Intro to Engineering Analysis, Studio/Laptop Version
Test #1, Sept. 18, 1998
NAME_
Solution________________
If you wish, you can use Maple to do the problems below, but you must show your methodology on the
test pages. Remember that clarity is important.
Problem 1 (48 points, 8 points each)
Use Cartesian vectors to solve the following problems based on the attached sketch. Boldface indicates
a vector quantity.
a) Write the position vector from O to A. Also
write a position vector along OA that has half
the magnitude of the vector from O to A.
z
Solution:
•
A (2,3,7)m
C (3,-2,5)m
•
•
O
b) Write a force vector of magnitude 10 N that
acts along direction OA.
B
Y = 45
Solution:
X = 90
x
OA = (2i + 3j + 7k) m
HOA = (i + 1.5j + 3.5k) m
y
OA = (2i + 3j + 7k) m
eOA = {(2i + 3j + 7k)/(22 + 32 + 72)}
eOA = {(2i + 3j + 7k)/7.87
F = 10 N (0.254i + 0.381j + 0.889k)
F = (2.54i + 3.81j + 8.89k) N
c) Write a force vector F of magnitude F that acts along direction AO.
Solution:
eAO = -eOA = -0.254i - 0.381j - 0.889k
F = F (-0.254i - 0.381j - 0.889k)
d) Write the position vector OB, when the length of vector OB is 10 m, x = 90 degrees, y = 45
degrees, and OB is in the y-z plane.
Solution:
OB = 10 m (cos x i + cos Y j + cos Z k)
Where x = 90 , Y = 45 , Z = 45
OB = 10 m (0i + 0.707j + 0.707k)
1
OB = (7.07j + 7.07k) m
e) What is the cosine of the angle between position vector OC and the positive x-axis?
Solution:
eOC  eX = (eOC)(ex) cos 
(by the definition of the dot product)
cos  = (eOC  eX)
*Note that eX is the unit vector along the x-axis (eX = i).
OC = (3i - 2j + 5k) m; eOC = {(3i - 2j + 5k)/(32 + -22 + 52)}
cos  = (eOC  i)
cos  = (0.487i – 0.325j + 0.812k)  (i)
cos  = 0.487
f) Write a position vector BA that goes from point B to A.
Solution:
BA = OA – OB = {(2 - 0)i + (3 - 7.07)j + (7 – 7.07)k)} m
BA = (2i – 4.07j - 0.07k) m
2
Problem 2 (20 points)
Italian surveyors examine the Leaning Tower of Pisa (see diagram below). With respect to a coordinate
system, they measure point A to be exactly in the middle of the ground floor at (0,0,0) meters. With
respect to that origin, they measure the coordinates of a point B, which is exactly in the middle of the
“roof”, to be (1, 5, 35) meters. Thus, AB is the Tower’s centerline. What angle does the Tower’s
centerline make with the y- and z-axes? Use the dot product.
Diagrammatically, this looks like:
z
A (0,0,0) m
B (1,5,35) m
y
x
3
Solution:
AB = (i + 5j + 35k) m
eAB = {(i + 5j + 35k)/(12 + 52 + 352)} = {(i + 5j + 35k)/35.37}
eAB = (0.028i + 0.141j + 0.989k)
For the angle AB makes with respect to the y axis:
eAB  eY = (eAB)(eY) cos Y
cos Y = (eAB  eY)
*Note that eY is the unit vector along the y-axis (eY = j).
cos Y = (eAB  eY)
cos Y =(0.028i + 0.141j + 0.989k)  (j)
cos Y = 0.141
Y = 81.89
Similarly, for the angle AB makes with respect to the z axis:
eAB  eZ = (eAB)(eZ) cos Z
cos Z = (eAB  eZ)
*Note that eZ is the unit vector along the z-axis (eY = k).
cos Z = (eAB  eZ)
cos Z =(0.028i + 0.141j + 0.989k)  (k)
cos Z = 0.989
Z = 8.51
4
Problem 3 (32 points) Answer questions a, b and c (below) for the following problem (from Engineering
Mechanics - Statics by Hibbeler):
Romeo is trying to reach Juliet by climbing with a constant velocity up a rope that is knotted at point A.
Any of the three segments of the rope can sustain a maximum force of 2000 N before breaking. Romeo
has a mass of 65 kg and Juliet has a mass of 60 kg. Juliet joins Romeo and they are at rest on the rope at
Romeo’s position shown in the diagram. What is the tension in each rope, and will any of the ropes
break?
a) (10 points) Draw an appropriate free body diagram for this problem.
Solution:
y
TAB
TAC
60
x
W
W = (65+60)(9.81)N
5
b. (10 points) Write out the relevant vectors and identify the unknown quantities.
Solution:
TAB = TAB (-cos 60i + sin 60j)
TAC = TAC (i)
W = 1225 (-j) N
Unknowns: TAB and TAC
b) (12 points) Set up, but don’t solve, the scalar equations that would have to be solved for the unknown
quantities in this problem. NOTE: If you wish to solve the equations, correct numerical answers will
be worth up to 5 extra bonus points for the test.
Solution:
For static equilibrium, F = 0
FX = -TAB cos 60 + TAC = 0
FY = TAB sin 60 – 1225 N = 0
(1)
(2)
Extra Credit:
Solving equation (2) directly:
TAB = 1414.5 N
Substituting into equation (1) and solving:
TAC = 707.3 N
Both of these tensions are less than the limit (2000 N). Therefore, Romeo and Juliet will climb
away to safety and live happily ever after…
6