### review

```Exam 1, Fall 2014
CE 2200
Section 1…. FORCE VECTORS
1.1 F2 = 70 lb. What is the magnitude of
the resultant force acting on the bracket?
(5 pts)
A. 111 lb
B. 121 lb
C. 131 lb
D. 141 lb
E. 151 lb
1.2 F1 = 675 N. What is the i component of F1 ?
(5 pts)
A. 416 N
B. 421 N
C. 433 N
D. 450 N
E. 468 N
(4/5)(675)cos30 = 468 N
1.3 What is the k component of F1 ? (5 pts)
A. 405 N
B. 390 N
C. 382 N
D. 375 N
E. 360 N
(3/5)(675) = 405 N
1.4 A force is expressed as a Cartesian vector F = {35 i – 62 j + 87 k} N.
What is the coordinate direction angle beta β of the force? (5 pts)
A. 109º
B. 114º
C. 119º
D. 123º
E. 128º
β=



62
cos-1 

2
2
2
 35  62  87 
=123°
1.5 F = {2i + 5j + 10k} kN. What is the
angle θ between the force F and the pole?
(5 pts)
A. 77.6°
B. 80.3°
C. 83.3°
D. 86.5°
E. 88.8°
F=
2  5  10
2
2
2
rOA = (2i + 2j – 1k) m
= 11.36 kN
rOA =
2  2 1
2
2
2
=3m
ϴ = cos-1 [F ∙ r / F(r) ] = cos-1 4/34.08 = 83.3°
Section 2…. Particle Equilibrium
2.1 (5 pts) If the bucket weighs 90 lb,
determine the tension developed in wire ED.
A. 42.3 lb
B. 30.2 lb
C. 54.4 lb
D. 66.5 lb
E. 75.8 lb
+
F
x
0
FED cos30 - 3/5 (FEB) = 0
+
F
y
EB
ED
0
FED sin30 + 4/5 (FEB) -90 = 0
FED = 54.39 lb FEB = 78.5 lb
FBD of E
90 lb
2.2 (5 pts) Knowing the weight of the bucket and assuming no other unknowns
have been solved for previously, what is the minimum number of scalar
equations that are required to solve for the tension developed in cable BA? (Fill in
the blank on the cover answer sheet)
4 Equations…
ƩFx and ƩFy @ E
then
ƩFx and ƩFy @ B
2.3 (5 pts) If the particle shown is in
equilibrium and F = 1000 N, determine
the magnitude of F1.
A. 900 N
B. 1200 N
C. 1000 N
D. 1100 N
E. 800 N
ƩFz =0
-1000cos30+F1sin60 = 0
F1 = 1000 N
2.4 (5 pts) What is the magnitude of the resultant force of the particle
system shown above? (Fill in the blank on the cover answer sheet)
ZERO – Particle is at EQUILBRIUM
2.5 (5 pts) For the same system shown
above, how many scalar equations can be
written and what is the maximum number
of unknowns that can be solved?
A. 3 unknowns, 3 equations
B. 2 unknowns, 4 equations
C. 9 unknowns, 6 equations
D. 4 unknowns, 4 equations
E. none of the above
3 Equations… 3 Unknowns
Sum forces
ƩFx = 0 ƩFY = 0 ƩFZ = 0
Section 3…. Moments
3.1 (5 pts) If F = 60 N, determine the magnitude of
the equivalent resultant moment acting at point O.
A.
B.
C.
D.
E.
465.9 N∙m
531.9 N∙m
399.9 N∙m
333.9 N∙m
672.2 N∙m
åM
= MR
MR = – 50(2) +20(3sin30) – 60(4+3cos30)
MR = – 465.885 N-m
+
O
3.2 (5 pts) The direction of the resultant moment calculated above is:
A. clockwise
B. counterclockwise
C. unable to be determined
3.3 (5 pts) The x-component of F2 shown
A. the x-axis only
B. the y-axis only
C. the z-axis only
D. the x and y axes
E. the x and z axes
F. the y and z axes
G. all axes
H. none of the above
F2X :
F2Y
F2X
F2Z
- is parallel to the x-axis and cannot create a moment
- is not parallel to either the y or z axes.
