• Home
  • Textbooks
  • Schaum’s Outline of College Physics
  • Simple Machines

Schaum’s Outline of College Physics

Eugene Hecht

Chapter 7

Simple Machines - all with Video Answers

Educators

Chapter Questions

02:24

Problem 1

In a particular hoist system, the load is lifted $10 \mathrm{~cm}$ for each $70 \mathrm{~cm}$ of movement of the rope that operates the device. What is the smallest input force that could possibly lift a 5.0-kN load?
The most advantageous situation possible is that in which all the input work is used to lift the load- that is, in which friction and other loss mechanisms are negligible. In that case,
Work input = Lifting work
If the load is lifted a distance $l$, the lifting work is $(5.0 \mathrm{kN})(l)$. The input force $F$, however, must work through a distance $7.0$ l. The above equation then becomes
$$
(F)(7.0 l)=(5.0 \mathrm{kN})(l)
$$
which gives $F=0.71 \mathrm{kN}$ as the smallest possible force required.

Emily Anderson
Emily Anderson
Numerade Educator
03:33

Problem 2

A hoisting machine lifts a 3000 -kg load a height of $8.00 \mathrm{~m}$ in a time of $20.0 \mathrm{~s}$. The power supplied to the engine is $18.0 \mathrm{hp}$. Compute $(a)$ the work output, $(b)$ the power output and power input, and $(c)$ the efficiency of the engine and hoist system.
(b) Power output = $\frac{\text { Work outpat }}{\text { Time taken }}=\frac{235 \mathrm{~kJ}}{200 \mathrm{~s}}=11.8 \mathrm{~kW}$
(c) Efficiency $=\frac{\text { Poner outpot }}{\text { Poner input }}=\frac{11.8 \mathrm{~kW}}{13.4 \mathrm{~kW}}=0.881-88.1 \%$
or Efficiency $=\frac{\text { Work oulput }}{\text { Work input }}-\frac{235 \mathrm{~kJ}}{(13.4 \mathrm{~kJ} / \mathrm{s})(20.0 \mathrm{~s})}-0.877-87.7 \mathrm{~s}$.
The efficiency is $88 \%$; the differences arise from the rounding off
process.

Emily Anderson
Emily Anderson
Numerade Educator
02:44

Problem 3

What power in $\mathrm{kW}$ is supplied to a $12.0$ -hp motor having an efficiency of $90.0$ percent when it is delivering its full rated output?
From the definition of efficiency,
$$
\text { Power input }=\frac{\text { Power output }}{\text { Efficiency }}=\frac{(12.0 \mathrm{hp})(0.746 \mathrm{~kW} / \mathrm{hp})}{0.900}=9.95 \mathrm{~kW}
$$

Emily Anderson
Emily Anderson
Numerade Educator
07:11

Problem 4

For the three levers shown in Fig. $7-1$, determine the vertical forces $F_{1}, F_{2}$, and $F_{3}$ required to support the $\operatorname{load} F_{W}=90 \mathrm{~N}$. Neglect the weights of the levers. Also find the IMA, AMA, and efficiency for each system.
In each case, we take torques about the fulcrum point as axis. If we assume that the lifting is occurring slowly at constant speed, then the systems are in equilibrium; the clockwise torques balance the counterclockwise torques. (Recall that torque $=r_{F} \sin \theta .$ )
Clockwise torque = Counterclockwise torque
$\begin{array}{lll}\text { (a) }(2.0 \mathrm{~m})(90 \mathrm{~N})(1)=(4.0 \mathrm{~m})\left(F_{1}\right)(1) & \text { from which } & F_{1}=\end{array}$
$45 \mathrm{~N}$
(b) $(1.0 \mathrm{~m})(90 \mathrm{~N})(1)=(3.0 \mathrm{~m})\left(F_{2}\right)(1) \quad$ from which $\quad F_{2}=$
$30 \mathrm{~N}$
(c) $(2.0 \mathrm{~m})(90 \mathrm{~N})(1)=(5.0 \mathrm{~m})\left(F_{3}\right) \sin 60^{\circ}$ from which $\quad F_{3}=42$
$\mathbf{N}$
To find the IMA of the system in Fig. $7-1(a)$, we notice that the load moves only half as far as the input force, and so
$$
\mathrm{IMA}=\text { Distance ratio }=2.0
$$
Similarly, in Fig. $7-1(b) .$ IMA $=3 / 1=3 .$ In Fig. $7-1(c)$, however, the lever arm is $(5.0 \mathrm{~m}) \sin 60^{\circ}=4.33 \mathrm{~m}$ and so the distance ratio is $4.33 / 2=2.16$. To summarize:
The efficiencies are $1.0$ because we have neglected friction at the fulcrums.

