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Schaum’s Outline of College Physics

Eugene Hecht

Chapter 3

Newton's Laws - all with Video Answers

Educators

Chapter Questions

03:26

Problem 1

As an introduction to dealing with force vectors, consider the four coplanar forces acting on a body at point $O$ as shown in Fig. $3-1(a)$. Find their resultant graphically. Starting from $O$, the four vectors are plotted in turn as drawn in Fig. $3-1(b)$. Place the tail end of each vector at the tip end of the preceding one. The arrow from $O$ to the tip of the last vector represents the resultant of the vectors.Measure $R$ from the scale drawing in Fig. $3-1(b)$ and find it to be $119 \mathrm{~N}$. Angle $\alpha$ is measured by protractor and is found to be $37^{\circ}$. Hence, the resultant makes an angle $\theta=180^{\circ}-37^{\circ}=143^{\circ}$ with the positive $x$ -axis. The resultant is $119 \mathrm{~N}$ at $143^{\circ}$.

Paul Gabriel
Paul Gabriel
Numerade Educator
04:54

Problem 2

To gain some practice treating force vectors before we get into Newton's Laws, examine the five coplanar forces seen in Fig. $3-2(a)$ acting on an object at the origin. Find their resultant analytically.
(1) First we find the $x$ - and $y$ -components of each force. These components are as follows:
Notice the $+$ and $-$ signs to indicate direction.
(2) The resultant $\overrightarrow{\mathbf{R}}$ has components $R_{x}=\sum F_{x}$ and $R_{y}=\sum F_{y}$, where we read $\sum F_{x}$ as "the sum of all the $x$ -force components." We then have
$$
\begin{array}{l}
R_{x}=19.0 \mathrm{~N}+7.50 \mathrm{~N}-11.3 \mathrm{~N}-9.53 \mathrm{~N}+0 \mathrm{~N}=+5.67 \mathrm{~N} \text { or }+5.7 \mathrm{~N} \\
R_{y}=0 \mathrm{~N}+13.0 \mathrm{~N}+11.3 \mathrm{~N}-5.50 \mathrm{~N}-22.0 \mathrm{~N}=-3.2 \mathrm{~N}
\end{array}
$$
(3) The magnitude of the resultant is
$$
R=\sqrt{R_{x}^{2}+R_{y}^{2}}=6.5 \mathrm{~N}
$$
(4) Finally, sketch the resultant as shown in Fig. $3-2(b)$ and find its angle. We see that
$$
\tan \phi=\frac{3.2 \mathrm{~N}}{5.7 \mathrm{~N}}=0.56
$$
from which $\phi=29^{\circ}$. Then $\theta=360^{\circ}-29^{\circ}=331^{\circ} .$ The resultant is $6.5 \mathrm{~N}$ at $331^{\circ}\left(\right.$ or $-29^{\circ}$ ) or $\overrightarrow{\mathbf{R}}=6.5 \mathrm{~N}-331^{\circ} \mathrm{FROM}+X$ -AXIS.

Anurag Kumar
Anurag Kumar
Numerade Educator
02:24

Problem 3

Solve Problem $3.1$ by use of the component method. Give your answer for the magnitude to two significant figures.
The forces and their components are as follows:Notice the sign of each component. To find the resultant,
$$
\begin{array}{l}
R_{x}=\sum F_{x}=80 \mathrm{~N}+70.7 \mathrm{~N}-95.3 \mathrm{~N}-150.4 \mathrm{~N}=-95.0 \mathrm{~N} \\
R_{y}=\sum F_{y}=0+70.7 \mathrm{~N}+55.0 \mathrm{~N}-54.7 \mathrm{~N}=71.0 \mathrm{~N}
\end{array}
$$
The resultant is shown in Fig. $3-3$; there,
$$
R=\sqrt{(95.0 \mathrm{~N})^{2}+(71.0 \mathrm{~N})^{2}}=118.6 \mathrm{~N} \text { or } 119 \mathrm{~N}
$$
Further, tan $\alpha=(71.0 \mathrm{~N}) /(95.0 \mathrm{~N})$, from which $\alpha=37^{\circ}$. Therefore the resultant is $119 \mathrm{~N}$ at $180^{\circ}-37^{\circ}=143^{\circ}$ or $\overrightarrow{\mathbf{R}}=119 \mathrm{~N}-143^{\circ} \mathrm{FROM}+X$ -AXIS.

Supratim Pal
Supratim Pal
Numerade Educator
02:00

Problem 4

A force of $100 \mathrm{~N}$ makes an angle of $\theta$ with the $x$ -axis and has a scalar $y$ -component of $30 \mathrm{~N}$. Find both the scalar $x$ -component of the force and the angle $\theta$. (Remember that the number $100 \mathrm{~N}$ has three significant figures whereas $30 \mathrm{~N}$ has only two.)Begin your analysis by drawing a diagram. Here the data are sketched roughly in Fig. $3-4$. We wish to find $F_{x}$ and $\theta$. Since
$$
\sin \theta=\frac{30 \mathrm{~N}}{100 \mathrm{~N}}=0.30
$$
$\theta=17.46^{\circ}$, and thus, to two significant figures, $\theta=17^{\circ} .$ Then, using the $\cos \theta$
$$
F_{x}=(100 \mathrm{~N}) \cos 17.46^{\circ}=95 \mathrm{~N}
$$

Shelby Mohamed
Shelby Mohamed
Numerade Educator
01:00

Problem 5

A child pulls on a rope attached to a sled with a force of $60 \mathrm{~N}$. The rope makes an angle of $40^{\circ}$ to the ground. ( $a$ ) Compute the effective value of the pull tending to move the sled along the ground.
(b) Compute the force tending to lift the sled vertically.
As depicted in Fig. $3-5$, the components of the $60 \mathrm{~N}$ force are $39 \mathrm{~N}$ and $46 \mathrm{~N}$. ( $a$ ) The pull along the ground is the horizontal component, $46 \mathrm{~N}$. (b) The lifting force is the vertical component, $39 \mathrm{~N}$.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
01:39

Problem 6

A car whose weight is $F_{W}$ is on a ramp which makes an angle $\theta$ to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight?
As rendered in Fig. $3-6$, the car's weight is a force $\overrightarrow{\mathbf{F}}_{W}$ that pulls straight down on the car. We take components of $\overrightarrow{\mathbf{F}}$ along the incline and perpendicular to it. The ramp must balance the force component $F_{W} \cos \theta$ if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is $F_{N}$ and $F_{N}=F_{W} \cos \theta$.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:14

Problem 7

Three forces that act on a particle are given by $\overrightarrow{\mathbf{F}}_{1}=(20 \hat{\mathbf{i}}-36 \hat{\mathbf{j}}+73 \hat{\mathbf{k}}) \mathbf{N}, \overrightarrow{\mathbf{F}}_{2}=(-17 \hat{\mathbf{i}}+21 \hat{\mathbf{j}}-46 \hat{\mathbf{k}})$
$\mathrm{N}$, and $\overrightarrow{\mathbf{F}}_{3}=(-12 \hat{\mathbf{k}}) \mathrm{N}$. Find their resultant vector. Also find the magnitude of the resultant to two significant figures.We know that
$$
\begin{array}{l}
R_{x}=\sum F_{x}=20 \mathrm{~N}-17 \mathrm{~N}+0 \mathrm{~N}=3 \mathrm{~N} \\
R_{y}=\sum F_{y}=-36 \mathrm{~N}+21 \mathrm{~N}+0 \mathrm{~N}=-15 \mathrm{~N} \\
R_{z}=\sum F_{z}=73 \mathrm{~N}-46 \mathrm{~N}-12 \mathrm{~N}=15 \mathrm{~N}
\end{array}
$$
Since $\overrightarrow{\mathbf{R}}=R_{x} \hat{\mathbf{i}}+R_{y} \hat{\mathbf{j}}+R_{z} \hat{\mathbf{k}}$,
$$
\overrightarrow{\mathbf{R}}=3 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}
$$
To two significant figures, the three-dimensional Pythagorean theorem then gives
$$
R=\sqrt{R_{x}^{2}+R_{y}^{2}+R_{z}^{2}}=\sqrt{459}=21 \mathrm{~N}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
01:37

Problem 8

Find the weight on the surface of the Earth of a body whose mass is (a) $3.00 \mathrm{~kg}$, and $(b) 200 \mathrm{~g}$.
The general relation between mass $m$ and weight $F_{W}$ is $F_{W}=m g$. In this relation, $m$ must be in kilograms, $g$ in meters per second squared, and $F_{W}$ in newtons. On Earth, $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$. The acceleration due to gravity varies from place to place in the universe.
(a) $\quad F_{W}=(3.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=29.4 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}=29.4 \mathrm{~N}$
(b) $F_{W}=(0.200 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=1.96 \mathrm{~N}$

Shelby Mohamed
Shelby Mohamed
Numerade Educator
00:51

Problem 9

A $20.0 \mathrm{~kg}$ object that can move freely is subjected to a resultant force of $45.0 \mathrm{~N}$ in the $-x$ -direction. Find the acceleration of the object.

