POE 2.1-12 Statics Unit Review Answer Key
MULTIPLE CHOICE: Choose the best response to the question and write the letter on the line
provided.
C 1)
____
C
____
2)
The law of physics that states “For every action force there is an equal and opposite
reaction force” is known as Newton’s
A.
first law of motion
B.
C.
third law of motion
D. theory of Calculus
A vector is a quantity that possesses
A.
magnitude only
C. magnitude and direction
D
____
3)
opposite direction
C. differing amounts
____
5)
B
____
6)
A
B.
a turning motion
D. changing orientation
Forces always occur in pairs that have
A.
A
____
4)
second law of motion
B.
the same magnitude
D. both A and B
An object’s Modulus of Elasticity depends upon
A.
what the object is made of
B.
how far apart the supports are
C.
how much force is applied
D. the orientation of the object
In Statics, the Right-Hand Rule has to do with an object’s
A.
Moment of Inertia
B.
direction of rotation
C.
Modulus of Elasticity
D.
straight line motion
For moment calculations you must consider the distance to the pivot point that is
A.
perpendicular to the force
C. parallel to the force
B.
shortest
D. around the outside
Free Response: For each of the following problems, be sure to answer each question
completely and to show all of your work. All answers should be accurate to three decimal
places, and must include units. The following formulas are provided for you.

I
8
4
3
L
X
F A
M
Δ
7. Given the diagram at the right, calculate the area
and the centroid of each of the three simple
shapes, and then the area and centroid of the
complex shape as a whole. The location of each
centroid should be stated in relation to the given x
and y axes.
E

3 2
h
b1

I


i

i
y
A A
y

i


i
x
A A
x

0.75"
 
2.25"
23
0.75" y
1
1.50"
x
Area   base  height   1.50 in  3.00 in   4.50 in2
Shape 1:
1
1
base  offset   3.00 in   0 in  1.50 in
2
2
1
1
Centroid y  height  offset  1.50 in   0 in  0.75 in
2
2
Centroid x 
1
1
base  height    0.75 in 0.75 in  0.28125 in2
2
2
1
1
Centroid x  offset  base  0.75 in   0.75 in   0.50 in
3
3
1
1
Centroid y  height  offset   0.75 in  1.50 in  1.75 in
3
3
Area 
1
1
base  height    2.25 in 0.75 in  0.84375 in2
2
2
1
1
Centroid x  base  offset   2.25 in  0.75 in  1.50 in
3
3
1
1
Centroid y  height  offset   0.75 in  1.50 in  1.75 in
3
3
Area 
Area  5.625 in2
Centroid x 
Centroid y 
 
 Ax
i
A i
 
 Ax
A i
i
4.50 in2
Area = __________
1.50 in
x = __________
0.75 in
y = __________
Shape 2:
2
0.28125 in
Area = __________
0.50 in
x = __________
1.75 in
y = __________
Shape 3:
2
0.84375 in
Area = __________
1.50 in
x = __________
1.75 in
y = __________
Entire complex shape: Area = __________
5.625 in2
8.15625 in3

 1.45 in
5.625 in2

3
5.34375 in
 0.95 in
5.625 in2
1.45 in
x = __________
0.95 in
y = __________
8. A 2x4 board (whose actual dimensions are 1.5” x 3.5”) is positioned on the floor as shown
below. The distance between the supports is exactly 5 feet. When your 178 pound friend
stands on the board it deflects exactly 1.48”. Given this information, please calculate the
board’s Moment of Inertia and Modulus of Elasticity.
bh3  3.5 in 1.5 in 
11.8125 in4
I


 0.984 in4
12
12
12
3

Modulus of Elasticity = ______________
FBX = _________
b
l
0
0
1
b
l
0
0
1

FAY = _________
b
l
3
7
1
b
l
3
7
1
b
l
0
0
2
FBY = _________
b
l
0
1
b
l
0
1

ΣFX = _________
b
l
0
7
2
b
l
0
7
2



FAX = _________
b
l
0
7
1
b
l
0
7
1
b
l
0
0
2
0
3
n
i
s
 


 

b
l
3
8
1
b
l
0
5
2
0
3
s
o
c

Y

3
4
n
i
s
B
F
Y


b
l
0
0
1
b
l
0
7
1
A
F
X



b
l
3
7
1
b
l
3
8
1
A
F
FY

X
Fx


B
F





Y
 
A
F
0
3
n
i
s

X
 
B
F
b
b
Yl
Xl
B 0 B 0
0 F 0
F 2
2
0
3
s
o
c

  
 


Y
 
A
F
3
4
n
i
s




0
3
b
b
Yl
Xl
A 0 A 0
5 F 5
F 2
2
3
4
s
o
c
 
3
4
p j b
p d l
3
o a 8
1
n
a
t
b
l
0
j p 5
d
y 2
a h
n
i
s
s
o 3
c 4
s
o
c
p p
p y
X
o h A
F
vectors. Round your answers to the nearest
whole unit. Be sure that your calculator
mode is in degrees!




b
l
0
0
2

B

b
l
0
5
2
A

FR
9. Two cables pull on a shackle as indicated in
the diagram at the right. Please calculate the
x and y component vectors for both vector A
and vector B, and then calculate the resultant
force   by summing the x and y component

i
s
p
6
1
0
,
0
5
5

i
s
p
6
1
0
,
0
5
5

5
n
i
3
0
9
.
9
6

3
n
i
b
l
0
0
0
,
8
4
4
,
8
3
3

 
4
n
i
4
8
9
.
0
8
4
n
i
8
4
.
1
 


n
i
0
6
b
l
8
7
1
I
8
4
3
L
X
F A
M
Δ
E

4
0.984 in
Moment of Inertia = ______________
ΣFY = _________
Sense ____________________
b
l
0
7
2

b
l
0
7
2

2
b
l
0
7
2
2
Magnitude __________

w
c
8
8



 
8
8



1

0
7 0
2 1
n
a
t


b
l
0
1
e
d
u
t
i
n
g
a
m

t
f
e
l
e
h
t
o
t
d
n
a
p
u
What is the sense, magnitude, and direction above horizontal of the resultant force vector?

Direction above horizontal __________
10. Given the truss diagram at the right with
the pivot point at A, please calculate the
moments at joints B, C, and D. Then
determine reactionary forces at both
supports. Finally, determine the force for
each member, and use your answers to fill
in all the blanks in the diagram below.
G
E
10 ft
A
D
C
B
10 ft
Include units in all of your answers.
10 ft
1200 lb
10 ft
900 lb
Be sure to indicate if each moment is positive or negative.
MB   1200 lb 10 ft   12,000 ft lb
12,000 ft lb
Moment at Joint B = ____________
MC    900 lb  20 ft   18,000 ft lb
18,000 ft lb
Moment at Joint C = ____________
MD   MA  MB   30,000 ft lb
30,000 ft lb
Moment at Joint D = ____________
For each reactionary force, be sure to indicate the direction.
For the force on each member, be sure to include whether it is in tension (T) or
compression (C). You may leave your answer in radical (square root) form if you wish.
RFDY 
G
1100
2 lb C
_______
0 lb 
_______
A
1100 lb 
_______
1100 lb T
_______
1200 lb T
_______
B
1200 lb
MD
30,000 ft lb

 1,000 lb
30 ft
30 ft
1000
lb C
_______
E
100
2 lb C _______
1000 lb T
_______
1000
2 lb C
_______
1100 lb T
_______
1000
lb T
_______
C
900 lb
D
1000
lb 
_______