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Blueprint

Solution
Manual

Page 1 of 43

s0010 CHAPTER 1
o0010

1-1
...
Toppling torque Ta 5 Fa × 1
...

1-2
...
1
...
E
...
2

f0010

L1

d1
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 d 1
and
¼
L2 d2

Page 2 of 43

o0020

1-3
...
E
...
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 30 sin 80° ¼ 29:4 cm
θ

¼

y0

tan —1

θ ¼ tan

x0 + 4
29:4
—1
5:2+4

¼ 72:6

°

f0015

100°
y
w

q
4 cm

x
o0025

1-4
...
1-13
W¼

Fm
10:5

¼ 335 N ¼ 75 lb

Page 3 of 43

o0030

1-5
...
E
...
5

f0020

y
160°
a


w

q


b

Fm
x



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
...
Using results from
Exercises 1
...
(1-10)
x component: Fm cos(θ + δ) ¼ Fr cos ϕ
y component: Fm sin(θ + δ) ¼ Fr sin ϕ + W
Torque is: 4 cm Fm sin (θ) ¼ 40 cm W × sin γ
From these we obtain 3 equations
1
...
7 ¼ Fr cos ϕ
2
...
7 ¼ Fr sin ϕ + 137 N
3
...
7 ¼ 10 × 137 × sin 30o
From 3
...
4o
Fr ¼ 2,386 N ¼ 536 lb

Page 4 of 43

o0050

1-6
...
1-12 (or 1-13) θ ¼ 72
...
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
...
(a) 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 × 106N2
Fr ¼ 1, 900 N
1, 790
°
tan ϕ ¼
¼ 2:77;ϕ ¼ 70:2
646

li7890
o0065
p0155
p0160
p0165

(b) yes
1-8
...

Referring to Problem 1-6
Added force Fm ¼ 720
¼ 103 N
7
Added force Fr ¼ 590
¼
84 N
7

Page 5 of 43

o0070

1-10
...
E
...
10 θ ¼ 72
...
3° ¼ 19
...

3:4

p0185

Speed of muscle contraction ¼ 4 cm=s
19:6 cm
Speed of weight displacement ¼
0:5s
¼ 38 cm=s:
Ratio of speeds is approximately inverse of mechanical advantage
...


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
...
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

Page 6 of 43

o0080

1-12
...

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
...
Torque conditions:
‘
2
‘ × 356 + 320 cos 30 ° ¼ Fm ‘sin 12°
3
2
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
p0235
p0240
p0250
f5780

1-13
...

From Fig
...
5 W
x component of force:
FT sin ϕ ¼ FA sin 15° ¼ 0
...
42 W
;F2T ¼ 12:1 N2

f0025

p0290

and FT ¼ 3
...
(a) Ff ¼ μ Fn Fn ¼ m g
μ ¼ 0
...
33 meters/sec
Work done by frictional force bringing the skater to a halt is Ffd
1
Ff d ¼ mv2
2
1 2
μmgd
mv
¼
2
ð8:33Þ2
v2
d¼
¼
¼ 354 meters
2μg 2 × 0:01 × 9:8
(b) The distance of coasting is independent of mass
...
As in Chapter 1 Eq
...
5 ¼ W × 0
...
(a) Let the force at point be B ¼ Fb
Let the force at point be C ¼ Fc
Equilibrium torque conditions about point A ¼ 0
FB × 2:5 cm=cos 45° ¼ ðFC × 3 cmÞ=cos 45°
2:5
;FC ¼
FB
3
Sum of forces perpendicular to the fin are equal to zero:
F sin 25° + FC ¼ FB

p0370
p0375

p0380

With F ¼ 0
...
1 by Eq
...
01,
;45° — θ ¼ 44:4° and θ ¼ 0:6°

2-4
...
Volume of residual saliva in the mouth ¼ 0
...

p0410
Let radius of the tongue (assumed to be a sphere) be rt and radius of the
oral cavity be rc
...
17 cm3 and (rc)3 ¼ 9
...
93 cm and rc ¼ 2
...
8¼cm3
p0435
Let the thickness of the coating¼Δr, then the volume of the saliva coating on the tongue and the oral cavity respectively is 4 π (rt)2 x Δr and 4 π
(rc)2 x Δr respectively
...
2 Δr + 56
...
8 cm3 or Δr ¼ 0
...
0775 mm
This is approximately ~0
...