We make use of the second law in component form, $\sum F_{x}=m a_{x}$, with $\sum F_{x}=-45.0 \mathrm{~N}$ and $m=20.0 \mathrm{~kg}$. Then
$$
a_{x}=\frac{\sum F_{x}}{m}=\frac{-45.0 \mathrm{~N}}{20.0 \mathrm{~kg}}=-2.25 \mathrm{~N} / \mathrm{kg}=-2.25 \mathrm{~m} / \mathrm{s}^{2}
$$
where we have used the fact that $1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}$. Because the resultant force on the object is in the $-x$ -direction, its acceleration is also in that direction.

Ipsita Mandal
Ipsita Mandal
Numerade Educator
01:27

Problem 10

The object in Fig. $3-7(a)$ weighs $50 \mathrm{~N}$ and is supported by a cord. Find the tension in the cord.
We mentally isolate the object for discussion. Two forces act on it, the upward pull of the cord and the downward pull of gravity. We represent the pull of the cord by $F_{T}$, the tension in the cord. The pull of gravity, the weight of the object, is $F_{W}=50 \mathrm{~N}$. These two forces are seen in the free-body diagram of Fig. $3-7(b)$.The forces are already in component form and so we can write the first condition for equilibrium at once, taking $u p$ and to the right as positive directions:
$$
\begin{array}{lll}
\pm \sum F_{x}=0 & \text { becomes } & 0=0 \\
+\uparrow \sum F_{y}=0 & \text { becomes } & F_{T}-50 \mathrm{~N}=0
\end{array}
$$
from which $F_{T}=50 \mathrm{~N}$. Thus, when a single vertical cord (called a hanger) supports a body at equilibrium, the tension in the cord equals the weight of the body.

Anurag Kumar
Anurag Kumar
Numerade Educator
02:29

Problem 11

A $5.0$ -kg object is to be given an upward acceleration of $0.30 \mathrm{~m} / \mathrm{s}^{2}$ by a rope pulling straight upward on it. What must be the tension in the rope?
The free-body diagram for the object is shown in Fig. $3-8(b)$. The tension in the rope is $F_{T}$, and the weight of the object is $F_{W}=m g=(5.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=49.1 \mathrm{~N}$. Using $\sum F_{y}=m a_{y}$ with $u p$ taken as positive,
$$
+\uparrow \sum F_{y}=F_{T}-m g=m a_{y} \quad \text { or } \quad F_{T}-49.1 \mathrm{~N}=(5.0 \mathrm{~kg})\left(0.30 \mathrm{~m} / \mathrm{s}^{2}\right)
$$
from which $F_{T}=50.6 \mathrm{~N}=51 \mathrm{~N}$. As a check, we notice that $F_{T}$ is larger than $F_{W}$, as it must be if the object is to accelerate upward.

Anurag Kumar
Anurag Kumar
Numerade Educator
02:10

Problem 12

A horizontal force of $140 \mathrm{~N}$ is needed to pull a $60.0-\mathrm{kg}$ box across a horizontal floor at constant speed. What is the coefficient of friction between floor and box? Determine it to three significant figures even though that's quite unrealistic.
The free-body diagram for the box is rendered in Fig. $3-9 .$ Because the box does not move up or down, $a_{y}=0$. Therefore,
$$
+\uparrow \sum F_{y}=m a_{y} \quad \text { yields } \quad F_{N}-m g=(m)\left(0 \mathrm{~m} / \mathrm{s}^{2}\right)
$$
from which we find that $F_{N}=m g=(60.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=588.6 \mathrm{~N}$. Further, because the box is moving horizontally at constant speed, $a_{x}=0$ and so
$$
\pm \sum F_{x}=m a_{x} \quad \text { leads to } \quad 140 \mathrm{~N}-F_{\mathrm{f}}=0
$$
from which the friction force is $F_{\mathrm{f}}=140 \mathrm{~N}$. Then
$$
\mu_{k}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{140 \mathrm{~N}}{588.6 \mathrm{~N}}=0.238
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
02:58

Problem 13

The only force acting on a $5.0-\mathrm{kg}$ object has components $F_{x}=20 \mathrm{~N}$ and $F_{y}=30 \mathrm{~N}$. Find the acceleration of the object.
Use $\sum F_{x}=m a_{x}$ and $\sum F_{y}=m a_{y}$ to obtain
$$
\begin{array}{l}
a_{x}=\frac{\sum F_{x}}{m}=\frac{20 \mathrm{~N}}{5.0 \mathrm{~kg}}=4.0 \mathrm{~m} / \mathrm{s}^{2} \\
a_{y}=\frac{\sum F_{y}}{m}=\frac{30 \mathrm{~N}}{5.0 \mathrm{~kg}}=6.0 \mathrm{~m} / \mathrm{s}^{2}
\end{array}
$$
These components of the acceleration are shown in Fig. $3-10$. From the figure,
$$
a=\sqrt{(4.0)^{2}+(6.0)^{2}} \mathrm{~m} / \mathrm{s}^{2}=7.2 \mathrm{~m} / \mathrm{s}^{2}
$$
and $\theta=\arctan (6.0 / 4.0)=56^{\circ}$.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
01:53

Problem 14

A $600-\mathrm{N}$ object is to be given an acceleration of $0.70 \mathrm{~m} / \mathrm{s}^{2} .$ How large an unbalanced force must act upon it?

Notice that the weight, not the mass, of the object is given. Assuming the weight was measured on the Earth, use $F_{W}=m g$ to find
$$
m=\frac{F_{W}}{g}=\frac{600 \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=61.2 \mathrm{~kg}
$$
Now that we know the mass of the object $(61.2 \mathrm{~kg})$ and the desired acceleration $\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)$, the force is
$$
F=m a=(61.2 \mathrm{~kg})\left(0.70 \mathrm{~m} / \mathrm{s}^{2}\right)=42.8 \mathrm{~N} \text { or } 43 \mathrm{~N}
$$

Shelby Mohamed
Shelby Mohamed
Numerade Educator
01:45

Problem 15

A constant force acts on a $5.0 \mathrm{~kg}$ object and reduces its velocity from $7.0 \mathrm{~m} / \mathrm{s}$ to $3.0 \mathrm{~m} / \mathrm{s}$ in a time of $3.0 \mathrm{~s}$. Determine the force.

We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2
$$
a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2}
$$
Use $F=m a$ with $m=5.0 \mathrm{~kg}$ :
$$
F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N}
$$
The minus sign indicates that the force is a retarding force, directed opposite to the motion.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
03:12

Problem 16

A $400-g$ block with an initial speed of $80 \mathrm{~cm} / \mathrm{s}$ slides along a horizontal tabletop against a friction force of $0.70 \mathrm{~N}$. (a) How far will it slide before stopping? $(b)$ What is the coefficient of friction between the block and the tabletop?
(a) Take the direction of motion as positive. The only unbalanced force acting on the block is the friction force, $-0.70 \mathrm{~N}$. Therefore,
$$
\sum F=m a \quad \text { becomes } \quad-0.70 \mathrm{~N}=(0.400 \mathrm{~kg})(a)
$$
from which $a=-1.75 \mathrm{~m} / \mathrm{s}^{2}$. (Notice that $m$ is always in kilograms.) To find the distance the block slides, make use of $v_{i x}=0.80 \mathrm{~m} / \mathrm{s}, v_{f x}=0$, and $a=-1.75 \mathrm{~m} / \mathrm{s}^{2}$. Then $v_{f x}^{2}-v_{i x}^{2}=2 a x$ yields
$$
x=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 a}=\frac{(0-0.64) \mathrm{m}^{2} / \mathrm{s}^{2}}{(2)\left(-1.75 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.18 \mathrm{~m}
$$
(b) Because the vertical forces on the block must cancel, the upward push of the table $F_{N}$ must equal the weight $m g$ of the block. Then
$$
\mu_{k}=\frac{\text { Friction force }}{F_{N}}=\frac{0.70 \mathrm{~N}}{(0.40 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.178 \text { or } 0.18
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
03:00

