20.1104 Intro. To Engng Analysis
Thursday AM, 14 December 1995
Final Exam
20 Points per problem
1.
Name _______________________
Section______________________
The three cables are used to support the 8-kg lamp. Determine the force developed in each cable for
equilibrium.
FBD: Isolate A as a particle.
Define Vectors:
W = -(8kg)(9.81 m/s2) k = -78.48 N k
FAB = FAB (AB/|AB|) = FAB j
FAC = FAC (AC/|AC|) = FAC i
FAD = FAD (AD/|AD|) = FAD [(-2 i - 4 j + 4 k) / ((-2)2 + (-4)2 + (4)2))1/2]
= -2/6 FAD i -4/6 FAD j +4/6 FAD k
Equilibrium:
F = 0 (for A):  Fx = 0: FAC - 2/6 FAD = 0
Fy = 0: FAB - 4/6 FAD = 0
Fz = 0: -78.48 N + 4/6 FAD = 0
Solution: Solving eq. (1) - (3) simultaneously yields,
FAD = 118 N
FAC = 39.2 N
FAB = 78.5 N
ANSWER
(1)
(2)
(3)
20.1104 Intro. To Engng Analysis
Thursday AM, 14 December 1995
Final Exam
20 Points per problem
2.
Name _______________________
Section______________________
Determine the horizontal and vertical components of force which the pins at C and D exert on ECD of
the frame.
FBD’s: Members ECD, CB and AD.
Equilibrium:
From FBD (I): MD = 0: rDE x (800 lb i) + r DC x FCB = 0
i
6
800
j
k
6 0 
0
0
i
j
2
2
k
0 0
FCBcos(26.56) FCBsin(26.56) 0
-(-6)(800) k + [(-2)(FCBsin(26.56) - (-2)(-FCBcos(26.56)] k = 0
FCB = 1788.9 lb
ANSWER
Now go to FBD (II): Here FCB is (1788.9 lb)(cos(26.56) i - sin(25.56) j)
Fx = 0: - Dx + (1789 lb)cos26.56 - Ax = 0
Fy = 0: - Dy - (1789 lb)sin26.56 = 0
MD = 0: (3 ft)(1789 lb)cos26.56 - (Ax)(6) = 0
Solution: Solving eq. (1) - (3) simultaneously yields,
Ax = 800 lb
Dx = 800 lb
Dy = -800 lb
ANSWER
(1)
(2)
(3)
20.1104 Intro. To Engng Analysis
Thursday AM, 14 December 1995
Final Exam
20 Points per problem
3.
Name _______________________
Section______________________
A and B are matrices, b, c and x are column matrices.
1 0 2
A  4 3 5 


 1 3 1
  1 3 2
B
 5 3 1
5
b  11
 
0
3
c 
 7
 x1 
 
x   x2
 x 3
You must show your work in the following parts to get full credit.
(a) Calculate BT. (2 pts)
(b) Calculate det(A). (5 pts)
For the system of equations A x = b.
(c)
(d)
(e)
(f)
Define the augmented matrix. (2 pts)
Using elementary row operations , calculate the row echelon form. (5 pts)
Calculate the reduced row echelon form. (4 pts)
Calculate the solution, x. (2 pts)
Solution:
(a)
 1 5
B =  3 3


 2 1
T
(b)
(c)
det(A) = 1(3 - 15) + 2(12 - 3) = 6
1 0 2 5 
4 3 5 11


1 3 1 0 
(d) Row 2 = Row 2 - 4(Row 1)
Row 3 = Row 3 - Row 1
Row 3 = Row 3 - Row 2
(e) Divide Row 3 by 2
Divide Row 2 by 3
Row 2 = Row 2 + Row 3
Row 1 = Row 1 - 2(Row 3)
1 0 2 5 
0 3  3  9 


0 3  1  5 
1 0 2 5 
0 3  3  9   Row Echelon Form


0 0 2 4 
1
0

0
1
0

0
0
1
0

 1  3

1
2 
2
0
0
1
0
0
1
5

 1

2
1
 Reduced Row Echelon Form
1
(f) Reduced row echelon form matrix is in the form I x , therefore x =   1 .
 
