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Unformatted text preview: 
 
 
 
 
 
 2.5
hour
exam.
Closed
Book
 Write
only
on
these
pages
 
 Mid­Term
Exam
#
1
 Date
 Sat
06
Feb
2010
 Name
 S
O
L
U
T
I
O
N
 ID
 J.
Salinas
 Section
 B
 Braimah
 C
 Zalok
 D
 Salinas
 E
 Majeed
 Thur AM E1: 4332 ME E2: 4494 ME E3: B149 LA E4: 501 SA
 Fri AM Lab
 B1: 4342 ME (Circle
 B2: 4332 ME one)
 B3: 3165 ME B4: 4494 ME
 Mon PM Tue AM C1: 516 SA D1: 4332 ME C2: 180 UC D2: 4494 ME C3: 317 SA D3: 502 SA C4: 4342 ME
 D4: B146 LA
 Instructions
 
 1. Attempt
all
questions
 2. Work
in
the
space
provided
 3. You
may
use
the
back
of
these
pages
 4. If
you
feel
that
not
enough
information
has
 been
provided,
make
an
assumption,
write
 it
down
and
proceed
with
your
solution
 
 
 
 
 
 
 
 Problem
 1
(20)
 2
(20)
 3
(20)
 4
(20)
 5
(20)
 
 
 
 
 
 Mark
 Total
 
 
 Problem
1
(20
marks)

 For

 A = 12 i + 5 j + 2 k ; B = 3 i − 2 j + 0 k ; C = 4 i +1 j - 9 k 
 (a) Calculate

 A × 
 B (b) Calculate

 B • C 
 (c) Calculate

 A × B • C 
 € 
 12 3 4 € A = 5 € B = −2; C = 1 ; 2 0 −9 12 3 5(0) − 2(−2) 4 (a) A × B = 5 × −2 = 2( 3) − 12(0) = 6 2 0 12(−2) − 5( 3) −39 
 3 4 (b) B • C = −2 • 1 = 3( 4 ) + (−2)1 + 0(−9) = 10 0 −9 12 3 4 4 4 (c) A × B • C = 5 × −2 • 1 = 6 • 1 = 373 2 0 −9 −39 −9 
 
 € SolutionMT1a.docx
 
 






Page 1 of 6
 € Problem
2

(20
Marks)

 Resolve
the
horizontal
600
lb
force
into
components
acting
along
the
v
and
u
axes
and
determine
 the
magnitude
of
these
components.
 u
 
 
 
 Fu
 
 
 
 600
 
 o
 120 30o
 
 
 Fv
 
 30o
 Fu
 
 
 
 
 v
 
 
 
 
 Sine Law 600 Fu Fv = = Sin 30 Sin 120 Sin 30 Fv = 600 Sin120 = 1039 lb Sin 30 Sin 30 = 600 lb Sin 30 Fv = 600 € SolutionMT1a.docx
 
 






Page 2 of 6
 Problem
3
(20
Marks)
 Determine
the
magnitude
and
direction
of
the
resultant
force
acting
on
the
pipe
assembly
 
 R = F + F2 1 
 Fx ( 4 )600 480 5 F = Fy = 0 = 0 1 3 Fz ( 5 )600 360 cos 60 1 2 1 1 F2 = 400 y = y → ( 2 )2 + y2 + (− 2 )2 = 1 1 cos 120 − 2 y= 1 2 
 
 1 200 2 F2 = 400 1 = 283 21 − 2 −200 
 
 
 
 
 
 
 
 
 
 
 
 
 
 480 200 680 R = 0 + 283 = 283 lb → R = 680 2 + 2832 + 160 2 = 754 lb Magnitude € 360 −200 160 680 0.902 R 1 uR = = 283 = 0.375 R 754 160 0.212 Direction € SolutionMT1a.docx
 
 






Page 3 of 6
 Problem
4
(20
marks)
 The
150‐lb
crate
is
supported
by
cables
AB,
AC
and
AD.
Determine
the
tension
in
these
cables.
 TAB + TAC + TAD + W = 0 −6 3 RAB T 2 TAB = TABuAB = TAB = TAB = AB 2 2 2 RAB 7 (−6) + 3 + 2 −6 3 2 
 Coordinates
 A
(6,0,0)
 B
(0,3,2)
 C
(
0,
‐2,
3)
 D
(x,
0,
0)
 TAC TAD 
 
 
 
