Physics, 6th Edition
Chapter 12. Simple Machines
Chapter 12. Simple Machines
Simple Machines, Efficiency, and Mechanical Advantage
12-1. A 25 percent efficient machine performs external work of 200 J. What input work is
required?
E
Work out
;
Work in
Work in 
Work out 200 J

;
E
0.25
Work in = 800 J
12-2. What is the input work of a 30 percent efficient gasoline engine if during each cycle it
performs 400 J of useful work?
E
Work out
;
Work in
Work in 
Work out 400 J

;
E
0.30
Work in = 1333 J
12-3. A 60-W motor lifts a 2-kg mass to a height of 4 m in 3 s. Compute the output power.
P
Work Fs

;
t
t
P
(2 kg)(9.8 m/s 2 )(4 m)
;
3s
P = 26.1 W
12.4. What is the efficiency of the motor in Problem 12-3? What is the rate at which work is
done against friction?
Pin  60 W; E 
Power out 26.1 W

;
Power in
60 W
Ploss = 60 W – 26.1 W;
E = 43.5%
Ploss = 33.9 W
12-5. A 60 percent efficient machine lifts a 10-kg mass at a constant speed of 3 m/s. What is the
required input power?
Pout = Fv = mgv = (10 kg)(9.8 m/s2)(3 m/s);
Pin 
Pout 294 W

;
E
0.60
160
Pout = 294 W
Pin = 490 W
Physics, 6th Edition
Chapter 12. Simple Machines
12-6. During the operation of a 300-hp engine, energy is lost to friction at the rate of 200 hp.
What is the useful output power and what is the efficiency of the engine?
Pout = 300 hp – 200 hp = 100 hp ;
E
Pout 100 hp

;
Pin 300 hp
E = 33.3%
12-7. A frictionless machine lifts a 200-lb load through a vertical distance of 10 ft. The input
force moves through a distance of 300 ft. What is the ideal mechanical advantage of the
machine? What is the magnitude of the input force?
MI 
si 300 ft

= 30 ;
so
10 ft
Fi 
Fo 200 lb

=
30
30
6.67 lb
Applications of the Lever Principle
12-8. One edge of a 50-kg safe is lifted with a 1.2-m steel rod. What input force is required at
the end of the rod if a fulcrum is placed 12 cm from the safe? (Hint: To lift one edge, a
1.2 m
force equal to one-half the weight of the safe is required.)
W = mg = (50 kg)(9.8 m/s2) = 490 N;
MI 
ri 1.08 m

 9;
ro 0.12 m
F = W/2 = 245 N
Fi 
Fo 490 N

MI
9
Fi = 54.4 N
12-9. For the nutcracker in Fig. 12-4a, the nut is located 2 cm from the fulcrum and an input
force of 20-N is applied at the handles which are 10 cm from the fulcrum. What is the
force applied to crack the nut?
Fulcrum
r 10 cm
MI  i 
 5;
ro
2 cm
Fo  M I Fi  5(20 N) ;
Fo = 100 N
161
Fo
Fi
Physics, 6th Edition
Chapter 12. Simple Machines
12-10. For the wheelbarrow in Fig. 12-4b, the center of gravity of a net load of 40 kg is located
50 cm from the wheel. What upward lift must be applied at a point on the handles that is
1.4 m from the wheel?
Fi
r 1.4 m
MI  i 
 2.8;
ro 0.5 m
Fi 
(40 kg)(9.8 m/s2 )
;
2.8
Fo
F
Fi  o
MI
Fulcrum
Fi = 140 N
12-11. What is the ideal mechanical advantage of the wheelbarrow in Problem 12-11?
MI 
ri 1.4 m

 2.8 ;
ro 0.5 m
Mi = 2.8
12-12. Find the ideal mechanical advantage of the crowbar in Fig. 12-4c if the input force is
applied 30 cm from the nail and the fulcrum is located 2 cm
from the nail?
MI 
Fo
Fi
ri 30 cm - 2 cm

 14 ;
ro
2 cm
Mi = 14.0
Fulcrum
12-13. The input force exerted by a muscle in the forearm (see Fig. 12-4d) is 120 N and acts a
distance of 4 cm from the elbow. The total length of the forearm is 25 cm. What weight
Fi
is being lifted?
MI 
ri Fo
 ;
ro Fi
Fo 
ri Fi (4 cm)(120 N)