Samuel Smith
Samuel Smith
Numerade Educator
04:53

Problem 5

Determine the force $F$ required to hold a $100-\mathrm{N}$ load $F_{W}$ in equilibrium for each of the pulley systems shown in Neglect friction and the weights of the pulleys. In each case determine the net force on the ceiling. (a) Load $F_{W}$ is supported by two ropes; each rope exerts an upward pull of $F_{T}=\frac{1}{2} F_{W} .$ Because the rope is continuous and the pulleys are frictionless, $F_{T}=F$. Then
$$
F=F_{T}=\frac{1}{2} F_{W}=\frac{1}{2}(100 \mathrm{~N})=50 \mathrm{~N}
$$
The net downward force on the ceiling is $50 \mathrm{~N}+100 \mathrm{~N}$.
(b) Here, too, the load is supported by the tensions in two ropes, $F_{T}$ and $F$, where $F_{T}=F$. Then
$$
F_{T}+F=F_{W} \quad \text { or } \quad F=\frac{1}{2} F_{W}=50 \mathrm{~N}
$$
The net downward force on the ceiling is $50 \mathrm{~N}$. The net upward force on the load is $100 \mathrm{~N}$.
(c) Let $F_{T 1}$ and $F_{T 2}$ be tensions around pulleys- $A$ and $-B$, respectively. Pulley-A is in equilibrium, and
$$
F_{T 1}+F_{T 1}-F_{W}=0 \quad \text { or } \quad F_{T 1}=\frac{1}{2} F_{W}
$$
Pulley-B, too, is in equilibrium, and
$$
F_{T 2}+F_{T 2}-F_{T 1}=0 \quad \text { or } \quad F_{T 2}=\frac{1}{2} F_{T 1}=\frac{1}{4} F_{W}
$$
But $F=F_{T 2}$ and so $F=\frac{1}{4} F_{W}=25 \mathrm{~N}$.
The ceiling pulls up on the pulley system with a net force of 50 $\mathrm{N}+25 \mathrm{~N}+50 \mathrm{~N}$, or $125 \mathrm{~N}$. Meanwhile the total downward force on the system (and so on the ceiling) is $100 \mathrm{~N}+25 \mathrm{~N}$, or $125 \mathrm{~N}$.
(d) Four ropes, each with the same tension $F_{T}$, support the load $F_{W} .$ Therefore,
$$
4 F_{T 1}=F_{W} \quad \text { and } \quad F=F_{T 1}=\frac{1}{4} F_{W}=25 \mathrm{~N}
$$
Since there are 5 ropes pulling down on the ceiling bracket, the ceiling must pull up on the pulley system with a net force of 5 $\times 25 \mathrm{~N}$, or $125 \mathrm{~N}$. The net downward force on the pulley system (and so on the ceiling) is the load $100 \mathrm{~N}$ plus $F=25 \mathrm{~N}$.
(e) We see at once $F=F_{T 1}$. Because the pulley on the left is in
equilibrium,
$$
F_{T 2}-F_{T 1}-F=0
$$
But $F_{T 1}=F$ and so $F_{T 2}=2 F$. The pulley on the right is also in equilibrium, and therefore,
$$
F_{T 1}+F_{T 2}+F_{T 1}-F_{W}=0
$$
Recalling that $F_{T 1}=F$ and that $F_{T 2}=2 F$ gives $4 F=F_{W}$; hence, $F=25 \mathrm{~N}$
The uppermost pulley supports a downward force of $2 \times 50 \mathrm{~N}$, and so the net upward force exerted by the ceiling is $100 \mathrm{~N}+$ $25 \mathrm{~N}$. Again, the net downward force on the pulley system (and so on the ceiling) is the load $100 \mathrm{~N}$ plus $F=25 \mathrm{~N}$, or 125 $\mathrm{N}$.