Problem 17

A $600-\mathrm{kg}$ car is coasting along a level road at $30 \mathrm{~m} / \mathrm{s}$. (a) How large a retarding force (assumed constant) is required to stop it in a distance of $70 \mathrm{~m} ?(b)$ What is the minimum coefficient of friction between tires and roadway if this is to be possible? Assume the wheels are not locked, in which case we are dealing with static friction-there's no sliding.
(a) First find the car's acceleration from a constant- $a$ equation. It is known that $v_{i x}=30 \mathrm{~m} / \mathrm{s}, v_{f x}=0$, and $x=70 \mathrm{~m}$. Use $v_{f x}^{2}=v_{i x}^{2}+2 a x$ to find
$$
a=\frac{v_{f x}^{2}-v_{i x}^{2}}{2 x}=\frac{0-900 \mathrm{~m}^{2} / \mathrm{s}^{2}}{140 \mathrm{~m}}=-6.43 \mathrm{~m} / \mathrm{s}^{2}
$$
Now write
$$
F=m a=(600 \mathrm{~kg})\left(-6.43 \mathrm{~m} / \mathrm{s}^{2}\right)-3858 \mathrm{~N} \text { or }-3.9 \mathrm{kN}
$$
(b) Assume the force found in ( $a$ ) is supplied as the friction force between the tires and roadway. Therefore, the magnitude of the friction force on the tires is $F_{\mathrm{f}}=3858 \mathrm{~N}$. The coefficient of friction is given by $\mu_{s}=F_{\mathrm{f}} / F_{N}$, where $F_{N}$ is the normal force. In the present case, the roadway pushes up on the car with a force equal to the car's weight. Therefore,
$$
F_{N}=F_{W}=m g=(600 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5886 \mathrm{~N}
$$
so that
$$
\mu_{s}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{3858}{5886}=0.66
$$
The coefficient of friction must be at least $0.66$ if the car is to stop within $70 \mathrm{~m}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:32

Problem 18

An $8000-\mathrm{kg}$ engine pulls a 40000 -kg train along a level track and gives it an acceleration $a_{1}=1.20 \mathrm{~m} / \mathrm{s}^{2}$. What acceleration $\left(a_{2}\right)$ would the engine give to a $16000-\mathrm{kg}$ train?
For a given engine force, the acceleration is inversely proportional to the total mass. Thus,
$$
a_{2}=\frac{m_{1}}{m_{2}} a_{1}=\frac{8000 \mathrm{~kg}+40000 \mathrm{~kg}}{8000 \mathrm{~kg}+16000 \mathrm{~kg}}\left(1.20 \mathrm{~m} / \mathrm{s}^{2}\right)=2.40 \mathrm{~m} / \mathrm{s}^{2}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
02:40

Problem 19

As shown in Fig. $3-11(a)$, an object of mass $m$ is supported by a cord. Find the tension in the cord if the object is $(a)$ at rest, $(b)$ moving at constant velocity, $(c)$ accelerating upward with acceleration $a=3 g / 2$, and $(d)$ accelerating downward at $a=0.75 g$.

Two forces act on the object: the tension $F_{T}$ upward and the downward pull of gravity $m g$. They are shown in the free-body diagram in Fig. $3-11(b)$. As a rule begin your analysis with a diagram. Take $u p$ as the positive direction and write $+\uparrow \sum F_{y}=m a_{y}$ in each case.
$F_{T}-m g=m a_{y}=0 \quad$ or $\quad F_{T}=m g$
(a) $a_{y}=0: \quad F_{T}-m g=m a_{\mathrm{v}}=0 \quad$ or
(b) $a_{y}=0$ :
$F_{T}-m g=m a_{y}=0 \quad$ or $\quad F_{T}=m g$
$\begin{array}{llll}\text { (c) } a_{y}=3 g / 2: & F_{T}-m g=m(3 g / 2) & \text { or } & F_{T}=2.5 m g \\ \text { (d) } a_{y}=-3 g / 4: & F_{T}-m g=m(-3 g / 4) & \text { or } & F_{T}=0.25 m g\end{array}$
Notice that the tension in the cord is less than $m g$ in part $(d) ;$ only then can the object have a downward acceleration. Can you explain why $F_{T}=0$ if $a_{y}=-g$ ?

Anurag Kumar
Anurag Kumar
Numerade Educator
01:19

Problem 20

A tow rope will break if the tension in it exceeds $1500 \mathrm{~N}$. It is used to tow a $700-\mathrm{kg}$ car along level ground. What is the largest acceleration the rope can give to the car? (Remember that 1500 has four significant figures; see Appendix A.)
The forces acting on the car are shown in Fig. $3-12$. Only the $x$ -directed force is of importance, because the $y$ -directed forces balance each other. Indicating the positive direction with a $+$ sign and a little arrow, we write,
$$
\pm \sum F_{x}=m a_{x} \quad \text { becomes } \quad 1500 \mathrm{~N}=(700 \mathrm{~kg})(a)
$$
from which $a=2.14 \mathrm{~m} / \mathrm{s}^{2}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:27

Problem 21

Compute the least acceleration with which a $45-\mathrm{kg}$ woman can slide down a rope if the rope can withstand a tension of only $300 \mathrm{~N}$.
The weight of the woman is $m g=(45 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=441 \mathrm{~N}$. Because the rope can support only $300 \mathrm{~N}$, the unbalanced downward force $F$ on the woman (i.e., the accelerating force) must be at least $441 \mathrm{~N}-300 \mathrm{~N}=141 \mathrm{~N}$. Her minimum downward acceleration is then
$$
a=\frac{F}{m}=\frac{141 \mathrm{~N}}{45 \mathrm{~kg}}=3.1 \mathrm{~m} / \mathrm{s}^{2}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
02:08

Problem 22

A $70-\mathrm{kg}$ box is slid along the floor by a $400-\mathrm{N}$ force as shown in Fig. 3-13. The coefficient of friction between the box and the floor is $0.50$ when the box is sliding. Find the acceleration of the box.
Since the $y$ -directed forces must balance,
$$
F_{N}=m g=(70 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=686.7 \mathrm{~N}
$$
But the friction force $F_{\mathrm{f}}$ is given by
$$
F_{\mathrm{f}}=\mu_{k} F_{N}=(0.50)(687 \mathrm{~N})=343.4 \mathrm{~N}
$$
Now write $\sum F_{x}=m a_{x}$ for the box, taking the direction of motion as positive:
$$
\pm \sum F_{x}=400 \mathrm{~N}-343.4 \mathrm{~N}=(70 \mathrm{~kg})(a) \quad \text { or } \quad a=0.81 \mathrm{~m} / \mathrm{s}^{2}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
04:52

Problem 23

Suppose, as depicted in Fig. 3-14, that a $70-\mathrm{kg}$ box is pulled by a $400-\mathrm{N}$ force at an angle of $30^{\circ}$ to the horizontal. The coefficient of kinetic friction is $0.50$. Find the acceleration of the box.Because the box does not move up or down, we have $\sum F_{y}=m a_{y}=0 .$ From Fig. $3-14$, this equation is
$$
+\uparrow \sum F_{y}=F_{N}+200 \mathrm{~N}-m g=0
$$
But $m g=(70 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=686.7 \mathrm{~N}$, and it follows that $F_{N}=486.7 \mathrm{~N}$.
Next find the friction force acting on the box:
$$
F_{\mathrm{f}}=\mu_{k} F_{N}=(0.50)(486.7 \mathrm{~N})=243.4 \mathrm{~N}
$$
Now write $\sum F_{x}=m a_{x}$ for the box. It is
$$
(346-243.4) \mathrm{N}=(70 \mathrm{~kg})\left(a_{x}\right)
$$
from which $a_{x}=1.466 \mathrm{~m} / \mathrm{s}^{2}$ or $1.5 \mathrm{~m} / \mathrm{s}^{2}$

