Solution Manual An Introduction to Biomechanics Solids and Fluids, Analysis and Design Humphrey O'Rourke 2nd Edition - Updated 2024
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Course
An Introduction to Biomechanics
Institution
An Introduction To Biomechanics
Solution Manual An Introduction to Biomechanics Solids and Fluids, Analysis and Design Humphrey O'Rourke 2nd Edition - Updated 2024
Complete Solution Manual With Answers
For synthesis, the k in d [C ] dt = − k [C ] is equal to −k s . For degradation, k is kd . Thus
we have
d [C ]s d [C ]d
= ks [C ] and = − kd [ C ] ,
dt dt
which if
d [C ] d [ C ]s d [C ]d
= +
dt dt dt
d [C ]
⇒ = k s [C ] − kd [C ]
dt
d [C ]
⇒ = ( k s − kd ) [C ]
dt
d [C ]
⇒ = ko [ C ]
dt
where ko is the overall reaction rate. Integrating with respect to time, we have
d [C ]
∫ dt = ∫ ko [C ] dt
dt
⇒ ∫ d [C ] = ko [C ] t + c1
⇒ [C ] = ko [C ] t + c1
⇒ (1 − ko t ) [C ] = c1
c1 c1
⇒ [C ] = = ,
1 − ko t 1 − ( k s − kd ) t
where we could calculate c1 if we were given an initial concentration. Note that if
, c1
k s > kd , ks − kd > 0, ∴ [C ] = ↑
1 − ko t
c1
k s < kd , ks − kd < 0, ∴ [C ] = ↓
1 − ko t
c1
k s = kd , ks − kd = 0, ∴ [C ] = = c1.
1 − 0t
1.17
(xx-Insert figure showing set-up)
Let point p be a point halfway between the applied forces. For convenience, let the point
be located at the origin of a 2-D Cartesian coordinate system, with the applied forces located at
x = d 2 and x = − d 2, and the forces oriented in the y direction such that F1 = Fˆj and
F2 = − Fˆj. Computing the moments at point p, we have ∑M) p
= r1 × F1 + r2 × F2 , where at p,
r1 = 1 2 diˆ and r2 = − 1 2 diˆ.
∑M) p
d
2
( ⎛ d⎞
)
F iˆ × ˆj + ⎜ − ⎟ ( − F ) iˆ × ˆj
=
⎝ 2⎠
( )
(
⇒ ∑ M ) p = dF iˆ × ˆj = dFkˆ. )
If we choose an arbitrary point a located by the vector ra = r cos α iˆ + r sin α ˆj ,
⎛1 ⎞ ⎛1 ⎞
r1 = ⎜ d − r cos α ⎟ iˆ + r sin α ˆj , r2 = − ⎜ d + r cos α ⎟ iˆ + r sin α ˆj
⎝2 ⎠ ⎝2 ⎠
and thus
∑M) a
⎛1
⎝2
⎞
( ) (
= ⎜ d − r cos α ⎟ F iˆ × ˆj + ( r sin α ) F ˆj × ˆj …
⎠
)
⎛ 1 ⎞
( ) (
+ ⎜ − d − r cos α ⎟ ( − F ) iˆ × ˆj + ( r sin α )( − F ) ˆj × ˆj
⎝ 2 ⎠
)
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