;
ro
25 cm
Fo =19.2 N
162
Fulcrum
Fo
Physics, 6th Edition
Chapter 12. Simple Machines
12-14. A wheel 20 cm in diameter is attached to an axle with a diameter of 6 cm. If a weight of
400 N is attached to the axle, what force must be applied to the rim of the wheel to lift the
Fo
weight at constant speed? Neglect friction.
F
R
MI  o  ;
Fi r
F r (400 N)(3 cm)
Fi  o 
R
10 cm
R
Fi
r
Fi = 120 N
W
12-15. A 20-kg mass is to be lifted with a rod 2 m long. If you can exert a downward force of
40 N on one end of the rod, where should you place a block of wood to act as a fulcrum?
ro
Fo (20 kg)(9.8 m/s 2 )
MI  MA 

; M I  4.9
Fi
(40 N)
MI 
ri
 4.9; ri  4.9ro ;
ro
ro + ri = 2 m;
Solving for r0 gives: ro = 0.339 m
We could also find ri:
ri
mg
40 N
Substitute: ro + (4.9ro) = 2 m
The fulcrum should be 33.9 cm from the weight.
ri = 4.9(0.339 m)
or
ri = 1.66 m
12-16. Determine the force F required to lift a 200-N load W with the pulley shown in Fig. 16a.
M I  2;
MI  MA 
Fo
;
Fi
Fi 
Fo
200 N

;
MA
2
Fi = 100 N
12-17. What input force is needed to lift the 200-N load with the arrangement in Fig. 12-16b?
M I  4;
MI  MA 
Fo
;
Fi
Fi 
Fo
200 N

;
MA
4
Fi = 50 N
12-18. What input forces are needed to lift the 200-N load for Fig. 12-16c and d?
M I  5;
MI  MA 
Fo
;
Fi
Fi 
Fo
200 N

;
MA
5
Fi = 40 N
M I  4;
MI  MA 
Fo
;
Fi
Fi 
Fo
200 N

;
MA
4
Fi = 50 N
163
Physics, 6th Edition
Chapter 12. Simple Machines
12-19. What is the mechanical advantage of a screwdriver used as a wheel and axle if its blade is
0.3 in. wide and the handle is 0.8 in?
0.3 in.
0.8 in.
MI 
0.8 in.
;
0.3 in.
MI = 2.67
*12-20. The chain hoist in Fig. 12-17 is a combination of the wheel and axle and the block and
tackle. Show that the ideal mechanical advantage of such a device is given by
Work out = Work in;
Fisi = Foso
Fi
si = 2R
so 
2 ( R  r )
2
MI 
 2 ( R  r ) 
Fi (2 R)  Fo 

2

Fo
2R

;
Fi R  r
MI 
Fi
2R
Rr
*12-21. Assume that the larger radius in Fig. 12-17 is three times the smaller radius. What input
force is required to lift a 10-kg load with no friction.
Fo = mg = (10 kg)(9.8 m/s2) = 98.0 N;
MI 
2R
2(3r )

 3;
R  r 3r  r
MI 
R = 3r
F0
98 N
; Fi 
;
Fi
3
Fi = 32.7 N
The Transmission of Torque
12-22. A 1500 rpm motor has a drive pulley 3 in. in diameter, and the driven pulley is 9 in. in
diameter. Find the ideal mechanical advantage and the rpm for the output pulley.
MI 
o 
Do 9 in.

 3;
Di 3 in.
i
3

(1500 rpm)
3
164
MI 
Do  i

 3;
Di  o
o = 500 rpm
Physics, 6th Edition
Chapter 12. Simple Machines
12-23. A 30 cm diameter input pulley turns at 200 rpm on a belt drive connected to an output
pulley 60 cm in diameter. What is the ratio of the output torque to the input torque? What
is the output rpm?
 o Do 60 cm


;
 i Di 30 cm
MI 
Do i

 2;
Di  o
o 
i
2

o
2
i
200 rpm
;
2
o = 100 rpm
12-24. A V-belt pulley system has output and input drives of diameters 6 in. and 4 in.,
respectively. A torque of 200 lb in. is applied to the input drive. Find the output torque?
 o Do 6 in.