Shelby Mohamed
Shelby Mohamed
Numerade Educator
03:15

Problem 24

A car coasting at $20 \mathrm{~m} / \mathrm{s}$ along a horizontal road has its brakes suddenly applied and eventually comes to rest. What is the shortest distance in which it can be stopped if the friction coefficient between tires and road is $0.90 ?$ Assume that all four wheels brake identically. If the brakes don't lock, the car stops via static friction.
The friction force at one wheel, call it wheel 1 , is
$$
F_{\mathrm{fl}}=\mu_{5} F_{N 1}=\mu_{s} F_{W 1}
$$
where $F_{W}$ is the weight carried by wheel 1 . We obtain the total friction force $F_{\mathrm{f}}$ by adding such terms for all four wheels:
$$
F_{\mathrm{f}}=\mu_{s} F_{W 1}+\mu_{s} F_{W_{2}}+\mu_{s} F_{W 3}+\mu_{s} F_{W 4}=\mu_{s}\left(F_{W_{1}}+F_{W 2}+F_{W 3}+F_{W_{4}}\right)=\mu_{s} F_{w}
$$
where $F_{W}$ is the total weight of the car. (Notice that we are assuming optimal braking at each wheel.) This friction force is the only unbalanced force on the car (we neglect air friction). Writing $F=m a$ for the car with $F$ replaced by $-\mu_{s} F_{W}$ gives $-\mu_{s} F_{W}=m a$, where $m$ is the car's mass and the positive direction is taken as the direction of motion. However, $F_{W}=m g$; so the car's acceleration is
$$
a=-\frac{\mu_{s} F_{W}}{m}=-\frac{\mu_{s} m g}{m}=-\mu_{s} g=(-0.90)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=-8.829 \mathrm{~m} / \mathrm{s}^{2}
$$
We can determine how far the car went before stopping by solving the constant- $a$ motion problem. Knowing that $v_{i}=20 \mathrm{~m} / \mathrm{s}, v_{f}=0$, and $a=-8.829 \mathrm{~m} / \mathrm{s}^{2}$, we find from $v_{f}^{2}-v_{i}^{2}=2 a x$ that
$$
x=\frac{(0-400) \mathrm{m}^{2} / \mathrm{s}^{2}}{-17.66 \mathrm{~m} / \mathrm{s}^{2}}=22.65 \mathrm{~m} \quad \text { or } \quad 23 \mathrm{~m}
$$
If the four wheels had not all been braking optimally, the stopping distance would have been longer.

Anurag Kumar
Anurag Kumar
Numerade Educator
03:21

Problem 25

As seen in Fig. $3-15$, a force of $400 \mathrm{~N}$ pushes on a $25-\mathrm{kg}$ box. Starting from rest, the box uniformly speeds up and achieves a velocity of $2.0 \mathrm{~m} / \mathrm{s}$ in a time of $4.0 \mathrm{~s}$. Compute the coefficient of kinetic friction between box and floor.The box experiences an unbalanced horizontal force which is the $x$ -component of the applied force minus the friction force. This resultant force, $\pm \sum F_{x}$, accelerates the box horizontally in the $+x$ -direction. We can find $a$ from the uniformly accelerated motion and, with that and Newton's Second Law, determine $\sum F_{x}$.
To determine $a$, make use of the fact that $v_{i}=0, v_{f}=2.0 \mathrm{~m} / \mathrm{s}, t=4.0 \mathrm{~s}$, and $v_{f}=v_{i}+a t$. From which it follows that
$$
a=\frac{v_{f}-v_{i}}{t}=\frac{2.0 \mathrm{~m} / \mathrm{s}}{4.0 \mathrm{~s}}=0.50 \mathrm{~m} / \mathrm{s}^{2}
$$
and so $a_{x}=a=0.50 \mathrm{~m} / \mathrm{s}^{2}$. From Fig. $3-15$,
$$
\pm \sum F_{x}=257 \mathrm{~N}-F_{\mathrm{f}}=(25 \mathrm{~kg})\left(0.50 \mathrm{~m} / \mathrm{s}^{2}\right) \quad \text { or } \quad F_{\mathrm{f}}=245 \mathrm{~N}
$$
To find the coefficient of friction, recall that $\mu_{k}=F_{\mathrm{f}} / F_{N}$. We need $F_{N}$, which can be obtained from $+\uparrow \sum F_{y}=m a_{y}=0$, since no vertical motion occurs. From Fig. $3-15$,
$$
+\uparrow \sum F_{y}=F_{N}-306 \mathrm{~N}-(25)(9.81) \mathrm{N}=0 \quad \text { or } \quad F_{N}=551 \mathrm{~N}
$$
Finally,
$$
\mu_{k}=\frac{F_{\mathrm{f}}}{F_{\mathrm{N}}}=\frac{245}{551}=0.44
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
02:59

Problem 26

A 200-N wagon is to be pulled up a $30^{\circ}$ incline at constant speed. How large a force parallel to the incline is needed if friction effects are negligible?

The situation is shown in Fig. $3-16(a)$. Because the wagon moves at a constant speed along a straight line, its velocity vector is constant. Therefore, the wagon is in translational equilibrium, and the first condition for equilibrium applies to it.
We isolate the wagon as the object. Three non-negligible forces act on it: (1) the pull of gravity $F_{W}$ (its weight), directed straight down; (2) the applied force $F$ exerted on the wagon parallel to the incline to pull it up the incline; (3) the push $F_{N}$ of the incline that supports the wagon. These three forces are shown in the free-body diagram in Fig. $3-16$.
For situations involving inclines, it is convenient to take the $x$ -axis parallel to the incline and the $y$ -axis perpendicular to it. After taking components along these axes, we can write the first condition for equilibrium:
$$
\begin{array}{lll}
+\sum F_{x}=0 & \text { becomes } & F-0.50 F_{W}=0 \\
+\sum F_{y}=0 & \text { becomes } & F_{N}-0.866 F_{W}=0
\end{array}
$$
Solving the first equation and recalling that $F_{W}=200 \mathrm{~N}$, we find that $F=0.50 F_{W}$. The required pulling force to two significant figures is $0.10 \mathrm{kN}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
03:25

Problem 27

A $20-\mathrm{kg}$ box sits on an incline as illustrated in Fig. $3-17$. The coefficient of kinetic friction between box and incline is $0.30$. Find the acceleration of the box down the incline.

In solving inclined-plane problems, take the $x$ - and $y$ -axes as shown in the figure, parallel and perpendicular to the incline. We find the acceleration by writing $\sum F_{x}=m a_{x}$. But first determine the friction force $F_{\mathrm{f}}$. Using the fact that $\cos 30^{\circ}=0.866$,
$$
{ }_{+} \Sigma F_{y}=m a_{y}=0 \text { gives } \quad F_{N}-0.866 \mathrm{mg}=0
$$
from which $F_{N}=(0.866)(20 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=169.9 \mathrm{~N}$. Now find $F_{\mathrm{f}}$ from
$$
F_{\mathrm{f}}=\mu_{k} F_{N}=(0.30)(169.9 \mathrm{~N})=50.97 \mathrm{~N}
$$
Writing $+\nearrow \sum F_{x}=m a_{x}$,
$$
F_{\mathrm{f}}-0.50 m g=m a_{x} \quad \text { or } \quad 50.97 \mathrm{~N}-(0.50)(20)(9.81) \mathrm{N}=(20 \mathrm{~kg})\left(a_{x}\right)
$$
from which $a_{x}=-2.36 \mathrm{~m} / \mathrm{s}^{2}$. The box accelerates down the incline at $2.4 \mathrm{~m} / \mathrm{s}^{2}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
03:21

Problem 28

When a force of $500 \mathrm{~N}$ pushes on a $25-\mathrm{kg}$ box as shown in Fig. $3-18$, the resulting acceleration of the box up the incline is $0.75 \mathrm{~m} / \mathrm{s}^{2}$. Compute the coefficient of kinetic friction between the box and the incline.

The acting forces and their components are shown in Fig. $3-18$. Notice how the $x$ - and $y$ -axes are taken. Since the box moves up the incline, the friction force (which always acts to retard the motion) is directed down the incline.
First find $F_{\mathrm{f}}$ by writing $\sum F_{x}=m a_{x} .$ From Fig. $3-18$, using $\sin 40^{\circ}=0.643$,
$$
+>\sum F_{x}=383 \mathrm{~N}-F_{\mathrm{f}}-(0.643)(25)(9.81) \mathrm{N}=(25 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right)
$$
from which $F_{\mathrm{f}}=206.6 \mathrm{~N}$.
We also need $F_{N}$. Writing $\sum F_{y}=m a_{y}=0$, and using $\cos 40^{\circ}=0.766$,
$$
+\nwarrow \sum F_{y}=F_{N}-321.4 \mathrm{~N}-(0.766)(25)(9.81) \mathrm{N}=0 \quad \text { or } \quad F_{N}=509.3 \mathrm{~N}
$$
Then
$$
\mu_{k}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{206.6}{509.3}=0.41
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
03:33

Problem 29

Two blocks, of masses $m_{1}$ and $m_{2}$, moving in the $x$ -direction are pushed by a force $F$ as shown in Fig. 3-19. The coefficient of friction between each block and the table is $0.40 .(a)$ What must be the value of $F$ if the blocks are to have an acceleration of $200 \mathrm{~cm} / \mathrm{s}^{2} ?$ How large a force does $m_{1}$ then exert on $m_{2}$ ? Use $m_{1}=300 \mathrm{~g}$ and $m_{2}=500 \mathrm{~g}$. Remember to work in SI units.