 −6 T = TAC = AC −2 7 (−6) 2 + (−2) 2 + 32 3 1 0 = TAD 0 W = 0 0 -150 −6 −2 3 € Scalar Equations −6 −6 1 0 0 T TAB 3 + AC −2 + TAD 0 + 0 = 0 7 7 2 3 0 -150 0 (1) − 6 TAB − 6 TAC + TAD = 0 7 7 3 (2) + 7 TAB − 2 TAC + 0 = 0 7 3 (3) + 2 TAB + 7 TAC + 0 = 150 7 3 From (2) TAC = 2 TAB and substitute this into (3) then (3) → 2 TAB + 7 33 72 ( )T AB + 0 = 150; from here 13 14 TAB = 150 and TAB = 162 lb TAC = 242 lb TAD = 346 lb 3 3 From TAC = 2 TAB we can now solve for TAC = 2 (162) → From (1) − 6 (162) − 6 (242) + TAD = 0 and 7 7 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 € SolutionMT1a.docx
 
 






Page 4 of 6
 Problem
5
(20
Marks)
 Replace
the
force
and
couple
moment
system
acting
on
the
overhang
beam
by
a
resultant
force
 alone,
and
specify
its
location
along
AB,
measured
from
A.
 Horizontal Force Components 
 
 −30 sin30 = −15 ← 
 5 + 13 26 = +10 → 
 
 Vertical Force Components 
 −30 cos 30 = − 26 ↓ 
 − 12 26 = −24 ↓ 
 13 
 Resultant Force 
 R x −15 +10 −5 
 FR = R = = = kN 24
€ 
 R y −26 − 24 −50 26
 
 45
 15
 10
 R = (-5) 2 + (−50) 2 = 50.3 kN 
 P
 
 
 R 
 −50 o B
 
 θ = tan -1 y = tan -1 = 84.3 R −5 x 
 x
 
 
 We
need
to
move
all
force
components
and
existing
moments
to
a
point,
P,
located
at
a
distance
x
 € from
point
A,
along
the
axis
of
the
beam
(0.3
m
from
the
top)
so
that
the
total
moment
is
zero.
 Moment
contribution
from
moving
forces:
Positive:
CCW,
Negative:
CW
 15(0.3)
‐
10(0.3)
+
26(x‐2)
‐
24(6‐x)
=
50x
–
194.5
 Existing
moment
=
‐
45
 Total
moment
=
50x
–
194.5‐
45
=
50x
‐239.5
=
0
 Solving
 x
=
4.79
m
 
 
 
 
 
 
 
 () () SolutionMT1a.docx
 
 






Page 5 of 6
 
 Mid
Term
1
Marking
guidelines
 Problem
1
 (a) 4
marks
 (b) 4
marks
 (c) 4
marks
 (d) 8
marks
 Correct
answer
=
full
marks










Wrong
answer
(including
sign)
=
0
marks
 
 Problem
2
 Sketch
=
8
marks
(perfect)
 Sine/Cosine
law/Other
Equations
=
6
marks
 Answer
(arithmetic)=
6
marks
 Missing
units
deduct
2
marks
 Wrong
significant
figures
deduct
2
marks
 
 Problem
3
(No
extra
drawing
needed)
 F1

Correct
3D
Cartesian
representation
3
MARKS
 F2

Determination
of
dir
cosine
for
y
component
4
MARKS
 F2

Correct
3D
Cartesian
representation
3
MARKS
 Force
Resultant
:

 Add
components,
2
MARKS
 calculate
magnitude
4
MARKS
 determine
direction
cosines
vector
4
MARKS
 Missing
units
deduct
2
marks
 Wrong
significant
figures
deduct
2
marks
 
 Problem
4
 TAB

Correct
3D
Cartesian
representation
3
MARKS
(position/unit
vectors)
 TAC

Correct
3D
Cartesian
representation
3
MARKS
(position/unit
vectors)
 TAD

Correct
3D
Cartesian
representation
2
MARKS
 Vector
and
scalar
equations
of
equilibrium
 Formulation/representation


4
MARKS
 Solution
8
MARKS
 Missing
units
deduct
2
marks
 Wrong
significant
figures
deduct
2
marks
 
 Problem
5
 Decompose
forces
into
Horizontal/Vertical

3
MARKS
 Calculate
resultant
Force
 Magnitude
3
MARKS
 Direction
3
MARKS
 Selection
of
point
to
move
forces
+
Moments
2
MARKS
 Contribution
of
moved
forces
to
moment
resultant
3
MARKS
 Contribution
of
existing
moment
1
MARKS
 Total
moment
resultant
2
MARKS
 Solving
for
location
of
point
of
zero
moment
3
MARKS
 Missing
units
deduct
2
marks
 Wrong
significant
figures
deduct
2
marks
 SolutionMT1a.docx
 
 






Page 6 of 6
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This note was uploaded on 07/17/2010 for the course ECOR 1001 taught by Professor Slinias during the Spring '10 term at Carleton CA.

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