Stephen Zaffke
Stephen Zaffke
Numerade Educator
03:24

Problem 6

Using the wheel and axle illustrated in We know that in one turn of the wheel-axle system, a length of cord equal to the circumference of the wheel or axle will be wound or unwound.
$$
\begin{aligned}
\mathrm{IMA} &=\frac{\text { Distance moved by } F}{\text { Distance moved by } F_{W}}=\frac{2 \pi R}{2 \pi r}=\frac{85 \mathrm{~cm}}{6.0 \mathrm{~cm}}=14.2=14 \\
\mathrm{AMA} &=\text { Force ratio }=\frac{400 \mathrm{~N}}{50 \mathrm{~N}}=8.0 \\
\text { Efficiency } &=\frac{\mathrm{AMA}}{\mathrm{IMA}}=\frac{8.0}{14.2}=0.56=56 \%
\end{aligned}
$$ a 400-N load can be raised by a force of $50 \mathrm{~N}$ applied to the rim of the wheel. The radii of the wheel $(R)$ and axle $(r)$ are $85 \mathrm{~cm}$ and $6.0 \mathrm{~cm}$, respectively. Determine the IMA, AMA, and efficiency of the machine.

Emily Anderson
Emily Anderson
Numerade Educator
04:12

Problem 7

The inclined plane depicted in long and rises $3.0$
m. ( $a$ ) What minimum force $F$ parallel to the plane is required to slide a 20 -kg box up the plane if friction is neglected? ( $b$ ) What is the IMA of the plane? ( $c$ ) Find the AMA and efficiency if a $64-\mathrm{N}$ force is actually required.
(a) There are several ways to approach this. Let's consider energy. Since there is no friction, the work done by the pushing force,
(F) $(15 \mathrm{~m})$, must equal the lifting work done, $(20 \mathrm{~kg})(9.81$
$\mathrm{m} / \mathrm{s}^{2}$ ) $(3.0 \mathrm{~m})$. Equating these two expressions and solving for $\mathrm{F}$ gives $F=39 \mathrm{~N}$
$(b)$
$\mathrm{IMA}=\frac{\text { Distance moved by } F}{\text { Distance } F_{w} \text { is lifted }}=\frac{15 \mathrm{~m}}{3.0 \mathrm{~m}}=5.0$
(c) $\mathrm{AMA}=$ Force ratio $=\frac{F_{W}}{F}=\frac{196 \mathrm{~N}}{64 \mathrm{~N}}=3.06=3.1$
Efficiency $=\frac{A M A}{I M A}=\frac{3.06}{5.0}=0.61=61 \%$
$\mathrm{O} r$, as a check.
Efficiency $=\frac{\text { Work output }}{\text { Work input }}=\frac{\left(F_{W}\right)(3.0 \mathrm{~m})}{(F)(15 \mathrm{~m})}=0.61=61 \%$

Emily Anderson
Emily Anderson
Numerade Educator
04:39

Problem 8

As seen in Fig. $7-5$, a jackscrew has a lever arm of $40 \mathrm{~cm}$ and a pitch of $5.0 \mathrm{~mm}$. If the efficiency is 30 percent, what horizontal force $F$ applied perpendicularly at the end of the lever arm is required to lift a load $F_{W}$ of 270 kg?
When the jack handle is moved around one complete circle, the input force moves a distance
$$
2 \pi r=2 \pi(0.40 \mathrm{~m})
$$
while the load is lifted a distance of $\mathrm{m}$. The IMA is therefore
$$
\mathrm{IMA}=\text { Distance ratio }=\frac{2 \pi(0.40 \mathrm{~m})}{0.0050 \mathrm{~m}}=0.50 \times 10^{3}
$$Since efficiency = AMA / IMA, we have
$$
\text { AMA }=(\text { Efficiency })(\mathrm{IMA})=(0.30)(502)=0.15 \times 10^{3}
$$
But AMA = (Load lifted)/(Input force) and therefore
$$
F=\frac{\text { Load lifted }}{\mathrm{AMA}}=\frac{(270 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)}{151}=18 \mathrm{~N}
$$