;
 i Di 4 in.
o
 1.5 ; o = 1.5(200 lb in.);
i
o = 300 lb ft
12-25. The ratio of output speed to input speed for a gear drive is 2 to 1. What is the mechanical
advantage?
MI 
Do i

;
Di  o
o
1
2

 ;
i M i 1
MI = ½
12-26. A set of two spur gears has 40 teeth and 10 teeth. What are the possible ideal mechanical
advantages.
MI 
Do N o

;
Di N i
Possible ratios :
40
10
and
10
40
MI = 4.0 and 0.250
12-27. For the spur gears in Problem 12-27, what is the rotational speed of the smaller gear, if
the speed of the larger gear is 200 rpm?
DL N L  s


;
Ds N s  L
s 
N L L 40(200 rpm)

;
Ns
10
165
s = 800 rpm
Physics, 6th Edition
Chapter 12. Simple Machines
Applications of the Inclined Plane
11-28. What must be the thickness of the base, if the wedge is 20 cm long and it is desired that
L
the input force be one-tenth of the output force?
MI 
L
 10;
t
t
L 20 cm

;
10
10
t
t = 2.00 cm
11-29. What should be the apex angle of a wedge if it is to have a mechanical advantage of 10?
tan  
t
;
L
MI 
L
 10;
t
tan  
1
;
10
 = 5.710
11-30. A 10-kg crate is moved from the ground to a loading platform by means of a ramp 6 m
long and 2 m high. Assume that k =0.25. What are the ideal and actual mechanical
advantages of the ramp?
MI 
N
[ sin  = 2/6 ;  = 19.50 ]
L 6m

;
t 2m
MI = 3.00
6m
Fi
F
2m


Fo = mg = (10 kg)(9.8 m/s2); Fo = 98.0 N
Fo = mg
Fk = kN = kmg cos  ; Fk = (0.25)(98 N)cos 19.5 = 23.1 N
0
Fi – Fk – mg sin  = 0; Fi – 23.1 N – (98 N)sin 19.50 = 0;
MA 
Fo
98.0 N

;
Fi 55.77 N
Fi = 55.77 N
MA = 1.76
11-31. For the ramp in Problem 11-31, what is the efficiency of the ramp?
E
M A 1.76

;
Mi
3
E = 0.586;
166
E = 58.6%
mg cos 
Physics, 6th Edition
Chapter 12. Simple Machines
*11-32. An input force of 20 lb is applied to the 6-in. handle of a wrench. used to tighten a ¼ in.diameter bolt? An actual output force of 600 lb is produced. If the bolt has 10 threads per
inch, what is the ideal mechanical advantage., and what is the efficiency?
P = 1/10 in/thread = 0.167 in.;
MA 
MI 
Fo 600 lb

 30 ;
Fi
20 lb
2 R 2 (6 in.)

 377;
P
0.1 in.
E
MA
30

;
M I 377
MI = 377
E = 8.00%
11-33. The lever of a screw jack is 24 in. long. If the screw has six threads per inch. What is the
ideal mechanical advantage?
P = 1/6 in/thread = 0.167 in.;
MI 
2 R 2 (24 in.)

 904;
P
0.167 in.
MI = 904
11-34. If the screw jack in Problem 11-34 is 15 percent efficient, what force is needed to lift
2000-lb?
MA =E MA = (0.15)(904) = 136;
MA 
Fo
 136;
Fi
Fi 
2000 lb
;
136
Fi = 14.7 lb
Challenge Problems
*11-35. An inclined plane is 6 m long and 1 m high. The coefficient of kinetic friction is 0.2.
What force is required to pull a weight of 2400 N up the incline at constant speed? What
N
is the efficiency of the inclined plane? [ sin  = 1/6 ;  = 9.590 ]
L 6m
MI  
;
t 1m
MI = 6.00
6m
Fo = 2400 N
Fo
98.0 N

;
Fi 55.77 N
1m

MA = 1.76
167
E
M A 1.76

;
MI
6
W cos 
Fo = 2400 N
Fi – Fk – mg sin  = 0; Fi – 473 N – (2400 N)sin 9.590 = 0;
MA 
F

Fk = kN = kW cos  ; Fk = (0.2)(2400 N)cos 9.590 = 473 N
Fi
Fi = 873 N
E = 29.3%
Physics, 6th Edition
Chapter 12. Simple Machines
*11-36. A wheel and axle is used to raise a mass of 700-kg. The radius of the wheel is 0.50 m,
and the radius of the axle is 0.04 m. If the actual efficiency is 60%, what input force
must be applied to the wheel?
Fo = W = mg = (700 kg)(9.8 m/s2); Fo = 6860 N;
MI 
R 0.5
M