- has a line of action which does not pass through the
y or z axes.
3.4 (5 pts) It is desired to calculate the moment
due to the force about point O. Determine the
position vector(s) that could be used in the
moment cross product calculation (r x F).
A. rOB
B. rBO
C. rOC
D. rCO
E. rAB
F. rBA
G. rAC
H. rCA
I. A. and C. only
J. E. and G. only
K. B. and D. only rOC
L. F. and H. only
The position vector used MUST:
Go FROM the point of interest TO a point on the line of
action of the force.
rOB
3.5 (5 pts) The cutting tool on the lathe exerts
a force F on the shaft as shown. When finding
the moment due to F about the x-axis using
the triple scalar product u∙(r x F), determine
the vector u.
A. [6, -4, -7]
B. [1, 0, 0]
C. [0.597, -0.398, -0.697]
D. [0, 1, 0]
E. [23.0, 0, 19.3]
F. [0, 0, 1]
r
u
MX
MO
1) MO = r x F (lies perpendicular to the plane incl. r & F)
2) MX = MO u
u represents the DIRECTION of the vector component
Section 4 … Couples, Equivalent
4.1 Determine the sum of the moments
exerted on the pipe by the two couples shown.
The magnitude of F = 20 N and the magnitude
of P = 30 N. (5 pts)
A. {40 i + 60 j – 103.9 k} N∙m
B. {60 i + 40 j – 69.3 k} N∙m
C. {– 40 i – 60 j + 103.9 k} N∙m
D. {– 60 i – 40 j + 69.3 k} N∙m
åM
åM
åM
X
: MRX = – (20)(2) = – 40 N-m
Y
: MRY = (30cos60)(4) – (30cos60)(8) = – 60 N-m
MRZ = – (30sin60)(4) + (30sin60)(8) = 103.9 N-m
:
Z
MR = { – 40i – 60j + 103.9k } N-m
Replace the force and couple system acting on
the beam by an equivalent force and couple
system at point O. (8 pts)
F1 = 30 kN
a=3m
F2 = 40 kN
b=2m
M = 210 kN-m
c=5m
4.2 Express the resultant equivalent force in
Cartesian vector format.
A. {34.6 i + 50.0 j} kN
B. {50.0 i + 34.6 j} kN
C. {– 34.6 i – 50.0 j} kN
D. {– 50.0 i – 34.6 j} kN
+ ® å Fx : FRX = 40cos30 = 34.64 kN
+ ­ å Fy : FRY = 30 + 40sin30 = 50.0 kN
Replace the force and couple system acting on
the beam by an equivalent force and couple
system at point O. (8 pts)
F1 = 30 kN
a=3m
F2 = 40 kN
b=2m
M = 210 kN-m
c=5m
4.3 Express the resultant moment in Cartesian
vector format.
A. {400 k} kN-m
B. {2290 k} kN-m
C. {– 400 k} kN-m
D. {– 2290 k} kN-m
+
åM
O
= MR
MR = (30)(3) + (40sin30)(5) + 210 = 400 kN-m
4.4 Replace the force and couple system acting
on the beam by only a single equivalent force
and find the location, x, where this force acts on
the beam measured (in meters) from point O.
(Fill in the blank on the cover answer sheet)
(5 pts)
400 kN-m = (50.0 kN)(x)
x = 8.0 m
equivalent resultant force. If a = 3 ft, b = 2 ft
and z = 600 lb/ft determine the magnitude of
that force. (5 pts)
A. 11,280 lb
B. 4,720 lb
C. 3,720 lb
D. 6,240 lb
E. 1,480 lb
F. 2,560 lb
åF
F
=
(1/2)(480)(3)
+
(1/2)(600)(6)
+
(600)(2)
:
R
y
FR = 720 + 1800 + 1200
FR = 3720 lb
```