<Force
EA Review sheet
<Unit conversion]
=
US
SI
F=Ea+Eparallel
·
mass
(kg)
slugs
force
(N)
(b)
magnitude
ft.1b
5280ft
smile
& perpendicular
o
(N.m)
energy
=
1609.34m
=
<Vector>
3.281ft(1fT 0,3048m)
=
↑16
1ft
IN/m2
4.448N
Direction
=
0.4535kg
e 2xi eyy
+
xz
=
0.0209/b/fTz
Idleia
CX
6 6.67x10
8
9.81m(s
32,2
x-axis
parallel
↑
C0S8F 2y
=
By
f/52
extention
e
COS8z ez
=
B
·
·
Units
=
EAs, given
to
↓
+
re
x
+
Fre
JEec (Fr.EnB) Een
S
I.e
·
1
=
atthe end of vectors, coordinates
magnitude
always
FF Ep
=
<87
·
VAP:
with
ABLAC, ABLAD
Determining the vector componentofF
·
20SOX ex
ex
"N.m2/kg2
=
the
<cosox:
-
=
plane containing
e
and
/82
=
k
ez
+
Thinkaboutox and the
souce
0.3048m
12in
=
cosine
=
=
z
=
"
-
=
11b
faxcollinear
Vector
·
10.76 fT2
=
1m
->
"Always"take Sgrt [/7)
=
1m
P
Es lEGB.
·
of force
express
with
ex.
F1
N
=
-
Vector
Parallel
components
Perpendicular to line
and
at
1, j
L
getparallel
To
Up
use
-
E.E
/
ndpT
(/elcs
=
(E
=
.
e.
wh
ment
cost
=
parallel on
a
multiply by
to getvector,
Ep
component,
dotproduct
e)e
↑
paellel
(normalvectorone
perpendicular
E
Op (.e).e
=
↳
In
On 0 rp
-
=
B WA
Ep
=
IFX Tcos(45-307
=
C1
+w
=
=
D
D
T
T
④
EFy =Ts.n (4530)
h3
=
+
w+
=
=
-100 Sn30
Ex)
-
100c0s0
B(2,8.6)
Find
18
components of E parallel
the
g(mA 7n)
·
EAB
=
)
Fix
A
W
Nc (N.)
=
NS
=F
w
=
Ep
-
0i
(103225i+sin2255)
Nc Ns F
+
+
+
0
=
0.8323 0.55
=
+
+
0.832.12
+
(6.654N)(0.832;
0.555.-6]
0.55543
+
Fp (5.545 3.094)N
=
EN
F
=
+
Ep
-
10i
12y
=
EN ((0i
=
&Fx, Fz
[0.10
+
6.463
-
+
-
64
16
+
+
(ens.E) IAP
=
(Ns(C0570-t Sin7.j)
-
+
+
=
=
=
18 2);
6y 44
( 2 2)"
=
=
zp
BTN
the line
A(2,2,2)m
B
a
to
X
>
I
+
perpendicular
E-TBCBG S
-
0
=
and
-
Ep
9.69k)N
eAB
2)
-
through
AB
m
Scalar
E. (E XW)
①
#x)
a)
vvev
5Fy
=
ris
volume parallelpiped
a
unitrector
thati s
perpendicular
minimum
pointAt o the
from
line
l
ioIsY
I
rExcor:
(6
=
-
-
e
+
+
160
+
12)k
+
INecoPacNNcCOSIN
4501N/
Sin
1Ne)
=
sin (x 45)
+
B(6,6,
-
3)
12(W)
=
25i(9
↓
X
z
483 72k(m"
45)(b)
+
xx
!A
((0, -2,3)m
+
+
0
=
X
o
18)4 (18 30)5
127
ina(N1
=
1
6
=
zEz
O
↑
as
=
(N2)
,
distance
0
=
INilsinx (N2/sin 45
·
to
OB,
the
Whatis
of
of
Determine the components
both OAand
b)
cm
(as
o.cix
1I
=
e
50(b
=
Product
Triple
(b)
-w
zFy
0
=
IN_lsin (x 45) 1W1
Sin45
=
+
FoAxFoR:-12i +485
I FonxfoB(
277
+
0.1373 0.5495
=
+
-
87.36
+
0.824%
e
=
rivia
*
72
18+
1
take the
magnitude of
Ion xFobl
no
(a)
Fx
Exz
Egn
F
2
EpE
18) 105) (2)
=
-
=
=
0.212i + 0.9545
-
0.21247
0
=
N W 0
=
+
+
W
Vector
parallel
=by
dIfob)
the
to
slope
(65) (8) (0)
=
-
=
carcels
foB)
36m
=
=
In,
,
Foo)EpETNNINFOnora
=
Find friction
FDECEso.ExE>=
force;
Ar
jy6
↑en
P
-
=I+N+w
N
N
0
-
(VbAX B)
en.(I
A w)
0
=
+
+
InN inw
+
0
=
En.SNenCtEnI
zc
2034580360K)
V (r) (Sn45: (06450.60
I
+
+
E.Ez-zismovinet
o
v 27
=
+
L
NA
0
nb
=
f
(f)
=
2 (0.3)2
5
=
+
jbT
<
=
-
F+
+
:10,1
the
-robineen n]
a
·
FBAX FB2
=
Bn +B e
=
=
=
In
>4
-niiNRoen
E
I
1
solar
ene
N NIn
-
=
EDE
FDE
=
=
=
IDE
·
(d1fobl el
A
o n y -z
is
plane,
components.
interms of
10(
=
FPE
+
(b)
the vector
,
=
m
Find:IFE)
I
both sides
100 |
Express
E 4p (13)
(Ne 1
3)
=
=
=
Point
+
W
dIfob()1
a)
(N21+ IN11COS (1 480) 1N/203450
=
=
d
Fx 0
I
=
87.36
=
-1roAlropsitoa
T
rxrop
3)
+xsy
mag
qT
&mg, koa