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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
...
(b) Assume as before that center of mass (m) of arm is at mid-length
of the arm
...
45 m
Let the force along the arm be Fa; this force is at 60° with respect to the
horizontal
...
Following Eq
...
The length of the step is 1m therefore, the runner does 100 steps in 10 sec
...
1 sec
Using the pendulum analogy,
T ¼ 2 × 0
...
2sec
A ¼ length of step (See Fig
...
4-11
amax ¼

o0215

1=2

1=2

2 0:9
¼ 2π 3 × 9:8

1=2

¼ 1:6 sec:

As discussed in text, time for 1 step is T2 ¼ 0:80 sec
Length of step is 0
...
The period of the arm swing is
T ¼ 2π

p0865
o0225
p0875

2

0:9
×

1=2

¼ 1:6 sec:

3 9:8
Same as that for the leg
4-8
...
of steps/sec 5 speed/length of step 5 vA(from Fig
...
of steps
...
4-12
T

Page 16 of 43

I ¼ m‘3 from Appendix A;
‘ ¼ length of leg
vmax
ωmax ¼
‘
Maximum energy/step 5 Er
2

p0880
p0885

p0890

1 2
Er ¼ Iωm see text
2
1 m‘2 π2v2 π2 2
¼
2 ¼ 6 mv
2 3 ‘2
¼ 1:64 mv
o0230

4-9
...
4
...
2 is
m‘2
I¼
‘ ¼ 2m
3
m‘2
T ¼ 2π
mg‘=2
¼ 2π

2‘
g

1=2

1=2

¼ 6:28 ×

4
9:8

1=2

¼ 4:0 sec:
p0905
o0235
p0915

Time of falling is T4 ¼ 1 sec
4-10
...
45 5 26
...
7° 5 0
...
)
The distance the center of mass is raised in the course of one step
is: 1m – 0
...
11m 5 11cm
...
From Eq
...
1)
E¼
SB ¼ 10 dyn=cm
2 y
14 × 100 × 1018
¼
2 14 × 1010
¼ 143 J

p0930

8

¼ 14:3 × 10 erg

Therefore the kinetic energy of runner that may cause fracture is
obtained from
1 2
mv ¼ 143 Jm ¼ 50 kg
2
v ¼ 2:39 m=sec ¼ 8:6 km=hr
¼ 5:3 m:p:h

o0245
p0940

5-2
...
However,
if the area of impact is smaller the maximum speed of fracture would correspondingly decrease
...
An impact force FB that will fracture the skull is
FB ¼ 1 cm2 × 109dyn=cm2 ¼ 109dyn
4

¼ 10 N
p0970

The impulsive force of the falling body is
F¼

mv
1=2
;v ¼ ð2ghÞ
Δt

Page 18 of 43

p0975
p0980
p0985

Therefore F ¼ (m(2gh)1/2)/Δt
Fracture will occur when F ¼ 104 N
or
F2Δt2

108 × 10—6

h¼

o0255

p0995

o0260

¼
¼ 5:1m
m22g 1 × 2 × 9:8
5-4
...
A1-4)
v v2
¼
t 2s
Therefore t ¼ 2sv
s ¼ 30 cm ¼ 0:3m
v ¼ 70 km=hr ¼ 19:4 m= sec ~ 20 m=sec
0:6
t¼
¼ 3 × 10—2 sec
20
5-5
...
1 rupture strength on stretching, (as in whiplash) is
83 × 107 dyn/cm2
FB ¼ 1 cm2 × 83 × 107 ¼ 83 × 102N
mv

p1005
p1010

Impulsive force is F ¼ Δt
Therefore the collisi

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