Physics in biology and medicine,
6th Edition by Davidovits, Chapter 1-18
SOLUTION MANUAL
,Table of contents
1. Static Forces
2. Friction
3. Translational Motion
4. Angular Motion
5. Elasticity and Strength of Materials
6. Insect Flight
7. Fluids
8. The Motion of Fluids
9. Heat and Kinetic Theory
10. Thermodynamics
11. Heat and Life
12. Waves and Sound
13. Electricity
14. Electrical Technology
15. Optics
16. Atomic Physics
17. Nuclear Physics
18. Nanotechnology in Biology and Medicine
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B978-0-12-813716-1.09983-3, 09983
c49915 Instructors Solution Manual
s0010 CHAPTER 1
o0010 1-1. (b). Toppling torque Ta 5 Fa × 1.2
0:9
Tw ¼ Restoring torque ¼ W ¼ 686 0:45
2
p0020 On the verge of toppling Ta ¼ Tw
686 0:45
;Fa ¼ ¼ 254 N
1:2
¼ 57:8 lb
p0025 Note that this force is about 6 times greater than required to topple
the person with feet together.
o0015 1-2. Referring to Fig. 1.10, and balancing torques around the fulcrum
W d1 ¼ F d2 or F ¼ W dd12
W d2
; ¼
F d1
p0035 Referring to Fig. E. 1.2
f0010
d1
L1
q
q
L2
d2
p0040 The magnitudes of the two angles of the lever arm with respect to the
horizontal are equal therefore,
L1 ¼ d1 sin θ L2 ¼ d2 sin θ
L1 d1
and ¼
L2 d2
e1
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e2 Instructors Solution Manual
o0020 1-3. Referring to Fig. E. 1.3, the sum of the two angles ω + 100o ¼ 180o
∴ω ¼ 80o
x0 ¼ 30 cos 80° ¼ 5:2 cm
y0 ¼ 30sin 80° ¼ 29:4 cm
1 y0
θ ¼ tan
x0 +4
29:4
θ ¼ tan 1
¼ 72:6°
5:2 + 4
f0015
c m
30
100°
y¢
w
q
x¢ 4 cm
o0025 1-4. Assuming that the diameter of the bicept is 8 cm (as in the text),
2
the muscle area is πd4 ¼ 50:3 cm2
Fm ¼ 50:3 cm2 7 106 dyn=cm2
¼ 3:52 108 dyn
¼ 3:52 103 N
p0055 From Eq. 1-13
Fm
W¼ ¼ 335 N ¼ 75 lb
10:5
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Instructors Solution Manual e3
o0030 1-5. Following Exercise 1-3 and referring to Fig. E. 1.5
f0020
y
160°
a a
w Fm
b q
γ x
α
w
ω + 160° ¼ 180° ;ω ¼ 20°
a ¼ 30 sin 20 ¼ 10:3cm
b ¼ 30 cos 20 ¼ 28:2cm
a 10:3
θ ¼ tan 1 ¼ tan 1
b+4 28:2 + 4
°
θ ¼ 17:7
p0065 The upper arm is at the same angle as in Fig. 1-12. Using results from
Exercises 1.3
1
1x 1 5:2
α ¼ tan ¼ tan ¼ 10°
y1 29:4
γ ¼ α + ω ¼ 10 + 20 ¼ 30°
δ ¼ 90 γ ¼ 60°
p0070 Following Eq. (1-10)
p0075 x component: Fm cos(θ + δ) ¼ Fr cos ϕ
p0080 y component: Fm sin(θ + δ) ¼ Fr sin ϕ + W
p0085 Torque is: 4 cm Fm sin (θ) ¼ 40 cm W sin γ
p0090 From these we obtain 3 equations
o0035 1. Fm cos 77.7 ¼ Fr cos ϕ
o0040 2. Fm sin 77.7 ¼ Fr sin ϕ + 137 N
o0045 3. Fm sin 17.7 ¼ 10 137 sin 30o
p0110 From 3. Fm ¼ 2,253 N (508 lb)
p0115 From 2 & 3 ϕ ¼ 78.4o
p0120 Fr ¼ 2,386 N ¼ 536 lb
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e4 Instructors Solution Manual
o0050 1-6. As in Fig. 1-12 (or 1-13) θ ¼ 72.6° and following Eq. 1-12
4 cm Fm sin θ ¼ 20cm W
W ¼ 14 9:8 ¼ 137 N
Fm ¼ 720 N ¼ 162 lb
p0130 Following Eqs. 1-10 and 1-11
Fm cos θ ¼ Fr cos ϕ
Fm sin θ ¼ 137 N + Fr sin ϕ
Fr cos ϕ ¼ 215N
Fr sin ϕ ¼ 550N
F2r ¼ 3:49 105 N2
Fr ¼ 590 N
550
tan ϕ ¼ ¼ 2:56;ϕ ¼ 68:6°
215
o0055 1-7. (a) As in Eq. 1-12
4 cm Fm sin θ ¼ ð20 cm + 40 cmÞW
W ¼ 14 9:8 ¼ 137 N
Fm ¼ 2,160 N
p0140 As in Eq. 1-15
Fr cos ϕ ¼ 646 N
Fr sin ϕ ¼ 2,060 2 137 ¼ 1, 790 N
F2r ¼ 3:61 106 N2
Fr ¼ 1, 900 N
1, 790
tan ϕ ¼ ¼ 2:77;ϕ ¼ 70:2°
646
li7890 (b) yes
o0065 1-8. Weight of arm is 2 kg or 17 of the 14 lb weight hanging from the arm in
Problem 1-6.