The friction forces on the blocks are $F_{f 1}=0.40 m_{1} g$ and $F_{f 2}=0.40 m_{2} g$. Take the two blocks in combination as the object for discussion; the horizontal forces on the object from outside (i.e., the external forces on it) are $F$, $F_{f 1}$, and $F_{f 2}$. Although the two blocks do push on each other, those pushes are internal forces; they are not part of the unbalanced external force on the two-mass object. For that object,
$$
\pm \sum F_{x}=m a_{x} \text { becomes } F-F_{f 1}-F_{\mathrm{f} 2}=\left(m_{1}+m_{2}\right) a_{x}
$$
(a) Solving for $F$ and substituting known values
$$
F=0.40 g\left(m_{1}+m_{2}\right)+\left(m_{1}+m_{2}\right) a_{x}=3.14 \mathrm{~N}+1.60 \mathrm{~N}=4.7 \mathrm{~N}
$$
(b) Now consider block $m_{2}$ alone. The forces acting on it in the $x$ -direction are the push of block $m_{1}$ on it (which we represent by $F_{b}$ ) and the retarding friction force $F_{12}=0.40 m_{2} g$. Then, for it,
$$
\pm \sum F_{x}=m a_{x} \quad \text { becomes } \quad F_{b}-F_{f 2}=m_{2} a_{x}
$$
We know that $a_{x}=2.0 \mathrm{~m} / \mathrm{s}^{2}$ and so
$$
F_{b}=F_{\mathrm{f} 2}+m_{2} a_{x}=1.96 \mathrm{~N}+1.00 \mathrm{~N}=2.96 \mathrm{~N}=3.0 \mathrm{~N}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
01:46

Problem 30

A cord passing over a light frictionless pulley has a $7.0$ -kg mass hanging from one end and a $9.0-\mathrm{kg}$ mass hanging from the other, as seen in Fig. 3-20. (This arrangement is called Atwood's machine.) Find the acceleration of the masses and the tension in the cord.

Because the pulley is easily turned, the tension in the cord will be the same on each side. The forces acting on each of the two masses are drawn in Fig. 3-20. Recall that the weight of an object is $m g$.

It is convenient in situations involving objects connected by cords to take the overall direction of motion of the system as the positive direction. That's often indicated by the direction of motion of the pulley when the system is let free to move. In the present case, the pulley would turn clockwise, and so we take $u p$ positive for the $7.0$ -kg mass, and down positive for the $9.0$ -kg mass. (If we do this, the acceleration will be positive for each mass. Because the cord doesn't stretch, the accelerations are numerically equal.) Writing $\sum F_{y}=m a_{y}$ for each mass in turn,
$$
+\uparrow \sum F_{y A}=F_{T}-(7.0)(9.81) \mathrm{N}=(7.0 \mathrm{~kg})(a) \text { and }+\downarrow \sum F_{y B}=(9.0)(9.81) \mathrm{N}-F_{T}=(9.0 \mathrm{~kg})(a)
$$
Add these two equations and the unknown $F_{T}$ drops out, giving
$$
(9.0-7.0)(9.81) \mathrm{N}=(16 \mathrm{~kg})(a)
$$
for which $a=1.23 \mathrm{~m} / \mathrm{s}^{2}$ or $1.2 \mathrm{~m} / \mathrm{s}^{2}$. Now substitute $1.23 \mathrm{~m} / \mathrm{s}^{2}$ for $a$ in either equation and obtain $F_{T}=77 \mathrm{~N}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
04:08

Problem 31

In Fig. 3-21, the coefficient of kinetic friction between block-A and the table is $0.20$. Also, $m_{A}=25 \mathrm{~kg}$, and $m_{B}=15 \mathrm{~kg}$. How far will object- $B$ drop in the first $3.0 \mathrm{~s}$ after the system is released?To find how far object- $B$ falls, we will need to determine the acceleration of the system.
Since, for block- $A$, there is no motion vertically, the normal force is
$$
F_{N}=m_{A g}=(25 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=245 \mathrm{~N}
$$
and
$$
F_{\mathrm{f}}=\mu_{k} F_{N}=(0.20)(245 \mathrm{~N})=49.1 \mathrm{~N}
$$
To calculate the acceleration of the system, apply $F=m a$ to each block in turn. Taking the direction of motion to be positive, as indicated in Fig. $3-21(a)$,
$$
\pm \sum F_{x A}=F_{T}-F_{\mathrm{f}}=m_{A} a \quad \text { or } \quad F_{T}-49.1 \mathrm{~N}=(25 \mathrm{~kg})(a)
$$
and $\quad+\downarrow \sum F_{y B}=m_{B} g-F_{T}=m_{B} a$
or $-F_{T}+(15)(9.81) \mathrm{N}=(15 \mathrm{~kg})(a)$
Eliminate $F_{T}$ by adding the two equations. Then, solving for $a$, we find $a=2.45 \mathrm{~m} / \mathrm{s}^{2}$. We now have to deal with a constant-acceleration motion problem where $a=2.45 \mathrm{~m} / \mathrm{s}^{2}, v_{i}=0$, and $t=3.0 \mathrm{~s}$. Hence,
$y=v_{i y} t+\frac{1}{2} a t^{2}$, which leads to $y=0+\frac{1}{2}\left(2.45 \mathrm{~m} / \mathrm{s}^{2}\right)(3.0 \mathrm{~s})^{2}=11 \mathrm{~m}$
This is the distance $B$ falls in the first $3.0 \mathrm{~s}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
02:10

Problem 32

How large a horizontal force in addition to $F_{T}$ must pull on block- $A$ in Fig. $3-21$ to give it an acceleration of $0.75 \mathrm{~m} / \mathrm{s}^{2}$ toward the left? Assume, as in Problem $3.31$, that $\mu_{k}=0.20, m_{A}=25 \mathrm{~kg}$, and $m_{B}=15 \mathrm{~kg}$. Redraw Fig 3-21 for this case, including a force $F$ pulling toward the left on $A$. In addition, the retarding friction force $F_{\mathrm{f}}$ must be reversed in direction. As in Problem 3.31, $F_{\mathrm{f}}=49.1 \mathrm{~N}$.
Write $F=m a$ for each block in turn, taking the direction of motion (to the left and up) to be positive. We have
$$
\begin{aligned}
\pm \sum F_{x A}=F-F_{T}-49.1 \mathrm{~N} &=(25 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \text { and }+\uparrow \sum F_{y B}=F_{T}-(15)(9.81) \mathrm{N} \\
&=(15 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right)
\end{aligned}
$$
Solve the last equation for $F_{T}$ and substitute in the previous equation. Then solve for the single unknown $F$, and find it to be $226 \mathrm{~N}$ or $0.23 \mathrm{kN}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
03:37

Problem 33

The coefficient of static friction between a box and the flat bed of a truck is $0.60$. What is the maximum acceleration the truck can have along level ground if the box is not to slide?
The box experiences only one $x$ -directed force, the friction force. When the box is on the verge of slipping, $F_{\mathrm{f}}=\mu_{s} F_{W}$, where $F_{W}$ is the weight of the box.
As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck:
otherwise, the box will slip. When the box is not slipping, $\sum F_{x}=m a_{x}$ applied to the box gives $F_{\mathrm{f}}=m a_{x}$ However, if the box is on the verge of slipping, $F_{\mathrm{f}}=\mu_{s} F_{W}$ so that $\mu_{s} F_{W}=m a_{x} .$ Because $F_{W}=m g$,
$$
a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2}
$$
as the maximum acceleration without slipping.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:17

