RENSSELAER POLYTECHNIC INSTITUTE
TROY, NY
EXAM NO. 2 INTRODUCTION TO ENGINEERING ANALYSIS
(ENGR-1100) – Fall 14
NAME: Solution
Section: ___________
RIN:
_______________________________
Wednesday, October 15, 2014
8:00 – 9:50
Please state clearly all assumptions made in order for full credit to be given.
Problem
Points
1
25
2
25
3
25
4
25
Total
100
Score
Problem #1 (25)
1
4
7
2 3
5 6 ,
8 9
1 2 3
,
4 5 6
1
3
5
Determine (if not possible, explain why):
2
4
6
(a)
(b)
(c)
and
(d) Determinant of using any method.
(e) If
, what is the dimension of ? Write explicitly this identity matrix.
1
4
1 2
2 3
3 4
5 6
5 6
22
49
1
4
7
1
4
7
1
4
7
28
64
64
49
2 3
5 6
8 9
2 3
5 6
8 9
1 1
3 1
5 1
9
19
29
28
22
22
49
1 2 3
4 5 6
7 8 9
2 3
5 6
8 9
1 1
4 1
12
26
40
64
49
28
64
1 2
1
3 4
4
5 6
2 4
4 4
6 4
15
33
51
2 3
5 3
3 5
6 5
1 2
4 2
28
22
2 3
5 6
1 9
4 19
7 29
2
5
8
2 5
4 5
6 5
12
26
40
3 6
6 6
1.778
1.361
is not possible, since
1 2
3 2
5 2
2 4
5 4
(4)
(4)
(10)
(5)
(2)
3
6
9
and
1 3
3 3
5 3
15
33
51
28
64
(a)
0.7778
0.6111
(b)
22
49
do not the same dimension
(c)
2 6
4 6
6 6
10 14
23 31
36 48
18
39
60
(c)
det
1
1
1 2
4 5
7 8
5 9
3
2
3
6
9
8 6
is a square matrix.
6
5
8
1
3
2
6
9
4 9
3
2
3
→
4 6
7 9
7 6
12
3
3
9
3
4 5
7 8
4 8
7 5
det
Size of is 3
3 and
1
0
0
0
0 0
1 0
0 1
(d)
(e)
Problem #2 (25)
A 100-kg cylinder is supported by two cables AC and BC, which are attached to the top of
vertical posts. A horizontal force P, making an angle  with the x axis, holds the cylinder in
the position shown in the figure below.
1. Identify the particle that needs to be analyzed to determine the tensions in the cables and the
force P. (1 points)
2. Draw a complete and separate FBD showing the particle and all the forces acting on the
particle. (8 points)
3. Express all forces in a Cartesian vector form (6 points)
4. Write the (scalar) equilibrium equations. (9 points)
5. Determine the magnitude of the force P. (2 point)
6. Determine the magnitude of the forces in the cables CA and CB. (4 points)
(a calculator may be used for questions 5 and 6).
Particle to be analyzed: C
FBD
z
TCA
4.8
TCB
P

W
x
A=(0.0,-3.6,5.0); B=(0.0,4.8,5.0); C=(1.2,0.0,2.4)
W  981k N
y
TCA  TCAu CA  TCA (0.2609i  0.7826 j  0.5652k)
u CA 
 1.2i  3.6 j  2.6k
( 1 . 2 )  ( 3 . 6 )  ( 2 . 6 )
2
2
 0.2609i  0.7826 j  0.5652k
2
TCB  TCB u CB  TCB (0.2147i  0.8588 j  0.4652k)
u CB 
 1.2i  4.8 j  2.6k
(1.2)  (4.8)  (2.6)
2
2
2
 0.2147i  0.8588 j  0.4652k
P  Pu p  P (cos(300 )i  sin(300 ) j  0.0k)
P  Pu p  P (0.8660i  0.5000 j  0.0k)
TCA  TCB  P  W  0
Equilibrium:
TCA (0.2609i  0.7826 j  0.5652k) 
TCB (0.2147i  0.8588 j  0.4652k) 
P (0.8660i  0.5000 j  0.0k)  981k N  0
 0.2609TCA  0.2147TCB  0.8660 P  0
 0.7826TCA  0.8588TCB  0.5000 P  0
 0.5652TCA  0.4652TCB  981 N
TCA  848.62 N
TCB  1077.73 N
P  522.86 N
Problem #3 (30)
A ball and a rod ABC are connected through a T junction. The ball and the rod weigh 20 lb. each
as shown in the figure. The length of the rod AC is 2 ft. The point B is in the middle of the rod.
Two 50-lb forces acting at points A and C, and a 50-lb-ft couple about the z-axis shown above
point O are also exerted on the system. If
3 ft. and
60° , find:
(a) The moment of the weight of the ball about point C.
(5)
(b) The moment of the weight of the ball about axis AC in Cartesian vector form.
(5)
(c) The resultant (both force and couple moment) of just the two 50-lb forces.
(5)
(d) The equivalent of the four forces and the 50-lb-ft couple moment at point O.
(10)
(e) Is it possible to find a point at which the equivalent of the four forces and the 50-lb-ft couple
is just a single force? If yes, find the point(s). If not, explain why.
(5)
Let’s point be center of the ball.
The moment of the weight of the ball about point :
6
1
1
6
0
20
0
60°
∙
10
120
0
50
2
100
0
60°
120
60°
20
0
;
120 ∙
60°
60°
60°
2
60°
0
60°
60°
2
2
60°
0
60°
0
100 60°
60°
60°
60°
60°
20
60°
20
6
0
60°
120
60°
0
2 60°
50
0
20
60°
120
0
50
2
0
6
0
60°
60°
10
0
60°
120 lb-ft
60°
30
51.96 lb-ft
(a)
(b)
(c)
20
20
20
20
20
50
20
50
86.60
50
86.60
50
86.60
50 ;
3
50 lb-ft
3
40 lb
86.60 lb-ft
is perpendicular to . Therefore, it is possible to find a point at which the resultant is just
a single force. Let’s call that point
, ,
.
0
40
0
0
0
40
40
40
86.60 . If
0, then
At any point on the line (
2.165 ft,
0,
40
0
0
(d)
86.60
86.60/40,
0, and is arbitrary.
:arbitrary) the resultant will be just
40 lb
(e)
Problem #4 (15)
Draw a separate and complete free body diagram for the following:
1) A bent bar of negligible mass is supported by a pin at A and a roller at B (the angle
=50o)
y
x
o
50
Ay
o
50
Ax
FB
2) A bar of negligible mass is supported by rollers at A and B and a rope AD.