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Solutions Manual For Engineering Mechanics Statics 3rd Edition By Michael Plesha, Gary Gray, Robert Witt, Francesco Costanzo (All Chapters, 100% Original Verified, A+ Grade) $20.49
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Solutions Manual For Engineering Mechanics Statics 3rd Edition By Michael Plesha, Gary Gray, Robert Witt, Francesco Costanzo (All Chapters, 100% Original Verified, A+ Grade)
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Course
Engineering Mechanics Statics
Institution
Engineering Mechanics Statics
This Is The Original 3rd Edition Of The Solution Manual From The Original Author All Other Files In The Market Are Fake/Old Editions. Other Sellers Have Changed The Old Edition Number To The New But The Solution Manual Is An Old Edition.
Solutions Manual For Engineering Mechanics Statics 3rd Ed...
Engineering Mechanics Statics
3e Michael Plesha, Gary Gray,
Robert Witt, Francesco
Costanzo
(Solutions Manual All
Chapters, 100% Original
Verified, A+ Grade)
Statics Part: CH 1-10
,1-0 Solutions Manual
Chapter 1
Solutions
,Statics 3e 1-1
Problem �.�
(a) Consider a situation in which the force F applied to a particle of mass m is zero. Multiply the scalar
form of Eq. (1.2) on page 7 (i.e., a = dv_dt) by dt, and integrate both sides to show that the velocity
v (also a scalar) is constant. Then use the scalar form of Eq. (1.1) to show that the (scalar) position
r is a linear function of time.
(b) Repeat Part (a) when the force applied to the particle is a nonzero constant, to show that the velocity
and position are linear and quadratic functions of time, respectively.
Solution
Part (a) Consider the scalar form of Eq. (1.3) on page 7 for the case with F = 0,
F = ma Ÿ 0 = ma Ÿ a = 0. (1)
Next, consider the scalar form of Eq. (1.2) on page 7,
dv
=a Ÿ dv = adt Ÿ dv = v = a dt. (2)
dt
Substituting a = 0 into Eq. (2) and evaluating the integral provides
v = constant = v0 , (3)
demonstrating that the velocity v is constant when the acceleration is zero. Next, consider the scalar form of
Eq. (1.1),
dr
= v Ÿ dr = vdt Ÿ dr = r = v dt. (4)
dt
For the case with constant velocity given by Eq. (3), it follows that
r= v0 dt = v0 dt = v0 t + c1 , (5)
where c1 is a constant of integration. Thus, the position r is a linear function of time when the acceleration
is zero. Note that in the special case that v0 = 0, then the position r does not change with time.
Part (b) When the force F is constant, then Newton’s second law provides
F = constant = ma Ÿ a = F _m = constant. (6)
Following the same procedure as used in Part (a), we find that
F F
v= a dt = dt = t + c2 , (7)
m m
where c2 is a constant of integration and v is shown to be a linear function of time. Likewise, recalling Eq. (4)
F F 2
⇠ ⇡
u= v dt = t + c2 dt = t + c2 t + c 3 , (8)
m 2m
which is a general quadratic function of time. To determine the constants of integration requires that initial
conditions be specified. That is, at some instant of time (usually t = 0), we must specify the position and
velocity of the particle.
, 1-2 Solutions Manual
Problem �.�
Using the length and force conversion factors in Table 1.2 on p. 10, verify that 1 slug = 14.59 kg.
Solution
kg m/s2
0 10 10 10 1
lb s2 /ft 4.448 N ft
1 slug = 1 slug = 14.59 kg. (1)
slug lb 0.3048 m N
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