Emily Anderson
Emily Anderson
Numerade Educator
04:14

Problem 9

A differential pulley (chain hoist) is drawn in . Two toothed pulleys of radii $r=10 \mathrm{~cm}$ and $R=11 \mathrm{~cm}$ are fastened together and turn on the same axle. A continuous chain passes over the smaller $(10 \mathrm{~cm})$ pulley, then around the movable pulley at the bottom, and finally around the $11-\mathrm{cm}$ pulley. The operator exerts a downward force $F$ on the chain to lift the $\operatorname{load} F_{W} .(a)$ Determine the IMA. (b) What is the efficiency of the machine if an applied force of $50 \mathrm{~N}$ is required to lift a load of $700 \mathrm{~N}$ ?

Stephen Zaffke
Stephen Zaffke
Numerade Educator
03:43

Problem 10

A motor furnishes 120 hp to a device that lifts a 5000 -kg load to a height of $13.0 \mathrm{~m}$ in a time of $20 \mathrm{~s}$. Find the efficiency of the machine.

Emily Anderson
Emily Anderson
Numerade Educator
03:15

Problem 11

Refer back to If a force of $200 \mathrm{~N}$ is required to lift a 50-kg load, find the IMA, AMA, and efficiency for the system.

Emily Anderson
Emily Anderson
Numerade Educator
04:03

Problem 12

In the $300-\mathrm{N}$ load is balanced by a force $F$ in both systems. Assuming efficiencies of 100 percent, how large is $F$ in each system? Assume all ropes to be vertical.

Stephen Zaffke
Stephen Zaffke
Numerade Educator
05:23

Problem 13

Consider a machine for which an applied force moves $3.3 \mathrm{~m}$ to raise a load $8.0 \mathrm{~cm}$. Find the ( $a$ ) IMA and (b) AMA if the efficiency is 60 percent. What load can be lifted by an applied force of $50 \mathrm{~N}$ if the efficiency is ( $c$ ) 100 percent and $(d) 60$ percent?

Emily Anderson
Emily Anderson
Numerade Educator
02:47

Problem 14

With a wheel and axle, a force of $80 \mathrm{~N}$ applied to the rim of the wheel can lift a load of $640 \mathrm{~N}$. The diameters of the wheel and axle are $36 \mathrm{~cm}$ and $4.0 \mathrm{~cm}$, respectively. Determine the AMA, IMA, and efficiency of the machine.

Emily Anderson
Emily Anderson
Numerade Educator
02:56

Problem 15

A hydraulic jack in a gas station lifts a 900 -kg car a distance of $0.25$ $\mathrm{cm}$ when a force of $150 \mathrm{~N}$ pushes a piston through a distance of 20 $\mathrm{cm}$. Find the IMA, AMA, and efficiency.

Emily Anderson
Emily Anderson
Numerade Educator
04:20

Problem 16

The screw of a mechanical press has a pitch of $0.20 \mathrm{~cm}$. The diameter of the wheel to which a tangential turning force $F$ is applied is $55 \mathrm{~cm}$. If the efficiency is 40 percent, how large must $F$ be to produce a force of $12 \mathrm{kN}$ in the press?

Emily Anderson
Emily Anderson
Numerade Educator
04:37

Problem 17

The diameters of the two upper pulleys of a chain hoist are $18 \mathrm{~cm}$ and $16 \mathrm{~cm}$. If the efficiency of the hoist is 45 percent,

Stephen Zaffke
Stephen Zaffke
Numerade Educator