p0155 Referring to Problem 1-6
p0160 Added force Fm ¼ 7207 ¼ 103 N
p0165 Added force Fr ¼ 590
7 ¼ 84 N
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Instructors Solution Manual e5
o0070 1-10. Referring to Fig. E. 1.10 θ ¼ 72.6° (from Exercise 1-3)
f7800
40 cm
Bicept
w
4 cm
b 2 cm
w q
a
b ¼ 2 sin 72:6° ¼ 1:91 cm
a ¼ 2 cos 72:6° ¼ 0:60 cm
p0180 The angle ω is:
1 b 1 1:91
ω ¼ tan ¼ tan ¼ 29:3°
4 a 3:4
p0185 Therefore the upward displacement of the weight due to 2 cm contrac-
tion of muscle is 40 sin 29.3° ¼ 19.6 cm.
Speed of muscle contraction ¼ 4 cm=s
19:6 cm
Speed of weight displacement ¼
0:5 s
¼ 38 cm=s:
p0190 Ratio of speeds is approximately inverse of mechanical advantage.
o0075 1-11. Using data given in the text, torque regarding hip, torque about the
insertion point is:
Fm 7cm ¼ W 3 + ð7 5:56Þ0:185 W
;Fm ¼ 0:47 W
p0200 As seen from Fig. 1-15, there are no forces in the x direction. The y
components of forces set to 0 leads to:
Fm + W ¼ Fr + 0:185 W
1:47 W ¼ Fr + 0:185 W
Fr ¼ 1:28 W
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o0080 1-12. (a) Torque about point A ¼ 0
‘ 2
‘ 160N + 320 cos 30° ¼ Fm ‘ sin 12°
2 3
p0210 Force exerted by muscle is Fm ¼ 2,000 N
p0215 From the geometry of figure angle of muscle with respect to y axis is 72°.
Angle of reaction force Fr at fifth lumbar with respect to y-axis is ϕ
x component of force ¼ 0
Fm sin 72° ¼ Fr sin ϕ
y comp of force ¼ 0
Fr cos ϕ ¼ 160 + 320 + Fm cos 72°
Fr sin ϕ ¼ 1, 902 N
Fr cos ϕ ¼ 1,098 N
;Fr 2 ¼ 4:82 106 N2 and Fr ¼ 2,200 N
o0085 (b) The added 20 kg mass is a force F ¼ 196 N. This force is added to
weight of arm and head. Torque conditions:
‘ 2
‘ 356 + 320 cos 30° ¼ Fm ‘ sin 12°
2 3
Fm ¼ 3, 220 N
p0225 Following 1-12 (a)
Fr sin ϕ ¼ 3,062
Fr cos ϕ ¼ 356 + 320 + Fm cos 72° ¼ 1,671 N
Fr 2 ¼ 12:17 106 N2 ;Fr ¼ 3,490 N
o0090 1-13. Let force of Achilles tendon ¼ FA
p0235 Let force on tibia ¼ FT
p0240 Let angle of FT with respect to vertical be ϕ ¼ 15°.
p0250 From Fig. 1-17 get
f5780
15°
FA
FT 15
y f
x
w
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6th Edition by Davidovits, Chapter 1-18
SOLUTION MANUAL
,Table of contents
1. Static Forces
2. Friction
3. Translational Motion
4. Angular Motion
5. Elasticity and Strength of Materials
6. Insect Flight
7. Fluids
8. The Motion of Fluids
9. Heat and Kinetic Theory
10. Thermodynamics
11. Heat and Life
12. Waves and Sound
13. Electricity
14. Electrical Technology
15. Optics
16. Atomic Physics
17. Nuclear Physics
18. Nanotechnology in Biology and Medicine
,These proofs may contain colour figures. Those figures may print black and white in the final printed book if a colour print product
has not been planned. The colour figures will appear in colour in all electronic versions of this book.
