W
7
Ff
8218
7v(
p
F
ENGINEERING MECHANICS: STATICS
-
- -
N
1...
a
Fi
with
MECHANICS-deals
INTO
the
bodies
and
~Fo
systems
and
them.
on
TWO:
STATIUS-deals
1)
DYNAMICS
2)
of
study
acting
forces
the
&
NIDED
the
with bodies at
deals
-
with
DIVIDED
FURTHER
moving
INTO
KINEMATICS
9)
rest
equilibrium.
bodies.
TWO:
without
-
or in
reference to the
forces
causing the motion
KINETICS
3)
2.
QUANTITY
VECTOR
VECTOR
1
has
-
-
has
the forces
A
Addition:
magnitude
as
well
+ B
A-B
=
=
as
A
B+A
S
direction
B
7
*
=
G
A + 1 B)
-
Dot Product:
·
Cross
·
IN A
A.B= ABCOSO
Product:AXB
PLANE
=
antB sint
CCARTESIANS
J
The
=
A
--------
=
iAx
AX
(Al
jAy
+
i
*E.......
to add
The
i
8:
A and B
d raw them
to
tail.
Multiplication:
VECTOR
causing
magnitude only
OPERATIONS
subtraction:
3.
to
QUANTITIES
1)SCALAR QUANTITY
2)
With reference
motion
the
↑
AYSICAL
-
magnitude of
=
vector A
NFAg
angle (8)
with the
tan")
horizontal
head
VECTOR
IN
SPACE
CCARTESIAN)
2
=
........
A jAx
=
+
jAy
+
kAz
The
(A)
Az
-
magnitude
of Vector A
AEE
rAy"
=
+
X
Note:
old letters
B
represent
a
vector
↓directional unit vector
along
x-axis
-directional unit
along
y-axis
J
K-directional
unit rector
Ax. magnitude of
Ay- magnitude
Az-magnitude
ORCE AND
#
Force
-
FORCE
any
vector
vector
along z-axis
in the x-axis
y-axis
vector
in
the
of vector
in
the z-axis
of
SYSTEMS
influence
that
can
change
colinearturned-actorte samenot
the state of the
resea
a
....
F
.
body
couple-pair
but
of
forces
parallel
the
of
same
magnitude
in direction.
opposite
=.......
-
·
Frictional Force
-
always acts
in
opposition
with
the
applied force
Y
friction
L
·
Coplanar Forces
-
lying
on
the
same
plane
"Fo
-
-
-
-
-
,
,
x
·
Concurrent Forces
-
meet
in
one
,
,
common
........
·
Forceis
Non-concurrent
do not
meet
EEEEEE
....
↑
F
Tr
,
,
point
,
y
-
FAN
I
Fa
61114H14
coplanar concurrent forces
in
one
common
point
non-coplanar non-concurrent forces
THE
RESULTANT
Resultant
-
the
as
Equilibrant
R
-
TWO
single force
of
the
the
single force
the
resultant
that
tanz
is of
the
the
one
same
same
body.
magnitude
effect
of
in direction
IFx=
ZFy
E
=
have
in
acting
where:
+
would
opposite
but
x(z
y)3
=
which
forces
many
COPLANAR FORIES
CONCURRENT
OR MORE
=
forces right-forces
forces upward
-
left
forces downward
EQUILIBRIUM CONDITION
Static
Equilibrium
when
-
of the
CONDITIONS
·
1.
FOR
Graphical
it will
at rest
IN
BODIES
Condition:
equilibrium
in
body
forces
The
The
resultant
is zero.
(vectors)
closed
a
polygon
Fa
Fu
T
v
Directional condition:
rest.
EQUILIBRIUM
↑
2.
remain at
F
The forces
(vectors) meet in
one
common
=......E
LAMI's
THEOREM
"When three forces
equilibrium,
then
acting
each
at
force
a
point
are
in
is proportional
the sine of the angle between the other
*
two forces
"
==
point.
3.Analytical
condition:
Forces (vectors)
must satisfy
three given
the
conditions.