 12.5; E  A  0.60 ;
r 0.04
MI
MA 
Fo
 7.5;
Fi
Fi 
EA = 0.60; R = 0.50 m
MA = 0.60(12.5) = 7.5
6860 N
;
7.5
Fi = 915 N
*11-37. A shaft rotating at 800 rpm delivers a torque of 240 N m to n output shaft that is rotating
at 200 rpm. If the efficiency of the machine is 70%, compute the output torque. What is
the output power?
E
[ o = 200 rpm = 20.94 rad/s ]
 o o
 0.70;
 ii
o 
0.70(240 N m)(800 rpm)
;
200 rpm
Po = oo = (672 N m)(20.94 rad/s);
 o = 672 N m
Po = 14.1 kW
*11-38. A screw jack has a screw whose pitch is 0.25 in. Its handle is 16 in. long, and a load of
1.9 tons is being lifted. Neglecting friction, what force is required at the end of the
handle? What is the mechanical advantage?
MI 
2 R 2 (16 in.)

;
P
0.25 in.
Neglecting friction: M A  M I 
Fo
;
Fi
Fi 
MI = 402
1.9 ton(2000 lb/ton)
; Fi = 9.45 lb
402
In absence of friction, MA = MI = 402;
168
MA = 402
Physics, 6th Edition
Chapter 12. Simple Machines
*11-39. A certain refrigeration compressor comes equipped with a 250-mm-diameter pulley and
is designed to operate at 600 rpm. What should be the diameter of the motor pulley if the
motor speed is 2000 rpm?
MI 
Do i

;
Di  o
Di 
 o Do (600 rpm)(250 mm)

; Di = 75.0 mm
i
2000 rpm
*11-40. In a fan belt, the driving wheel is 20 cm in diameter and the driven wheel is 50 cm in
diameter. The power input comes from a 4-kW motor that causes the driving wheel to
rotate at 300 rpm. If the efficiency is 80 percent, calculate the rpm and the torque
delivered to the driven wheel.
MI 
[ 300 rpm = 31.4 rad/s ]
Do 50 cm

 2.5 ;
Di 20 cm
Pi   i i ;
i 
Pi
i

MA = EMI = 0.8(2.5) = 2
4000 W
 127.3 N m
31.4 rad/s
o = MAi = 2(127.3 N m) ;
o 
o = 255 N m
Di i (20 cm)(300 rpm)

;
Do
50 cm
o = 120 rpm
11-41. A log-splitting wedge has a side length of 16 cm and the angle at the apex is 100. What is
the ideal mechanical advantage?
t = L sin  = (16 cm) sin 100 = 2.78 cm;
MI 
L 16 cm

;
t 2.78 cm
MI = 5.76
11-42. A machine has an efficiency of 72 percent. An input force of 500 N moves through
parallel distance of 40 cm. How much energy is lost in the process.
Work in = (500 N)(0.40 m) = 200 J; Work out = 0.72(200 J) = 144 J
Energy Lost = Work in – Work out = 200 J – 144 J; Loss = 56.0 J
169
Physics, 6th Edition
Chapter 12. Simple Machines
*11-43. A motor with an efficiency of 80 percent operates a winch with an efficiency of 50
percent. If the power supplied to the motor is 6 kW, how far will the winch lift a 400-kg
mass in a time of 4 s?
ET = (0.80)(0.50) = 0.40;
Work in = (6000 W)(4 s) = 24,000 J
Work out = 0.40(24,000 J) = 9600 J;
so 
Work out
9600 J

;
Fo
(400 kg)(9.8 m/s 2 )
Work out = Foso
so = 2.45 m
Critical Thinking Questions
12-44. A 60-N weight is lifted in the three different ways shown in Fig. 12-18. Compute the
ideal mechanical advantage and the required input force for each application.
MI 
Fi 
Fo
MI
MI 
Fi 
Fo
MI
MI 
Fi 
Fo
MI
ri 80 cm

ro 40 cm
60 N
MI = 2.00
Fi = 30 N
ri 120 cm

ro 40 cm
MI = 3.00
60 N
;
3.0
Fi = 20 N
Fi 
ri
40 cm

ro 120 cm
Fi 
60 N
;
0.333
40 cm
Fi
60 N
;
2.0
Fi 
80 cm
60 N
Fi
80 cm
MI = 0.333
60 N
40 cm
Fi
80 cm
40 cm
Fi = 180 N
The above are examples of the three classes of levers. Note the advantages
and disadvantages of each type of lever in terms of the mechanical
advantage and the required input forces.
170
Physics, 6th Edition
Chapter 12. Simple Machines
*12-45. A worm drive similar to that shown in Fig. 12-11 has n teeth in the gear wheel. (If n =
80, one complete turn of the worm will advance the wheel one-eightieth of a revolution.)
Derive an expression for the ideal mechanical advantage of the worm gear in terms of the
radius of the input pulley R, the radius of the output shaft r, and the number of teeth n in
the gear wheel.
Work in  Work out ;
Fi si  Fo so ;
 2 r 
Fi (2 R)  Fo 
;
 n 
MI 
si  2 R;
Fo nR