B978-0-12-813716-1.09983-3, 09983
c49915 Instructors Solution Manual
s0010 CHAPTER 1
o0010 1-1. (b). Toppling torque Ta 5 Fa × 1.2
0:9
Tw ¼ Restoring torque ¼ W ¼ 686 0:45
2
p0020 On the verge of toppling Ta ¼ Tw
686 0:45
;Fa ¼ ¼ 254 N
1:2
¼ 57:8 lb
p0025 Note that this force is about 6 times greater than required to topple
the person with feet together.
o0015 1-2. Referring to Fig. 1.10, and balancing torques around the fulcrum
W d1 ¼ F d2 or F ¼ W dd12
W d2
; ¼
F d1
p0035 Referring to Fig. E. 1.2
f0010
d1
L1
q
q
L2
d2
p0040 The magnitudes of the two angles of the lever arm with respect to the
horizontal are equal therefore,
L1 ¼ d1 sin θ L2 ¼ d2 sin θ
L1 d1
and ¼
L2 d2
e1
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e2 Instructors Solution Manual
o0020 1-3. Referring to Fig. E. 1.3, the sum of the two angles ω + 100o ¼ 180o
∴ω ¼ 80o
x0 ¼ 30 cos 80° ¼ 5:2 cm
y0 ¼ 30sin 80° ¼ 29:4 cm
1 y0
θ ¼ tan
x0 +4
29:4
θ ¼ tan 1
¼ 72:6°
5:2 + 4
f0015
c m
30
100°
y¢
w
q
x¢ 4 cm
o0025 1-4. Assuming that the diameter of the bicept is 8 cm (as in the text),
2
the muscle area is πd4 ¼ 50:3 cm2
Fm ¼ 50:3 cm2 7 106 dyn=cm2
¼ 3:52 108 dyn
¼ 3:52 103 N
p0055 From Eq. 1-13
Fm
W¼ ¼ 335 N ¼ 75 lb
10:5
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Instructors Solution Manual e3
o0030 1-5. Following Exercise 1-3 and referring to Fig. E. 1.5
f0020
y
160°
a a
w Fm
b q
γ x
α
w
ω + 160° ¼ 180° ;ω ¼ 20°
a ¼ 30 sin 20 ¼ 10:3cm
b ¼ 30 cos 20 ¼ 28:2cm
a 10:3
θ ¼ tan 1 ¼ tan 1
b+4 28:2 + 4
°
θ ¼ 17:7
p0065 The upper arm is at the same angle as in Fig. 1-12. Using results from
Exercises 1.3
1
1x 1 5:2
α ¼ tan ¼ tan ¼ 10°
y1 29:4
γ ¼ α + ω ¼ 10 + 20 ¼ 30°
δ ¼ 90 γ ¼ 60°
p0070 Following Eq. (1-10)
p0075 x component: Fm cos(θ + δ) ¼ Fr cos ϕ
p0080 y component: Fm sin(θ + δ) ¼ Fr sin ϕ + W
p0085 Torque is: 4 cm Fm sin (θ) ¼ 40 cm W sin γ
p0090 From these we obtain 3 equations
o0035 1. Fm cos 77.7 ¼ Fr cos ϕ
o0040 2. Fm sin 77.7 ¼ Fr sin ϕ + 137 N
o0045 3. Fm sin 17.7 ¼ 10 137 sin 30o
p0110 From 3. Fm ¼ 2,253 N (508 lb)
p0115 From 2 & 3 ϕ ¼ 78.4o
p0120 Fr ¼ 2,386 N ¼ 536 lb
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e4 Instructors Solution Manual
o0050 1-6. As in Fig. 1-12 (or 1-13) θ ¼ 72.6° and following Eq. 1-12
4 cm Fm sin θ ¼ 20cm W
W ¼ 14 9:8 ¼ 137 N
Fm ¼ 720 N ¼ 162 lb
p0130 Following Eqs. 1-10 and 1-11
Fm cos θ ¼ Fr cos ϕ
Fm sin θ ¼ 137 N + Fr sin ϕ
Fr cos ϕ ¼ 215N
Fr sin ϕ ¼ 550N
F2r ¼ 3:49 105 N2
Fr ¼ 590 N
550
tan ϕ ¼ ¼ 2:56;ϕ ¼ 68:6°
215
o0055 1-7. (a) As in Eq. 1-12
4 cm Fm sin θ ¼ ð20 cm + 40 cmÞW
W ¼ 14 9:8 ¼ 137 N
Fm ¼ 2,160 N
p0140 As in Eq. 1-15
Fr cos ϕ ¼ 646 N
Fr sin ϕ ¼ 2,060 2 137 ¼ 1, 790 N
F2r ¼ 3:61 106 N2
Fr ¼ 1, 900 N
1, 790
tan ϕ ¼ ¼ 2:77;ϕ ¼ 70:2°
646
li7890 (b) yes
o0065 1-8. Weight of arm is 2 kg or 17 of the 14 lb weight hanging from the arm in
Problem 1-6.