-EXEII
]
-
<
~
-
v
I
L
3
FRICTION
Friction
a
-
bodies
two
Friction
Static
motion
opposing the
force
-
impending
or
value
Dynamic
or
force
a
Rolling
called
between
starting
Kinetic friction
motion
Friction
between
-
two stationary surfaces in
-
them.
Its
RICTION
ON A
occurs
when
there
is
relative
when
the another surface
one
surface
but does not slip
rotates
or
as
BLOCK
.
t
-...
v
" on
overs
r
N
it
slide at the
W
P
(sliding
the surfaces in contact
occurs
"
maximum
friction.
contact
#
of
in contact
contact that prevents motion between
is
motion
moves
point
over
of
Ff
de
UN
=
tant
=
note:
or
F
where:
coefficient
n
=
#=
frictional force
N
normal force
=
frictions
of friction
of friction
= angle
8
applied
↑=
Ms
=
the following will be
coefficient of static
friction
Md coefficient of dynamic
Kinetic friction.
=
force
and
ls> And
BELT FRICTION
H
=
erp
"Go""<i
in
where;
&= angle
4
=
T
M
of contact
tension
=
=
&
OLLING
tension
(in
radians)
(fight side)
(slack side)
coefficient
of
friction
FRICTION CRESISTANCES
W
-
·Er
p
=
FaT
where:
a=
coefficient
length)
of
used:
reaction
R=
specific coefficient
of
rolling
resistance
(a
dimension
of
or
CABLES
↑
ARABOLI2 CABLES
The cable
parabolic
is
and
horizontally
the
if the
span-to-sag
uniformly
load is
ratio
greater
is
than
>10
L
K
<)
W
T
EEEFF
TI
8
Symmetrical
supports: (Parabolic cable)
(T) at
Tension
The
T
the
supports:
+H*
N2
=
tension
The
H
at
the
lowest point:
t
=
he
t
length
of
the
cable
·Approximate Formula
s L +
=
od--2d
Exact Formula
·
s
mz
ETmy-
=
+
(n)m +
vmt)]
where.
m=
O
H
O
For
W
~
+, tant-
distributed
T
10
or
For
unsymmetrical
Supports
(Parabolic
curve)
A
B
~
-
T
r
#
S1
d
> H
S1
-..........
-
-
K
tension
TH
at
Length
=
-
-
-
-
supports:
the lowest
i
=
S
the
UH2
=
Tension
H
at
dz
Se
2
TB
=
+H2
M
point:
H
i
=
where;
tension
↑
at the
support
H=
of the cable:
S,+ Se
Intensity at
point
the lowest
W intensity of the load
=
d
L
=
=
sag
span
or
between
distance
supports
CATENARY
The
cable
is
of the cable
or
a
catenary
and the
if the
uniformly
load is
is
span-to-say
distributed
lesser than
equal to 10
L
d
&
r
3.
O
-
*
-0
d
x
B
O
r
Turnerforfrie
Y
along
the
length
For
·
symmetrical supports (catenary)
point
lowest
T
Half
H
wy
=
·
and the
supports (T)
(H)
the
at
Tension
intensity
WC
=
of the cable
Length
ge yz- c2
=
·
between supports
Distance
L 2X
=
Note:
y c cosh(E)
x
For
s
=
=
osinn
(E)
cn))
=
unsymmetrical supports (Catenary)
L
d
&
-
*I
↓
x2
16
B
O
T
scrivvivrrit
O
be
5,
·
Tension at the
Y
Length of
·
(x)
ST
·
=
Distance
x,
L
=
the
Supports
T
=
WY,
wye
cable
(y)--ch
(S2)"
=
5,+32
between
cIn)()
x, + XG
=
=
=
supports
xz
yz)
cIn(
=
(y)" c
-
at
the
Note;
ccosh(1)
yz
y=
s
csinn
=
(t)
sz
ccosh/2)
=
csink()
=
where;
↑tension
weight
w
=
&span
#-tension at the
support
of the cable per unit
or
point
lowest
length
y -height
clearance
minimum
c=
the
at
distance between supports
S:
of the
support
half length of the
cable
CENTROID
centroid
center of
or
gravity
-
point
weight
the
where
of the
body
is
concentrated.
