IEAMakeupExam 2 S 13 Solution

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RENSSELAER POLYTECHNIC INSTITUTE
TROY, NY
MAKEUPEXAM NO. 2 INTRODUCTION TO ENGINEERING ANALYSIS
(ENGR-1100) – Spring 13
NAME: Solution
Section: ___________
RIN:
_______________________________
Wednesday, March 27, 2013
5:00 – 6:50
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 %)
For the following matrices:
3 8 10
0 
 4 8
2 0 101






A  0 7 0 ; B  1 5 3; C  5; D  
; E  8 2 
23
1
0


1 3 4 
4
5 4
Determine the following expressions. If an expression is not valid, state the reason:
a.
B
b.
(D  D )
c.
I 2  (E E)
T


 C B  C 
T
(10)
T T
T
(2)
1
(8)
d. Det (A) using the duplicate column method.
(5)
Note: You need to show all the work to receive full credit.
a.
1
0
0
[( + )( ∗ )] =
5 + 5 ∗ [1 5 3] ∗ 5
3
4
4
1
= 10 ∗ [1 ∗ 0 + 5 ∗ 5 + 3 ∗ 4]
7
1 ∗ 32
= 10 ∗ 32
7 ∗ 32
= [32
320
224]
b.
Invalid, dimensions are not similar
(2) (2)
(2)
(2)
(2)
(2)
c.
∗(
∗ )
1 0
∗
0 1
105
→ det
68
∗( ∗ )
0.02
=
−0.0162
=
1 0
=
∗
0 1
4 8
8 2
4
5
∗ 8
4
5
105 68
68 84
68
= 105 ∗ 84 − 68 = 4196
84
1 0
84 −68
=
∗
0 1
−68 105
−0.0162
0.025
8
2
4
(1)
(2)
(2)
=
84 −68
−68 105
(2)
(1)
d.
3 8 10 3 8
det( ) = 0 7 0 0 7
1 3 4 1 3
= 3 ∗ 7 ∗ 4 + 8 ∗ 0 ∗ 1 + 10 ∗ 0 ∗ 3 − 1 ∗ 7 ∗ 10 − 3 ∗ 0 ∗ 3 − 4 ∗ 0 ∗ 8
= 14
(2)
(2)
(1)
Problem #2 (25%)
A 75- kg container is suspended from a frictionless ring A, to which cables AC and AE are attached
and maintained at equilibrium. A force P is applied to end F of a third cable that passes over a
frictionless pulley at B and through ring A and then is attached to a support at D. (hint: cable FBAD
is a continuous cable passing through a frictionless pulley and a ring and two other cables: CA and
EA are tied to the ring)
a. Draw the complete and separate free body diagram of the identified particle to determine magnitude
of P.
(5)
b. Write the equilibrium equations for the FBD.
(12)
c. Determine the tension (force) in each of the three cables and obtain P.
(8)
Problem #3 (25 %)
The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the
position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine
the moment of that force about the line joining Points D and B.
(All vector answers should be given in the Cartesian coordinates shown)
First note:
TCF  TCF

CF
CF
CF  (0.6) 2  (0.9) 2  (0.2) 2  1.1 m
33 N
(0.6i  0.9 j  0.2k )
1.1
 3[(6 N)i  (9 N) j  (2 N)k ]
Then
TCF 
Also,
DB  (1.2) 2  (0.35) 2  (0) 2
Then
Now
 1.25 m

DB
 DB 
DB
1

(1.2i  0.35 j)
1.25
1

(24i  7 j)
25
M DB   DB  (rC/D  TCF )
where
Then
rC/D  (0.2 m) j  (0.4 m)k
M DB
24 7
0
1

(3) 0 0.2 0.4
25
6 9 2

or M DB  9.50 N  m
3
(9.6  16.8  86.4)
25
Problem #4 (25%)
Beam AD carries a 40-lb load and 300-lb-ft torque as shown. The beam is held by a fixed
support at D and by the cable BE that is attached to a counter weight W = 100 lb.
(1) Draw complete free-body-diagram for the beam AD.
(5)
(2) Write equilibrium equations for beam AD with all known and unknown forces and torques.
(10)
(3) Determine the magnitude and rotation direction of the unknown torque.
(5)
(4) Determine reactions at D in Cartesian vector form.
(5)
Solution:
(1)
(2)
∑ M = 0, ⇒ −M + 300 + ( 40 cos30) 4 = 0
∑ F = 0, ⇒ D − 40sin30 = 0 (2)
∑ F = 0, ⇒ D − 40 cos30 = 0 (3)
D
x
x
y
y
(3) From Eq.(1),
(1)
M D = 300 +138.56 = 438.56 lb ⋅ ft clockwise
(5) from Eqs (2) and (3):
Dx = 40sin30 = 20 lb
Dy = 40 cos30 = 34.64 lb
D = 20i + 34.64 j lb
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