Page 10 of 43

s0020 CHAPTER 3
o0125
p0470
p0475
p0480
p0485

o0130

p0495

3-1
...
3-12
Work done on the body ¼ W(c + H)
W ¼ 70 kg × 9
...
2 meters
∴ work done ¼ 823 J
work
823 J
Power ¼
¼
¼ 4, 120 W:
time 0:2 sec
3-2
...
Assuming as in the text that Fm W
¼
From Eq
...
Square Eq
...

Fr2 ¼ 5 W2 — 4W2 sin θ

p0510

Obtain
sin θ ¼

p0515

p0520

5W2 — F2r
4W2

Substitute into 3-20 and obtain quadratic equation Fr2 + 1
...
41 ∴ θ ¼ 65
...
From the text vo ¼ 3
...
7 × sin 45° ¼ 2
...
3-1
vy ¼ vy o + at

p0560

At maximum height of jump vy ¼ 0
Therefore

p0565

—a vy o
vy o
¼
a
g
2:62
¼
¼ 0:267 sec
9:8
Duration of jump ¼ 2t ¼ 0
...
(a) Eq
...

Because weight on the moon is
Eq
...
3-19 and Eq
...
1 and obtain
Fr2 + 0:236 Fr W — 3:97 W2 ¼ 0

p0595
p0600
p0605
p0610
p0615
p0620

Solve for Fr taking positive answer
Fr ¼ 1
...
707 × 1
...
665
θ ¼ 48
...
4 × 0
...
1 m2/sec2
From Eq
...
3-15
2 θ
2v
ðvo2Þ cos
22:1
o
H¼
¼ 0¼
¼ 3:39 m
2g0
4g 4 × 1:63
(c) From Eq
...
Duration of jump ¼ 2 × t ¼ 4
...

¼1:63

Page 12 of 43

o0160

3-6
...
3-24
vt ¼

W
CA

1=2

ρ ¼ density

4
W ¼ πr3ρ A ¼ πr2
3
C ¼ 0:88 kg=m3ðsee text section 3:7Þ
r ¼ 0:5 cm ¼ 5 × 10—3m
ρ ¼ 1g=cm3 ¼ 103kg=m3
vt ¼
¼

4rgρ
3c

1=2

4 × 5 × 10—3 × 9:8 × 103

1=2

3 × 0:88
¼ 8:6 m= sec
o0165

m=sec

3-7
...
3-24
W
A¼ 2
vt C
W ¼ 70 × 9:8N
v2t ¼ 196 m2 C ¼ 0:88 kg=m3

p0665
p0670
o0170
o0175

Therefore A ¼ 3
...
13 m
3-8
...
3-24
W
vt ¼
(a)
CA

1=2

ρ ¼ density ¼ 0:92 × 103kg=m3
4
W ¼ πr3ρ A ¼ πr2
3
C ¼ 0:88 kg=m3ðsee textÞ
r ¼ 5 × 10—3mρ ¼ 0:92 × 103kg=m3
vt ¼
¼

4rgρ
3C

1=2

4 × 5 × 10—3 × 9:8 × 0:92 × 103
3 × 0:98

1=2

m=sec

¼ 8:3 m= sec

Page 13 of 43

o0180

p0690
o0185

(b) Note from Eq
...
3 m/s × (4)1/2 ¼ 16
...
Work done in each step ¼ mv2 (see text Section 3
...
Assume area of runner is 0
...
7)
Wind velocity ¼ 30 km/hr ¼ 8
...
33 + 4
...
8 m/sec
Fa ¼ C Av2 ¼ 0
...
2 × 165 ¼ 29
...
Centrifugal force Fc
Mass of the arm ¼ (0
...
06)70 kg ¼ 9
...

R location of center of mass from shoulder ¼90 cm ¼ 0

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