3
---ar: oAnON-ArE E
ETF
y
V
.
Centroid
·
Wx
of volume.
V,x,
=
VaXe +VgYy
+
Centroid of Area:
·
·
-........
Y2
I
&
I
-x
r
A
x
E
=
A,x1
=
centroid
·
AzX2+AgXy
+
of Line:
L= LX +
Lexz+23X3
x
X3
MOMENT AND MOMENT OF INERTIA
Inertia
-
natural
o in
Moment
or
tendency
motion,
forque
-
to
the
of
an
object
continue
cross
perpendicular distance
to
moving
product
to
which
remain
at
of
at
rest
constant
force
at rest
speed.
and the
force
the
when
is
applied.
7
p
...
~F
M
=
FXr
where:
moment
M
=
or
torque
=applied force
F
r moment
=
arm
or
perpendicular distance
Moment of Inertia
or
tend to
volume
of the
due to the
rotate
elements
mass
or
tendency
natural
-
body
distribution
body.
of the
rotate
to
of area,
Also
know
the second moment-
as
PARALLEL-AXIS
of
moment
The
ITRANSFER AXIS INERTIA)
THEOREM
is
equal
to
inertia
with
respect
axis
to
parallel
and
the
#x
the
and
square
parallel
two
the
it,
of the
inertia
sum
to
the
of the
the
the
at
of
a
moment
centroidal
product
the
-
certain
of
axis
area
between
shortest distance
axes.
.............xo
Exo+ Ad
=
of
body
d
----------------.-
X
where:
Exo
A
centroidal
=
=
of inertia
moment
area
& =distance
Xo: centroidal
AND
CENTROID
or
MOMENT OF
neutral axis
INERTIA
OF COMMON
↓ TRIANOLE
Ex
=
n
Fx0-th
y
(about
(about NA)
r
n
C
O
....---------------
r
(centroid)
=
A
GEMETRIC FIGURES
gz................-
NA
x
2.RECTANOLE
Ex
h
=
(aboub)
Exo-he
(about NH
=
(centroid)
I
------.0--------th
r
-
·---
b--.-E
circle
3
Fx0-Ed
Fx0
=
(about
NA)
(about
NAS
9
fr
NA
---------------
No
..............
X
Ellipse
4.
Hab
I
OF
COMMON
sphere
raxis
...S
Emr2
=
GEOMETRIC SHAPES
Thin-walled hollow sphere
I
3.
solid
Homogenous
2.
↳
............al-...- x0
MOMENT OF INERTIA
MASS
1.
r
=
as
EMr
2
cylinder
=
Imr2
4.
Hollow
I
~
St,.........
.............
axis
~
cylinder
M(R
=
=
ax
+
rY
~
v
=
axis
Right
I:
cone
circular
5.
mrr
~
&
slender rod,
#=
8.
rod,
................
axis
through
one
mE
=
+
EXAMPLE
1. What
axis
through
b2)
center
axis
~
a
y
is
the
3
cross
+ 5K
Solution:
t]
=
...
i
PROBLEMS
2i +
B
=
AxB
end
2
m(az
=
axis
L
Rectangular plate,
I
-
through center
m(2
slender
7.
I
axis
StE........ax
product
Axt
of the
vectors,
A
=
i
+
4
+ 6K
and
AXB
i(4X5 4x3) )(X5 2xx) + k)(X3 2X4)
i(20 18) (5 12) + 17)3 8)
=
-
-
-
-
-
-
-
-
-
EFF-517,1
forces
2. Two
go
magnitude
units
of the
and 40 units
at
right angles.
What is the
resultant?
Solution:
7
R
R
30
3.
five
The
forces
magnitude
units
I
40
shown
402
=
act at
of the resultant
point
0.
What
is
force?
CON
-SON
-
30303
4ON
-30N
30
300
Solution:
R
11
*
- LON
A+(2Fy)
=
zFx
=
20 +
3010530+
40cos60"-60s1n30'
35.98
=>
Fy
=
=
=
1
30s1n30+40s1n60"
+ 50
+
6010s400
157.60
+ (151.60)2
05
=
12
=
155.8N
A
Herrate solution:
+ 30130 + 40X60
+ 50x900
120100
R
=
E8N/
+
6011200
the
4.