p0155 Referring to Problem 1-6
p0160 Added force Fm ¼ 7207 ¼ 103 N
p0165 Added force Fr ¼ 590
7 ¼ 84 N
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Instructors Solution Manual e5
o0070 1-10. Referring to Fig. E. 1.10 θ ¼ 72.6° (from Exercise 1-3)
f7800
40 cm
Bicept
w
4 cm
b 2 cm
w q
a
b ¼ 2 sin 72:6° ¼ 1:91 cm
a ¼ 2 cos 72:6° ¼ 0:60 cm
p0180 The angle ω is:
1 b 1 1:91
ω ¼ tan ¼ tan ¼ 29:3°
4 a 3:4
p0185 Therefore the upward displacement of the weight due to 2 cm contrac-
tion of muscle is 40 sin 29.3° ¼ 19.6 cm.
Speed of muscle contraction ¼ 4 cm=s
19:6 cm
Speed of weight displacement ¼
0:5 s
¼ 38 cm=s:
p0190 Ratio of speeds is approximately inverse of mechanical advantage.
o0075 1-11. Using data given in the text, torque regarding hip, torque about the
insertion point is:
Fm 7cm ¼ W 3 + ð7 5:56Þ0:185 W
;Fm ¼ 0:47 W
p0200 As seen from Fig. 1-15, there are no forces in the x direction. The y
components of forces set to 0 leads to:
Fm + W ¼ Fr + 0:185 W
1:47 W ¼ Fr + 0:185 W
Fr ¼ 1:28 W
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o0080 1-12. (a) Torque about point A ¼ 0
‘ 2
‘ 160N + 320 cos 30° ¼ Fm ‘ sin 12°
2 3
p0210 Force exerted by muscle is Fm ¼ 2,000 N
p0215 From the geometry of figure angle of muscle with respect to y axis is 72°.
Angle of reaction force Fr at fifth lumbar with respect to y-axis is ϕ
x component of force ¼ 0
Fm sin 72° ¼ Fr sin ϕ
y comp of force ¼ 0
Fr cos ϕ ¼ 160 + 320 + Fm cos 72°
Fr sin ϕ ¼ 1, 902 N
Fr cos ϕ ¼ 1,098 N
;Fr 2 ¼ 4:82 106 N2 and Fr ¼ 2,200 N
o0085 (b) The added 20 kg mass is a force F ¼ 196 N. This force is added to
weight of arm and head. Torque conditions:
‘ 2
‘ 356 + 320 cos 30° ¼ Fm ‘ sin 12°
2 3
Fm ¼ 3, 220 N
p0225 Following 1-12 (a)
Fr sin ϕ ¼ 3,062
Fr cos ϕ ¼ 356 + 320 + Fm cos 72° ¼ 1,671 N
Fr 2 ¼ 12:17 106 N2 ;Fr ¼ 3,490 N
o0090 1-13. Let force of Achilles tendon ¼ FA
p0235 Let force on tibia ¼ FT
p0240 Let angle of FT with respect to vertical be ϕ ¼ 15°.
p0250 From Fig. 1-17 get
f5780
15°
FA
FT 15
y f
x
w
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