;
Fi
r
so 
MI 
2 r
n
nR
r
*12-46. The worm drive of previous problem has 80 teeth in the gear wheel. If the radius of the
input wheel is 30 cm and the radius of the output shaft is 5 cm, what input force is
required to lift a 1200-kg load? Assume an efficiency of 80%.
MI 
nR (80)(30 cm)

 480;
r
5 cm
Fi 
M A  EM I  (0.8)(480)  384
(1200 kg)(9.8 m/s2 )
;
384
Fi = 30.6 N
*12-47. The oarlock on a 3.5 m oar is 1 m from the handle end. A person in a rowboat applies a
force of 50 N to the handle end. What is the ideal mechanical advantage and output force?
Does moving the oarlock closer to the handle end increase or decrease the mechanical
advantage? How far from the handle end should the oarlock be placed to produce a 20%
1m
Fo
2.5 m
increase in the output force?
Fi
ri
1m
MI  
MI = 0.400
ro 2.5 m
Moving oarlock closer to hands produces a smaller mechanical advantage: Smaller
171
Physics, 6th Edition
Chapter 12. Simple Machines
12-47. (Cont.)
A 20% increase in Fo means the new mechanical advantage is 1.2MA
M’A = 1.2(0.40) = 0.48;
ri
 0.48;
ro
ri  0.48ro  0.48(3.5 m - ri );
ri = 1.14 m
ri = 1.68 – 0.48ri
The oarlock should be placed 1.14 m from the hand.
*12-48. Sketch a block and tackle system with a mechanical advantage of 5.
The system drawn to the right has five lifting ropes pulling the
weight upward, and thus has a mechanical advantage of 5.
*12-49. A 60-W motor drives the input pulley of a belt drive at 150 rpm. The
diameters of the input and output pulleys are 60 cm and 20 cm, respectively. Assume an actual mechanical advantage of 0.25. (a) What is the
output torque? (b) What is the output power? (c) What is the efficiency?
Mi 
Do 20 cm

 0.333
Di 60 cm
E
o 
Di i (60 cm)(150 rpm)

Do
20 cm
MA
0.25

;
M I 0.333
E = 0.750 or 75.0%
Po = 0.75Pi = 0.75(60 W) = 45 W;
oo = 45 W;
o = 450 rpm
o = 450 rpm = 47.1 rad/s;
o 
Po = 45 W
45 W
;
47.1 rad/s
o = 0.955 N m
*12-50. A pair of step pulleys (Fig. 12-19) makes it possible to change output speeds merely by
shifting the belt. If an electric motor turns the input pulley at 2000 rpm, find the possible
angular speeds of the output shaft. The pulley diameters are 4, 6, and 8 cm.
172
Physics, 6th Edition
Chapter 12. Simple Machines
*12-50. (Cont.) First consider the 4-in. pulley, which can drive 4-in., 6 in.., and 8-in. output
pulley. Case 1 has di = 4 in. and do = 4 in.
(a) odo = idi
 di 
 i
 do 
o  
 4 in. 
i = 2000 rpm
o  
 (2000 rpm);
 4 in. 
o = 2000 rpm
 4 in. 
 (2000 rpm);
 6 in. 
o = 1333 rpm
o  
 4 in. 
o  
 (2000 rpm);
 8 in. 
o = 1000 rpm
(a) Next we consider the 6-in. pulley as the input pulley
 6 in. 
o  
 (2000 rpm);
 4 in. 
 6 in. 
o  
 (2000 rpm);
 6 in. 
 6 in. 
o  
 (2000 rpm);
 8 in. 
o = 3000 rpm
o = 2000 rpm
o = 1500 rpm
(c) Next we consider the 8-in. pulley as the input pulley
 8 in. 
o  
 (2000 rpm);
 4 in. 
 8 in. 
o = 4000 rpm
o  
 (2000 rpm);
 6 in. 
o = 2667 rpm
 8 in. 
 (2000 rpm);
 8 in. 
o = 2000 rpm
o  
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