20.1100
INTRODUCTION TO ENGINEERING ANALYSIS
STUDIO/LAPTOP VERSION
TEST #1-SOLUTION
Spring 1999
February 5
The exam is open book/open notes/open laptop. No internet connection is allowed,
however. You have one hour to complete the exam. You may use Maple to help with
the calculation, but the problem formulation and solution methodology still need to be
included.
1. A force is applied to an eye bolt as shown. (40%)
z
1m
50N
F
y
1m
1m
2m
x
(a) Find the unit vector, e, in the same direction as F (express e in Cartesian Coordinate)
(10%)
Solution: The position vector along the direction of the force is:
r = (1i – 2j + 1k) m
The magnitude of this position vector is:
r = {(1)2 + (-2)2 + (1)2)}1/2 = 2.45 m
Therefore, the unit vector is just the position vector divided by it’s own magnitude:
eF = (1i – 2j + 1k)/(2.45)
eF = 0.408i – 0.817j + 0.408k
(b) Express F in Cartesian Coordinate (5%)
Solution: The force vector is just the magnitude of F (50 N) times the unit vector (eF) along the force’s
direction:
F = 50N (0.408i – 0.817j + 0.408k)
F = (20.412i – 40.825j + 20.412k) N
(c) Draw the vector a = i - j in the following diagram. (5%)
z
a = i-j
i
y
-j
x
(d) Find F along the vector a = i - j. (Recall that a vector c along a unit vector d means
(c d ) d ) (10%)
Solution: First, we need to find the dot product of F with a unit vector along the direction of a.
That unit vector is:
ea = (i – j)/{(1)2 + (-1)2)}1/2 = 0.707i – 0.707j
So the dot product is:
F  ea = (20.412i – 40.825j + 20.412k) N  (0.707i – 0.707j)
F  ea = (20.412 N)(0.707) + (-40.825 N)(-0.707) = 43.302 N
Now to give this result a direction (along the direction of the vector a = i - j), just multiply the
scalar result we got with the unit vector ea:
(43.202 N)(0.707i – 0.707j) = (30.619i – 30.619j) N
(e) Find the angle between F and a (10%)
Solution: We can also use to dot product to find the angle between F and a:
 = cos-1(eF  ea) = cos-1{(0.408)(0.707) + (-0.817)(-0.707)} = cos-1(0.866)
 = 30
2. The hot air balloon shown below is tethered with three mooring cables. These cables
are tied to the ground on the x-y plane. Assume the net lift of the balloon is 500N.
Write down the force balance equations for all forces exerted on the balloon. (You
do not have to solve these equations.) (30%)
z
D (0,0,20)m
B (-10,0,0)m
y
C (5,-10,0)m
A (10,10,0)m
x
z
500 N
TDC
rDC = (5i – 10j - 20k) m
TDB
rDB = (-10i – 20k) m
y
TDA
rDA = (10i – 10j - 20k) m
x
Solution: First, write out each force vector in Cartesian components:
DC = (5i – 10j – 20k) m
eDC = (5i – 10j – 20k)/{(5)2 + (-10)2 + (-20)2}1/2 = 0.218i – 0.436j – 0.873k
TDC = TDC (0.218i – 0.436j – 0.873k)
DA = (10i – 10j – 20k) m
eDA = (10i – 10j – 20k)/{(10)2 + (-10)2 + (-20)2}1/2 = 0.408i – 0.408j – 0.817k
TDA = TDA (0.408i – 0.408j – 0.817k)
DB = (-10i – 20k) m
eDB = (-10i – 20k)/{(-10)2 + (-20)2}1/2 = -0.447i – 0.894k
TDB = TDB (-0.447i – 0.894k)
FL = 500k N
Now, write out the equilibrium equations:
F = 0
FX = 0, FY = 0, FZ = 0
FX = TDC (0.218) + TDA (0.408) + TDB (-0.447) = 0
FY = TDC (-0.436) + TDA (-0.408) = 0
FZ = TDC (-0.873) + TDA (-0.817) + TDB (-0.894) + 500 = 0
3. A cable is used to support blocks A and B as shown below. Block A has mass 10Kg
and Block B 20Kg. Block A is supported by a small wheel that is free to roll on the
cable. (30%)
10m
y
A
B
(a) Draw the free body diagram (there only needs to be one FBD) (10%)
10 m
T
(20)(9.8) N
T
(20)(9.8) N
y


y
x
W (10)(9.8)N
(b) Write down the Force balance equation (10%)
Solution: Write out the equilibrium equations:
FX = T (cos) - T(cos) = 0
So  = . If the two angles are equal, and the heights of the right triangles are equal, then the base
of each triangle must also be equal. In other words, the horizontal distance from the wall to pulley
A is 5 m and the horizontal distance from pulley A to pulley B is also 5 m.
FY = T (sin) + T(sin) – (10)(9.807) N = 0
(c) Solve for the displacement y (10%)
Solution: We know that the tension in the cable is equal to the weight of pulley B.
T = (20)(9.807) N
FY = (20)(9.807)(sin) + (20)(9.807)(sin) – (10)(9.807) = 0
FY = 4sin – 1 = 0
sin = 1/4
Based on the geometry,
sin = (y)/{(5)2 + (y)2}1/2
Combining these two equations:
1/4 = (y)/{(5)2 + (y)2}1/2
15y2 = 25
y = 1.291 m