GOON
block
rests
is
0.20, what
to
keep the
is
300
on
the
block
value
moving
of the force
up the
Solution:
N
600N
3001346
300sim35
3
Prosporin...
of friction
horizontally
applied
P
PS In 300 + 000 cos300
=
=
UN
cos300
↑
#f
coefficient
plane?
If
a
*
=
600sIn30" +
↑LOS300: 600514360
+
n 0.20
=
Ff
0.20/PS1380 + 600 COS380)
PCOS30: 600sIn200 +0.20PSIN300
120205300
no
is
If the
plane.
+
p
=
600s1n300
COS300
+ 120
-
3N
Cos300
0.20 s1n 300
=
527N1
*
5.
An
archer
arms
arrow
with
his
angle
string.
solution:
of
of 25
his
bow
upward
and
pulls
the
of
string
force
80N, the string
with the vertical. Find the tension
degrees
of
a
makes
an
of the
uninter
50
35
-
155box
=
T=
44N
-
3.
object
An
horizontal
a
weighing
drum.
and the drum
the force
400N
The
0.25.
is
raise
that will
held
is
by
a
rope
coefficient of friction
angle
If the
the
of
that passes
between the
contact
1500,
is
over
rope
determine
object.
solution:
1500
T
T,eM8,Tz
=
7
-
F
L
F
=
=
400N
u 0.25
400N
=
F
400210.25)/1508550
=
F 769.47N
=
#ON
11
7.A
in
at 1.4N
central horizontal force
diameter
rolling
on
resistance
a
is
level
is necessary
Assuming that
surface.
0.6y5rm,
solution:
to
is the
what
move
the
mass
drum 900mm
a
coefficient
of
the
of
drum.
W
p
1.4N
=
*
Er 100
=
-
0
FaT
a 0.635mm
=
P
=
Na+
w mg
=
m
-
t
g
=
m
45or
R)
=
k
to
,
=
450mm
A
say
of not
load
of
cable
be?
more
in
tension
allowable
8. The maximum
loft
than
to
is
foot.
10016 per horizontal
cable
a
3000lbs.
is
allowed
be
should
long
How
under
a
the
solution:
Since
·
the
then
load
the
w
-
- -
is
cable
L
K
uniformly distributed horizontally
parabolic.
<)
100th
=
is
T
Er Y
O
=
+
30002
Let
3000 lbs
T
1
He,
=
H
-
(ot(ol
=
L
=
X, shift solve
1
=
41.59 ft
SELtdtoaddto
47.11 ft
3
*
=
weighing 0.5 loft is stretched between two level supports
130ft apart. If the say of the cable is soft, determine the
9. Awive
&aximum
tension.
solution:
·
so
it
O
is
-=
d
&
load is
The
a
distributed throughout
c
160ft
80ft
x
-0
d
=
y
80f
B
=
4oft
y-
=
r
tuvvvvviveinte
Y
ccosh(()
y (y
y (y
=
Let y
y
d
=
=
o
33
3
length
of the
cable
catenary.
-
=
x
the
=
=
-
-
a)
d)cos(
40)
cosh
(
40)
80
X, shift solve,
125.945 ft
zw)(125.943+)
T
=
62.97
Tx63/b
-
10.Anisoscelestrianglehas a 10cmbase and a10cm altitudeorientative
to
line
a
parallel
in
vertex
the
to
base
and
through
the upper
am*
solution:
-A
Eh
-
----...-r-- X
10
d
A
-..........
ViTn
10cm
Ix
+x
Ex
=
Exo+ Adh
u (Eb)(En)+
=
=
(- (1)(10)(5(10))
+
=500cm"
1
EFERENCES:
R
1.
2001
2.
ROJAS
SOLVED
PROBLEMS
INESAS
BY
TIONG,
ROJAS
AND CAPOTE
POCKETBOOK CREP)
BY: ENOR. JAYMEL HILOMA