Problem 34

In Fig. 3-22, the two boxes have identical masses of $40 \mathrm{~kg}$. Both experience a sliding friction force with $\mu_{k}=0.15$. Find the acceleration of the boxes and the tension in the tie cord.
Using $F_{f}=\mu_{k} F_{N}$, the friction forces on the two boxes are
$$
F_{\mathrm{fA}}=(0.15)(\mathrm{mg}) \quad \text { and } \quad F_{\mathrm{fB}}=(0.15)(0.866 \mathrm{mg})
$$
But $m=40 \mathrm{~kg}$, so $F_{f A}=58.9 \mathrm{~N}$ and $F_{\mathrm{f} B}=51.0 \mathrm{~N}$.
Now apply $\sum F_{x}=m a_{x}$ to each block in turn, taking the direction of motion of the system as positive. We want to sum the forces parallel to each surface, and that's often indicated using a subscript $\|$ symbol. Accordingly
$$
\pm \sum F_{\| A}=F_{T}-58.9 \mathrm{~N}=(40 \mathrm{~kg})(a) \quad \text { and } \quad \searrow \sum F_{\| B}=0.5 m g-F_{T}-51 \mathrm{~N}=(40 \mathrm{~kg})(a)
$$
Solving these two equations for $a$ and $F_{T}$ gives $a=1.1 \mathrm{~m} / \mathrm{s}^{2}$ and $F_{T}=0.10 \mathrm{kN}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
03:52

Problem 35

In the system shown in Fig. $3-23(a)$, force $F$ accelerates block- 1 of mass $m_{1}$ to the right. Write an expression for its acceleration in terms of $F$ and the coefficient of friction $\mu_{k}$ at the contact surfaces.The horizontal forces on the blocks are shown in Fig. $3-23(b)$ and $(c) .$ Block- 2 of mass $m_{2}$ is pressed against block-1 by its weight $m_{2} g$. This is the normal force where $m_{1}$ and $m_{2}$ are in contact, so the friction force there is $F_{\mathrm{f} 2}=\mu_{k} m_{2} g$. At the bottom surface of $m_{1}$, however, the normal force is $\left(m_{1}+m_{2}\right) g$. Hence, $F_{\mathrm{f}}^{\prime}=\mu_{k}\left(m_{1}+m_{2}\right) g$. We now write $\sum F_{x}=m a_{x}$ for each block, taking the direction of motion of the system as positive (i.e., to the left on block- 2 and to the right on block-1):
$$
\pm \sum F_{x 2}=F_{T}-\mu_{k} m_{2} g=m_{2} a \quad \text { and } \quad \pm \sum F_{x 1}=F-F_{T}-\mu m_{2} g-\mu_{k}\left(m_{1}+m_{2}\right) \mathrm{g}=m_{1} a
$$
Eliminate $F_{T}$ by adding the two equations to obtain
$$
F-2 \mu_{k} m_{2} g-\mu_{k}\left(m_{1}+m_{2}\right)(g)=\left(m_{1}+m_{2}\right)(a)
$$
from which it follows that $a=\frac{F-2 \mu_{k} m_{2} g}{m_{1}+m_{2}}-\mu_{k} g$

Anurag Kumar
Anurag Kumar
Numerade Educator
01:58

Problem 36

In the system of Fig. 3-24, friction and the mass of the pulley are both negligible. Find the acceleration of $m_{2}$ if $m_{1}=300 \mathrm{~g}, m_{2}=500 \mathrm{~g}$, and $F=1.50 \mathrm{~N}$.Notice that $m_{1}$ has twice as large an acceleration as $m_{2} .$ (When the pulley moves a distance $d, m_{1}$ moves a distance $2 d$.) Also notice that the tension $F_{T 1}$ in the cord pulling $m_{1}$ is half $F_{T 2}$, that in the cord pulling the pulley, because the total force on the pulley must be zero. $(F=m a$ tells us that this is so because the mass of the pulley is zero.) Writing $\sum F_{x}=m a_{x}$ for each mass,$$
\pm \sum F_{x 1}=F_{T 1}=\left(m_{1}\right)(2 a) \text { and } \pm \sum F_{x 2}=F-F_{T 2}=m_{2} a
$$
However, we know that $F_{T 1}=\frac{1}{2} F_{T 2}$ and so the first equation gives $F_{T 2}=4 m_{1} a$. Substitution in the second equation yields
$$
F=\left(4 m_{1}+m_{2}\right)(a) \quad \text { or } \quad a=\frac{F}{4 m_{1}+m_{2}}=\frac{1.50 \mathrm{~N}}{1.20 \mathrm{~kg}+0.50 \mathrm{~kg}}=0.882 \mathrm{~m} / \mathrm{s}^{2}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
02:53

Problem 37

In Fig. 3-25, the weights of the objects are $200 \mathrm{~N}$ and $300 \mathrm{~N}$. The pulleys are essentially frictionless and massless. Pulley $P_{1}$ has a stationary axle, but pulley $P_{2}$ is free to move up and down. Find the tensions $F_{T 1}$ and $F_{T 2}$ and the acceleration of each body. Only do this problem if you are already familiar with the action of pulleys. Mass $B$ will rise and mass $A$ will fall. You can see this by noticing that the forces acting on pulley $P_{2}$ are $2 F_{T 2}$ up and $F_{T 1}$ down. Since the pulley has no mass, it can have no acceleration, and so $F_{T 1}=2 F_{T 2}$ (the inertialess object transmits the tension). Twice as large a force is pulling upward on $B$ as on $A$.

Let $a$ be the downward acceleration of $A$. Then $a / 2$ is the upward acceleration of $B$. (Why?) Now write $\sum F_{y}=m a_{y}$ for each mass in turn, taking the direction of motion as positive in each case. We have
$$
+\uparrow \sum F_{B}=F_{T 1}-300 \mathrm{~N}=\left(m_{B}\right)\left(\frac{1}{2} a\right) \quad \text { and } \quad+\downarrow \sum F_{A}=200 \mathrm{~N}-F_{T 2}=m_{\mathrm{A}} a
$$
But $m=F_{W} / g$ and so $m_{A}=(200 / 9.81) \mathrm{kg}$ and $m_{B}=(300 / 9.81) \mathrm{kg} .$ Further $F_{T 1}=2 F_{T 2} .$ Substitution of these values in the two equations allows us to compute $F_{T 2}$ and then $F_{T 1}$ and $a$. The results are
$$
F_{T 1}=327 \mathrm{~N} \quad F_{T 2}=164 \mathrm{~N} \quad a=1.78 \mathrm{~m} / \mathrm{s}^{2}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
01:41

Problem 38

The Moon, whose mass is $7.35 \times 10^{22} \mathrm{~kg}$, orbits the Earth, whose mass is $5.98 \times 10^{24} \mathrm{~kg}$, at a mean distance of $3.85 \times 10^{8} \mathrm{~m}$. It is held in a nearly circular orbit by the Earth-Moon gravitational interaction. Determine the force of gravity due to the planet acting on the Moon.
From the univesal law of gravitation
$$
F_{G}=G \frac{m M}{R^{2}}
$$we get
$$
F_{G}=6.673 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg} \frac{\left(7.35 \times 10^{22} \mathrm{~kg}\right)\left(5.98 \times 10^{24}\right)}{\left(3.85 \times 10^{8} \mathrm{~m}\right)^{2}}
$$
which yields
$$
F_{G}=1.98 \times 10^{20} \mathrm{~N}
$$
This is also the force on the Earth due to the Moon.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:34

Problem 39

Compute the mass of the Earth, assuming it to be a sphere of radius $6370 \mathrm{~km}$. Use the fact that $\mathrm{g}=9.81 \mathrm{~m} / \mathrm{s}^{2}$ and give your answer to three significant figures.
Let $M_{E}$ be the mass of the Earth, and $m$ the mass of an object. The weight of the object on the planet's surface is equal to $m g .$ It is also equal to the gravitational force $G\left(M_{E} m\right) R_{E}^{2}$, where $R_{E}$ is the Earth's radius. Hence,
$$
m g=G \frac{M_{E} m}{R_{E}^{2}}
$$
from which $\quad M_{E}=\frac{g R_{E}^{2}}{G}=\frac{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(6.37 \times 10^{6} \mathrm{~m}\right)^{2}}{6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}}=5.97 \times 10^{24} \mathrm{~kg}$

