m_{B} + m_{C}}$

Inserting the values

$\begin{array}{l} a = \frac{96}{(10 + 10 + 10)} \\ a = 3.2 m / s^{2} \end{array}$

The net force on block A, B and C is given as

$\begin{array}{l} F_{A} = m_{A} a = 10 \times 3.2 = 32 N \\ F_{B} = m_{B} a = 10 \times 3.2 = 32 N \\ F_{C} = m_{C} a = 10 \times 3.2 = 32 N \end{array}$

The contact force between A and B is given as

$\begin{array}{l} F_{A B} = F_{B A} = \frac{(m_{B} + m_{C}) F}{m_{A} + m_{B} + m_{C}} = \frac{(10 + 10) (96)}{10 + 10 + 10} = 64 N \\ F_{B C} = F_{C B} = \frac{m_{C} F}{m_{A} + m_{B} + m_{C}} = \frac{(10) (96)}{10 + 10 + 10} = 32 N \end{array}$

Chapter 4, Problem 31P

Chapter 4, Problem 33P

Chapter 4 Solutions

Physics: Principles with Applications

Ch. 4 - Why does a child in a wagon seem to fall backward...Ch. 4 - A box rests on the (frictionless) bed of a truck....Ch. 4 - Prob. 3Q Ch. 4 - If the acceleration of an object is zero, are no...Ch. 4 - Prob. 5Q Ch. 4 - Prob. 6Q Ch. 4 - Prob. 7Q Ch. 4 - (a) Why do you push down harder on the pedals of a...Ch. 4 - Prob. 9Q Ch. 4 - Prob. 10Q

Ch. 4 - Prob. 11Q Ch. 4 - Prob. 12Q Ch. 4 - Prob. 13Q Ch. 4 - Prob. 14Q Ch. 4 - Prob. 15Q Ch. 4 - Prob. 16Q Ch. 4 - Prob. 17Q Ch. 4 - Prob. 18Q Ch. 4 - A block is given a brief push so that it slides up...Ch. 4 - Prob. 20Q Ch. 4 - Prob. 21Q Ch. 4 - What force is needed to accelerate a sled (mass =...Ch. 4 - Prob. 2P Ch. 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. 8P Ch. 4 - Prob. 9P Ch. 4 - Prob. 10P Ch. 4 - Prob. 11P Ch. 4 - Prob. 12P Ch. 4 - Prob. 13P Ch. 4 - Prob. 14P Ch. 4 - Prob. 15P Ch. 4 - Prob. 16P Ch. 4 - Prob. 17P Ch. 4 - Prob. 18P Ch. 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. 21P Ch. 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. 25P Ch. 4 - A train locomotive is pulling two cars of the same...Ch. 4 - Prob. 27P Ch. 4 - A 27-kg chandelier hangs from a ceiling on a...Ch. 4 - Prob. 29P Ch. 4 - Figure 4-53 [shows a block (mass mA) on a smooth...Ch. 4 - Prob. 31P Ch. 4 - Prob. 32P Ch. 4 - 35. (Ill) Suppose the pulley in Fig. 4-55 is...Ch. 4 - Prob. 34P Ch. 4 - A force of 35.0 N is required to start a 6.0-kg...Ch. 4 - Prob. 36P Ch. 4 - Prob. 37P Ch. 4 - Prob. 38P Ch. 4 - Prob. 39P Ch. 4 - A box is given a push so that it slides across the...Ch. 4 - Prob. 41P Ch. 4 - Prob. 42P Ch. 4 - Prob. 43P Ch. 4 - 46. (II) For the system of Fig. 4-32 (Example...Ch. 4 - Prob. 45P Ch. 4 - Prob. 46P Ch. 4 - Prob. 47P Ch. 4 - A person pushes a 14.0-kg lawn mower at constant...Ch. 4 - Prob. 49P Ch. 4 - (a) A box sits at rest on a rough 33° inclined...Ch. 4 - Prob. 51P Ch. 4 - Prob. 52P Ch. 4 - Prob. 53P Ch. 4 - A 25.0-kg box is released on a 27° incline and...Ch. 4 - Prob. 55P Ch. 4 - Prob. 56P Ch. 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. 59P Ch. 4 - Prob. 60P Ch. 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. 64P Ch. 4 - Prob. 65P Ch. 4 - Prob. 66GP Ch. 4 - Prob. 67GP Ch. 4 - Prob. 68GP Ch. 4 - Prob. 69GP Ch. 4 - Prob. 70GP Ch. 4 - Prob. 71GP Ch. 4 - Prob. 72GP Ch. 4 - Prob. 73GP Ch. 4 - Prob. 74GP Ch. 4 - Prob. 75GP Ch. 4 - Prob. 76GP Ch. 4 - Prob. 77GP Ch. 4 - Prob. 78GP Ch. 4 - Prob. 79GP Ch. 4 - Prob. 80GP Ch. 4 - Prob. 81GP Ch. 4 - Prob. 82GP Ch. 4 - Prob. 83GP Ch. 4 - Prob. 84GP Ch. 4 - Prob. 85GP Ch. 4 - Prob. 86GP Ch. 4 - Prob. 87GP Ch. 4 - Prob. 88GP Ch. 4 - Prob. 89GP

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