Home Page Textbooks Physics: Principles and Problems Page 104: Practice Problems

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 104: Practice Problems

Exercise 28

Step 1

1 of 2

The ball is acted upon by two forces, the force of gravity by which the Earth attracts the ball, and the pushing force by which the hand pushes the ball away. Newton’s third law tells us that every force has an opposite force of the same amount and opposite direction. Since the hand and the Earth act with force on the ball, so the ball acts on the Earth attractively with force and on the hand with repulsive force.

Result

2 of 2

$$
begin{align*}
& F_{hand on ball}, F_{Earth on ball} tag{Forces on ball} \
& F_{Ball on hand}, F_{Ball on Earth} tag{Forces exerted by ball}
end{align*}
$$

Exercise 29

Step 1

1 of 2

The brick is acted upon by the force of gravity by which the Earth attracts the brick, also, the brick acts by the gravitational force on the Earth. These two forces are, according to Newton’s third law, interaction pair, they are of equal amounts and opposite directions. The reason why a brick falls to Earth is that a brick is far lighter than the Earth so the Earth rests.

Result

2 of 2

The brick is acted upon by the force of gravity by which the Earth attracts the brick, also, the brick acts by the gravitational force on the Earth.

Exercise 30

Step 1

1 of 2

The only force acting on the ball, thrown into the air is the force of gravity exerted from the Earth. Also the ball acts on the Earth with a force of equal amount but in the opposite direction. Exercise scan

Result

2 of 2

Force acting on the ball is force of gravity exerted by the Earth, and ball exerts force of gravity on the Earth. Free body diagram in explanation.

Exercise 31

Step 1

1 of 2

$$
begin{align*}
& F_g tag{Force of graviti on body by the Earth} \
& F_{s-c} tag{Force exerted by suitcase on cart} \
& F_{c-s} tag{Force exerted by cart on suitcase}\
& F_n tag{force exerted by floor on cart}
end{align*}
$$

Force pair is $F_{s-c} , F_{c-s}$ for which stands $F_{s-c}=-F_{c-s}$.

Exercise scan

Result

2 of 2

Force pair is $F_{s-c} , F_{c-s}$. For labels and free body diagrams see explanation.

Exercise 32

Step 1

1 of 4

In this task we want to calculate the maximum acceleration by which a worker is allowed to lift a bucket.

Known:

$$
begin{align*}
m&=42 mathrm{kg} \
F_{t max}&=450 mathrm{N} \
g&=9.8 mathrm{m/s^2}
end{align*}
$$

Unknown:

$$
begin{align*}
a_{max}&=?
end{align*}
$$

Step 2

2 of 4

In this example, the rope tension force $F_ {t max}$ is equal to the sum of the gravitational force and the force required to accelerate the bucket.

$$
begin{align*}
F_{t max}&=F_g+F \
F_{t max}&=mg+ma_{max} \
a_{max}&=dfrac{F_{t max}-mg}{m}
end{align*}
$$

Step 3

3 of 4

$$
begin{align*}
a_{max}&=dfrac{F_{t max}-mg}{m} \
&=dfrac{450 mathrm{N}-42 mathrm{kg}cdot 9.8 mathrm{m/s^2}}{42 mathrm{kg}} \
&=boxed{0.9143 mathrm{m/s^2}}
end{align*}
$$

The maximum acceleration by which a worker can lift a bucket without breaking the rope is $a_ {max} = 0.9143 mathrm {m/s^2}$.

Result

4 of 4

$$
a_ {max} = 0.9143 mathrm {m/s^2}
$$

Exercise 33

Step 1

1 of 3

In this task we want to calculate the magnitude of the strength of the force between the tire and the wheel.

Known:

$$
begin{align*}
F_{Diego}&=31 mathrm{N} \
F_{Mika}&=23 mathrm{N}
end{align*}
$$

Unknown:

$$
begin{align*}
F_{tire-wheel}&=?
end{align*}
$$

Step 2

2 of 3

Suppose Diego and Mika pull together then the total force is approximately equal to the force that holds the tire and wheel connected. the total force is the sum of the individual forces by which diego and mika act.

$$
begin{align*}
F_{net}&=F_{Mika}+F_{Diego} \
&=23 mathrm{N}+31 mathrm{N} \
&=54 mathrm{N} \
F_{tire-wheel}&=F_{net} \
&=boxed{54 mathrm{N}}
end{align*}
$$

Result

3 of 3

$$
F=54 mathrm{N}
$$

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Page 104: Practice Problems

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