Anurag Kumar
Anurag Kumar
Numerade Educator
01:52

Problem 40

Consider an essentially spherical homogeneous celestial body of mass $M$. The acceleration due to gravity in its vicinity beyond its surface at a distance $R$ from its center is $g_{R}$. Show that
$$
g_{R}=\frac{G M}{R^{2}}
$$
Notice that the acceleration drops off as $1 / R^{2}$.
Imagine an object of mass $m$ at a distance $R$ from the center of our celestial body. Its weight is $F_{W}=m g_{R}$, but that's also the gravitation force on it due to the mass $M$, that is, $F_{W}=F_{G}$. Hence,
$$
m g_{R}=G \frac{m M}{R^{2}}
$$

Shelby Mohamed
Shelby Mohamed
Numerade Educator
01:37

Problem 41

The mythical planet Mongo has twice the mass and twice the radius of Earth. Compute the acceleration due to gravity at its surface.
We know from Problem $3.40$ that in general
$$
g_{R}=\frac{G M}{R^{2}}
$$
Then for the Earth at its surface
$$
g_{E}=\frac{G M_{E}}{R_{E}^{2}}=g=9.81 \mathrm{~m} / \mathrm{s}^{2}
$$
where $R_{E}$ is the Earth's radius and $M_{E}$ is its mass. For Mongo
$$
g_{M}=\frac{G M_{M}}{R_{M}^{2}}=\frac{G\left(2 M_{E}\right)}{\left(2 R_{E}\right)^{2}}
$$
and
or
$$
\begin{array}{c}
g_{M}=\frac{1}{2} \frac{G M_{E}}{R_{E}^{2}}=\frac{1}{2}\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) \\
g_{M}=4.91 \mathrm{~m} / \mathrm{s}^{2}
\end{array}
$$

Anurag Kumar
Anurag Kumar
Numerade Educator
06:39

Problem 42

Two forces act on a point object as follows: $100 \mathrm{~N}$ at $170.0^{\circ}$ and $100 \mathrm{~N}$ at $50.0^{\circ} .$ Find their resultant.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
04:13

Problem 43

Compute algebraically the resultant of the following coplanar forces: $100 \mathrm{~N}$ at $30^{\circ}, 141.4 \mathrm{~N}$ at $45^{\circ}$, and $100 \mathrm{~N}$ at $240^{\circ}$. Check your result graphically.

Paul Gabriel
Paul Gabriel
Numerade Educator
04:07

Problem 44

Two forces, $80 \mathrm{~N}$ and $100 \mathrm{~N}$, acting at an angle of $60^{\circ}$ with each other, pull on an object. ( $a$ ) What single force would replace the two forces? (b) What single force (called the equilibrant) would balance the two forces? Solve algebraically.

Paul Gabriel
Paul Gabriel
Numerade Educator
04:39

Problem 45

Find algebraically the $(a)$ resultant and $(b)$ equilibrant (see Problem 3.44) of the following coplanar forces: $300 \mathrm{~N}$ at exactly $0^{\circ}, 400 \mathrm{~N}$ at $30^{\circ}$, and $400 \mathrm{~N}$ at $150^{\circ}$.

Paul Gabriel
Paul Gabriel
Numerade Educator
02:31

Problem 46

Having hauled it to the top of a tilted driveway, a child is holding a wagon from rolling back down. The driveway is inclined at $20^{\circ}$ to the horizontal. If the wagon weighs $150 \mathrm{~N}$, with what force must the child pull on the handle if the handle is parallel to and pointing up the incline?

Paul Gabriel
Paul Gabriel
Numerade Educator
03:08

Problem 47

Repeat Problem $3.46$ if the handle is now raised at an angle of $30^{\circ}$ above the incline.

Paul Gabriel
Paul Gabriel
Numerade Educator
01:26

Problem 48

Once ignited, a small rocket motor on a spacecraft exerts a constant force of $10 \mathrm{~N}$ for $7.80 \mathrm{~s}$. During the burn, the rocket causes the $100-\mathrm{kg}$ craft to accelerate uniformly. Determine that acceleration.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:32

Problem 49

Typically, a bullet leaves a standard 45 -caliber pistol (5.0-in. barrel) at a speed of $262 \mathrm{~m} / \mathrm{s}$. If it takes $1 \mathrm{~ms}$ to traverse the barrel, determine the average acceleration experienced by the $16.2-\mathrm{g}$ bullet within the gun, and then compute the average force exerted on it.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:22

Problem 50

A force acts on a $2-\mathrm{kg}$ mass and gives it an acceleration of $3 \mathrm{~m} / \mathrm{s}^{2}$. What acceleration is produced by the same force when acting on a mass of $(a) 1 \mathrm{~kg} ?$ (b) $4 \mathrm{~kg} ?(c)$ How large is the force?

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:26

Problem 51

An object has a mass of $300 \mathrm{~g}$. (a) What is its weight on Earth? ( $b$ ) What is its mass on the Moon? ( $c$ ) What will be its acceleration on the Moon under the action of a $0.500-\mathrm{N}$ resultant force?

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:51

Problem 52

A horizontal cable pulls a $200-\mathrm{kg}$ cart along a horizontal track. The tension in the cable is $500 \mathrm{~N}$. Starting from rest, (a) How long will it take the cart to reach a speed of $8.0 \mathrm{~m} / \mathrm{s}$ ? (b) How far will it have gone?

Shelby Mohamed
Shelby Mohamed
Numerade Educator
01:14

Problem 53

A $900-\mathrm{kg}$ car is going $20 \mathrm{~m} / \mathrm{s}$ along a level road. How large a constant retarding force is required to stop it in a distance of $30 \mathrm{~m}$ ? [Hint: First find its deceleration.]

Anurag Kumar
Anurag Kumar
Numerade Educator
02:43

Problem 54

A $12.0-\mathrm{g}$ bullet is accelerated from rest to a speed of $700 \mathrm{~m} / \mathrm{s}$ as it travels $20.0 \mathrm{~cm}$ in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force? [Hint: Be careful of units.]

Shelby Mohamed
Shelby Mohamed
Numerade Educator
03:56

Problem 55

A 20 -kg crate hangs at the end of a long rope. Find its acceleration (magnitude and direction) when the tension in the rope is $(a) 250 \mathrm{~N},(b) 150 \mathrm{~N},(c)$ zero, $(d) 196 \mathrm{~N}$.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
04:29

Problem 56

A $5.0-\mathrm{kg}$ mass hangs at the end of a cord. Find the tension in the cord if the acceleration of the mass is
(a) $1.5 \mathrm{~m} / \mathrm{s}^{2}$ up, $(b) 1.5 \mathrm{~m} / \mathrm{s}^{2}$ down,
(c) $9.81 \mathrm{~m} / \mathrm{s}^{2}$ down. Don't forget gravity.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
05:20

Problem 57

A 700-N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of ( $a$ ) $1.8 \mathrm{~m} / \mathrm{s}^{2}$ up? (b) $1.8 \mathrm{~m} / \mathrm{s}^{2}$ down?
(c) $9.8 \mathrm{~m} / \mathrm{s}^{2}$ down?

Shelby Mohamed
Shelby Mohamed
Numerade Educator
01:35

Problem 58

Using the scale described in Problem 3.57, a 65.0-kg astronaut weighs himself on the Moon, where $g=1.60 \mathrm{~m} / \mathrm{s}^{2} .$ What does the scale read?

Shelby Mohamed
Shelby Mohamed
Numerade Educator
04:36

Problem 59

A cord passing over a frictionless, massless pulley has a $4.0$ -kg object tied to one end and a $12-\mathrm{kg}$ object tied
to the other. Compute the acceleration and the tension in the cord.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
03:21

Problem 60

An elevator starts from rest with a constant upward acceleration. It moves $2.0 \mathrm{~m}$ in the first $0.60 \mathrm{~s}$. A passenger in the elevator is holding a $3.0-\mathrm{kg}$ package by a vertical string. What is the tension in the string during the accelerating process?

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:19

Problem 61

Just as her parachute opens, a $60-\mathrm{kg}$ parachutist is falling at a speed of $50 \mathrm{~m} / \mathrm{s}$. After $0.80 \mathrm{~s}$ has passed, the chute is fully open and her speed has dropped to $12.0 \mathrm{~m} / \mathrm{s}$. Find the average retarding force exerted upon the chutist during this time if the deceleration is uniform.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
07:51

Problem 62

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a $900-\mathrm{g}$ mass. ( $a$ ) Find the tension in each string when the masses are accelerating upward at $0.700 \mathrm{~m} / \mathrm{s}^{2}$. Don't forget gravity. (b) Find the tension in each string when the acceleration is $0.700 \mathrm{~m} / \mathrm{s}^{2}$ downward.