2
 
20.1104 Intro. To Engng Analysis
Thursday AM, 14 December 1995
Final Exam
20 Points per problem
Name _______________________
Section______________________
4.
The stiff-leg derrick used on ships is supported by a ball-and-socket joint at D and two cables BA and
BC. The Cables are attached to a smooth collar ring at B, which allows rotation of the derrick about the
Z-axis. The derrick supports a crate having a mass of 200 kg. Answer the following questions. All of
the questions can be answered using Maple – if you wish – as long as you write down the Maple
commands that you use and the numerical results.
a.
(2 points) Write BA as a position vector in Cartesian component form and find the length of cable BA.
BA = (2 i - 3 j - 6 k ) m.
|BA| = (2)2 + (-3)2 + (-6)2 m = 49 = 7 m.
b.
(2 points) Write a Cartesian vector representing the tension in cable BA, with a magnitude T BA.
TBA = TBA eBA =
c.
2 3 6 
 TBA i  j  k
7 7 7 
BA
BA
(2 points) Use a vector method to find the angle between cables BA and BC.
Use dot product:
|BA| |BC| cos  = BA  BC, where BA and |BA| are given above.
BC = (-6 i - 3 j - 6 k) m.
|BC| = (-6)2 + (-3)2 + (-6)2 m = 9 m.
 cos  =
(2 i - 3 j - 6 k )  (-6 i - 3 j - 6 k)
(7)(9)

33
63
 = 27.65 ANSWER
d.
(2 points) Use vectors to compute a unit vector that is perpendicular to the plane containing cables BA
and BC.
Use cross product of unit vectors along BA and along BC.
BA BC
16
8
eperp. = eBA x eBC =
x
 0i 
j k
21
21
BA BC
4.
(concluded)
20.1104 Intro. To Engng Analysis
Thursday AM, 14 December 1995
Final Exam
20 Points per problem
e.
Name _______________________
Section______________________
(12 points) Determine the tensions in the cables BA and BC as well as the x, y and z components of the
reaction at D.
FBD: Derrick
Define Vectors:
2 3 6 
TBA = TBA i  j  k
7 7 7 
  2 1 2 
i j k
 3
3
3 
TBC = TBC
W = -200kg (9.81 m/s2) k = 1962 N k.
D = (Dx i + Dy j + Dz k) N
Equilibrium:
Fx = 0: TBA(2/7) + TBC(-2/3) + Dx = 0.
Fy = 0: TBA(-3/7) + TBC(-1/3) + Dy = 0.
(ii)
Fx = 0: TBA(-6/7) + TBC(-2/3) + Dz -1962 N = 0.
(i)
(iii)
MD = 0: rB/D x TBA + rB/D x TBC + rE/D x W = 0.
rB/D = 6 m k.
rE/D = 1 i + 4 j m.
MD = 0: [(18/7) TBA + 2 TBC - 7848] i + [12/7 TBA - 4 TBC + 1962] j = 0
Solution: Generate a matrix from eq. (i) - (iii) and i and j moment equations.
aug :=
2
 18 / 7
 12 / 7  4
 2/7 2/3

 3 / 7  1 / 3
 6 / 7  2 / 3
0
0
1
0
0
0
0
0
1
0
0 7848 
0  1962 

0
0

0
0 
1 1962 
TBA = 2003 N
1 0 0 0 0 2003
TBC = 1349 N
0 1 0 0 0 1349 
Dx = 327 N
ANSWER
0 0 1 0 0 327 
D
y = 1308 N


Dz = 4578 N
0 0 0 1 0 1308 
0 0 0 0 1 4578
The winch at W on the truck is used to hoist the garbage
bin onto the bed of the truck. If the loaded bin
solution := rref(aug);
5.
weighs 8500 lb. and has a center of gravity at G, determine the tension in the cable needed to begin the
20.1104 Intro. To Engng Analysis
Name _______________________
Thursday AM, 14 December 1995
Section______________________
Final Exam
20 Points per problem
lift. The coefficients of static friction at A and B are A = 0.3 and B = 0.2, respectively. Neglect the
height of the support at A and the small distance between the cable and the bed of the truck.
FBD: Garbage Bin
F=N
AF = A AN
AF = (0.3) AN
BF = B BN
BF = (0.2) BN
Equilibrium:
Fx = 0: -BN(cos 60) - BN(0.2)(cos 30) - AN(0.3) + TW(cos 30) = 0.
Simplify: 0.6732 BN - 0.3 AN + 0.8660 TW = 0.
(i)
Fy = 0: AN - BN(0.2)(cos 60) - BN(cos 30) + TW(cos 60) - 8500 lb = 0.
Simplify: 0.7660 BN + AN + 0.5 TW - 8500 lb = 0.
(ii)
MB = 0: -AN(22) +(8500)(12) = 0.
Simplify: 102000 - 22 AN = 0
(iii)
Solving (i) - (iii) simultaneously yields,
AN = 4636 lb
BN = 2651 lb
TW = 3667 lb
ANSWER