INSTRUCTOR’S SOLUTIONS MANUAL
for
ENGINEERING MECHANICS: DYNAMICS
SI COMPUTATIONAL EDITION
by
Robert W. Soutas-Little
Daniel J. Inman
Daniel S. Balint
, INSTRUCTOR’S INTRODUCTION
This text introduces and encourages the use of computational software in teaching
dynamics. The software, whether it be Mathcad, Mathematica, MATLAB, or Maple,
can be used much like a calculator to enhance students’ understanding and interest in
the course material. This manual contains the solutions to the problems in the text.
Some problems in the text are marked with a single computer icon denoting that
software is useful in solving the problem. Others are marked with a double computer
icon denoting that some sort of computational software is required to obtain the
solution. Input syntax for computational solutions are given in this manual in many
instances, using MATLAB in some cases and Mathcad in others. However, all four
software packages (Mathematica, Maple, Mathcad, MATLAB) can be used equally
well to solve dynamics problems. Although some software-specific syntax is given
in this manual, the Computational Supplements should be used for instructions on
solving dynamics problems using computational software.
Some obvious questions arise in teaching a course for the first time using computa-
tional software. The major question is, “If students use computers in doing their
homework, how can they be examined?” Engineering computations come in roughly
two categories: those that students should be able to do with a few short lines of writ-
ten thought and those that are best done on a machine because they involve repeated
or complicated steps likely to induce errors. On an exam, it is most important to
measure the students’ understanding of dynamics, not arithmetic and algebra. Com-
putational software allows the teaching and examining to be focused on the mechanics
while the homework can be used to sharpen the students’ skills at obtaining correct
numerical answers to fairly complex problems. Thus, exams can focus on things like
making free body diagrams and writing the equations of motion rather than on solv-
ing the equations. By using computational software in the homework, the lecturers
and exams are free to focus on fundamentals.
Solutions of the homework problems are set up explicitly in this manual, but in
many cases numerical solutions are obtained using computational software. The input
syntax is presented in some cases, but the problems can often be solved numerically
using different combinations of input syntax. It is therefore up to the user to decide
how best to solve each problem using computational software. This manual provides
some guidance, but the primary instruction on how to solve dynamics problems using
computational software is given in the Computational Supplements.
, CHAPTER 1
1.1 Using the definition of acceleration:
∆v 100 km/hr − 0 27.8 m/sec − 0
α= − = = 3.02 m/sec2
∆t 9.2 − 0 9.2 sec
1
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deductions. Partners are also subject to self-employment taxes on their share of the income, which can significantly impact tax liability.- **Corporations**: - C-Corporations are taxed at the corporate tax rate (currently 21%). Dividends are taxed
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1.2 Differentiate x(t) to obtain the velocity:
v(t) = ẋ(t) = −2.4t + 28 m/sec.
Differentiating again yields the acceleration:
a(t) = ẍ ( t) = −2.4 m/sec2.
So v(t) = 0 = −2.4t + 28.
Solving this for t yields that:
v(t) = 0 at t = 11.7 sec.
1.3 Evaluating x at zero yields x(0) = 5 m.
Differentiating yields ẋ(t) = v(t) = 3t2 − 2
so that v(0) = −2 m/s.
Likewise
a(t) = ẍ ( t) = 6t
so that a(0) = 0.
Now at t = 3 sec,
x(3) = 27 − 2(3) + 5 = 26 m,
v(3) = 27 − 2 = 25 m/s
and a(3) = 18 m/s2.
Since the velocity changes sign during this interval, the particle has doubled
back and to compute the total distance traveled during the interval you must
compute how far it travels before it changes direction and then add this to the
distance traveled after the particle has changed direction. The particle changes
direction when the velocity is zero, or at the value of t for which
v(t) = 3t2 − 2 = 0,
or at time
t = 0.8165.
The particle first moves from
x(0) = 5 m to x(0.8165) = 3.9 m or a distance of 1.1 m.
It then changes direction and moves from 3.9 m to
x(3) = 26 m. Thus it travels a total distance of
(26 − 3.9) + 1.1 = 1.1 + 1.1 + 21 = 23.2 m.
2