Shelby Mohamed
Shelby Mohamed
Numerade Educator
02:21

Problem 63

A 20 -kg wagon is pulled along the level ground by a rope inclined at $30^{\circ}$ above the horizontal. A friction force of $30 \mathrm{~N}$ opposes the motion. How large is the pulling force if the wagon is moving with $(a)$ constant speed and
(b) an acceleration of $0.40 \mathrm{~m} / \mathrm{s}^{2}$ ?

Paul Gabriel
Paul Gabriel
Numerade Educator
01:42

Problem 64

A $12-\mathrm{kg}$ box is released from the top of an incline that is $5.0 \mathrm{~m}$ long and makes an angle of $40^{\circ}$ to the horizontal. A 60-N friction force impedes the motion of the box. ( $a$ ) What will be the acceleration of the box, and $(b)$ how long will it take to reach the bottom of the incline?

Anurag Kumar
Anurag Kumar
Numerade Educator
01:03

Problem 65

For the situation outlined in Problem $3.64$, what is the coefficient of friction between the box and the incline?

Anurag Kumar
Anurag Kumar
Numerade Educator
01:50

Problem 66

An inclined plane makes an angle of $30^{\circ}$ with the horizontal. Find the constant force, applied parallel to the plane, required to cause a $15-\mathrm{kg}$ box to slide $(a)$ up the plane with acceleration $1.2 \mathrm{~m} / \mathrm{s}^{2}$ and $(b)$ down the incline with acceleration $1.2 \mathrm{~m} / \mathrm{s}^{2} .$ Neglect friction forces.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:51

Problem 67

A horizontal force $F$ is exerted on a $20-\mathrm{kg}$ box to slide it up a $30^{\circ}$ incline. The friction force retarding the motion is $80 \mathrm{~N}$. How large must $F$ be if the acceleration of the moving box is to be $(a)$ zero and $(b) 0.75 \mathrm{~m} / \mathrm{s}^{2} ?$ The situation resembles that of Fig. $3-18$.

Anurag Kumar
Anurag Kumar
Numerade Educator
02:37

Problem 68

An inclined plane making an angle of $25^{\circ}$ with the horizontal has a pulley at its top. A $30-\mathrm{kg}$ block on the plane is connected to a freely hanging 20 -kg block by means of a cord passing over the pulley. Compute the distance the $20-\mathrm{kg}$ block will fall in $2.0 \mathrm{~s}$ starting from rest. Neglect friction.

Anurag Kumar
Anurag Kumar
Numerade Educator
02:27

Problem 69

Repeat Problem $3.68$ if the coefficient of friction between block and plane is $0.20 .$

Anurag Kumar
Anurag Kumar
Numerade Educator
02:20

Problem 70

A horizontal force of $200 \mathrm{~N}$ is required to cause a $15-\mathrm{kg}$ block to slide up a $20^{\circ}$ incline with an acceleration of $25 \mathrm{~cm} / \mathrm{s}^{2}$. Find $(a)$ the friction force on the block and $(b)$ the coefficient of friction.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:25

Problem 71

Find the acceleration of the blocks in Fig. 3-26 if friction forces are negligible. What is the tension in the cord connecting them?

Anurag Kumar
Anurag Kumar
Numerade Educator
02:37

Problem 72

Repeat Problem $3.71$ if the coefficient of kinetic friction between the blocks and the table is $0.30$.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:48

Problem 73

How large a force $F$ is needed in Fig. $3-27$ to pull out the $6.0$ -kg block with an acceleration of $1.50 \mathrm{~m} / \mathrm{s}^{2}$ if the coefficient of friction at its surfaces is $0.40 ?$

Anurag Kumar
Anurag Kumar
Numerade Educator
01:42

Problem 74

In Fig. $3-28$, how large a force $F$ is needed to give the blocks an acceleration of $3.0 \mathrm{~m} / \mathrm{s}^{2}$ if the coefficient of kinetic friction between blocks and table is $0.20 ?$ How large a force does the $1.50-\mathrm{kg}$ block then exert on the $2.0$ -kg block?

Anurag Kumar
Anurag Kumar
Numerade Educator
03:33

Problem 75

(a) What is the smallest force parallel to a $37^{\circ}$ incline needed to keep a $100-\mathrm{N}$ weight from sliding down the incline if the coefficients of static and kinetic friction are both $0.30 ?(b)$ What parallel force is required to keep the weight moving up the incline at constant speed? ( $c$ ) If the parallel pushing force is $94 \mathrm{~N}$, what will be the acceleration of the object? $(d)$ If the object in $(c)$ starts from rest, how far will it move in $10 \mathrm{~s}$ ?

Anurag Kumar
Anurag Kumar
Numerade Educator
View

Problem 76

A $5.0-\mathrm{kg}$ block rests on a $30^{\circ}$ incline. The coefficient of static friction between the block and the incline is $0.20$. How large a horizontal force must push on the block if the block is to be on the verge of sliding $(a)$ up the incline and $(b)$ down the incline?

Susan Hallstrom
Susan Hallstrom
Numerade Educator
02:21

Problem 77

Three blocks with masses $6.0 \mathrm{~kg}, 9.0 \mathrm{~kg}$, and $10 \mathrm{~kg}$ are connected as shown in Fig. $3-29 .$ The coefficient of friction between the table and the $10-\mathrm{kg}$ block is $0.20$. Find $(a)$ the acceleration of the system and $(b)$ the
tension in the cord on the left and in the cord on the right.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:15

Problem 78

Floating in space far from anything else are two spherical asteroids, one having a mass of $20 \times 10^{10} \mathrm{~kg}$ and the other a mass of $40 \times 10^{10} \mathrm{~kg}$. Compute the force of attraction on each one due to gravity when their center-to-center separation is $10 \times 10^{6} \mathrm{~m}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:24

Problem 79

Two cannonballs that each weigh $4.00 \mathrm{kN}$ on Earth are floating in space far from any other objects. Determine the mutually attractive gravitational force acting on them when they are separated, center-to-center, by $10.0 \mathrm{~m}$.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:52

Problem 80

A space station that weighs $10.0 \mathrm{MN}$ on Earth is positioned at a distance of ten Earth radii from the center of the planet. What would it weigh out there in space- that is, what is the value of the gravity force pulling it toward Earth?

Anurag Kumar
Anurag Kumar
Numerade Educator
01:15

Problem 81

An object that weighs $2700 \mathrm{~N}$ on the surface of the Earth is raised to a height (i.e., altitude) of two Earth radii above the surface. What will it weigh up there?

Anurag Kumar
Anurag Kumar
Numerade Educator
01:09

Problem 82

Imagine a planet having a mass twice that of Earth and a radius equal to $1.414$ times that of Earth. Determine the acceleration due to gravity at its surface.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:25

Problem 83

The Earth's radius is about $6370 \mathrm{~km}$. An object that has a mass of $20 \mathrm{~kg}$ is taken to a height of $160 \mathrm{~km}$ above the Earth's surface. ( $a$ ) What is the object's mass at this height? ( $b$ ) How much does the object weigh (i.e. how large a gravitational force does it experience) at this height?

Anurag Kumar
Anurag Kumar
Numerade Educator
01:42

Problem 84

A man who weighs $1000 \mathrm{~N}$ on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is $4.80$ times Earth's mass and a diameter which is $0.500$ times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?

Anurag Kumar
Anurag Kumar
Numerade Educator
02:18

Problem 85

The radius of the Earth is about $6370 \mathrm{~km}$, while that of Mars is about $3440 \mathrm{~km}$. If an object weighs $200 \mathrm{~N}$ on Earth, what would it weigh, and what would be the acceleration due to gravity, on Mars? The mass of Mars is $0.11$ that of Earth. Neglect planetary rotations and local mass variations.

Anurag Kumar
Anurag Kumar
Numerade Educator
01:40

Problem 86

The fabled planet Dune has a diameter eight times that of Earth and a mass twice as large. If a robot weighs $1800 \mathrm{~N}$ on the surface of (nonspinning) Dune, what will it weigh at the poles on Earth? Take our planet to be a sphere.

Anurag Kumar
Anurag Kumar
Numerade Educator
02:49

Problem 87

An astronaut weighs $480 \mathrm{~N}$ on Earth. She visits the planet Krypton, which has a mass and diameter each ten times that of Earth. Determine her weight at a distance of two Kryptonian radii above that fictional planet.

Anurag Kumar
Anurag Kumar
Numerade Educator