Engineering Mechanics Statics
and Dynamics 3e Michael Plesha,
Gary Gray, Robert Witt,
Francesco Costanzo (Solutions
Manual All Chapters, 100%
Original Verified, A+ Grade)
All Chapters Solutions Manual
Supplement files download link
at the end of this file.
,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 + c3 , (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
and Dynamics 3e Michael Plesha,
Gary Gray, Robert Witt,
Francesco Costanzo (Solutions
Manual All Chapters, 100%
Original Verified, A+ Grade)
All Chapters Solutions Manual
Supplement files download link
at the end of this file.
,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 + c3 , (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
