SWK 122 Practice
Jeff Hanson
, to
nting
F =
Fx
F 2500
=
X AB
, ~T
2500N
=
A)
-
(B
Mason,Filien
-
Bios 140 602
+
so,
8021
+
see
=
40 60 80k
-
- +
187.7
F FY=
F
2500( 0.5575 0.143)
0.3712 0.557 0.743
=
0.3712
= - + -
-
+ -
-
Fa ( = -
927.52 1392.5
+ -
1857.54)N
e
, Resultantof Vector forces:
① Blue
Z Triangle
is
-
② Directional
it
Cosines
③ U -
I method
Goal:Find of
the resultant vectors Ben:F x FsinOzcos4 =
Fy F sin
=
Ozs in
-y Fz=Fcos Oz
F, Fz, Fs Find the directional
Cosine the resultant.
angles
of
Directional Cosines:
F 80N
For F,: Blue
Using
=
Triangle 0 600 Fe Fcos- Fy Fcosty Fz FcosOz
= =
=
=
-
48.96 80cosfy
80cosfx -51.42= 80cos
Oz
Oz 1350
..-
..28.28
=
=
=
:. Ox 69.30,
=
:.fy=127.7, =.0z 130.0,
=
... F Fsinz
= cosd Fy Fsin
=
Oz sin
=
(80) sin 135°cos(-60) 80sin
=
135:sin (60) Fz Fcosz
=
=28. 28N =-
48.99N 88 cos 135
=
51.42A
Calculate Fresultant:
= -
48.995-51.42K)
minee
-
N
F, (28.28i 48.985 56.57)N
=
- -
For Fe: Directional ( +
42.43y
30.00k)N
Using
+
cosines. F2 =
-
30.00
Fz (20.86i 26.085 10.434)N
+ +
=
Fx Fcos c
=
Fy FcosOy
=
Fz Fcos 8z
=
68
=
cos 128 =
60 cos 45 -60 cos 60 Fres=(19.14 2
19.53y-16.141) N
+
30N 42.43 N 30A IFres)=31.75N
=-
=
-
30i) 3scosty cost is
"435
+
N coste =
-
:.x 52.93°:fy
=
52.04°:.0z 120.55°
=
=
For Fs:X/U-method
4A
(0.5963: 0.1454 0.29813)
=
+ +
AB (4i i
5j 2i)
=
=
+ +
1AB) =
6.708 N
E 1F =
(35)(0.5963i 0.74525 0.2981k)
=
+ +
i) N
ne i
10.43
26.095
+
+
, Dot Product
of vectors:
I main
equations:
0.
coso-l
for 'f'
finding angle between 2 rectors
⑳I Finel Fo =
line
for
finding the
magnitude
ofthe
projection
of a force on
any
line.
F, 125N
=
E
a rA
...in
a) Find the between E, and Fr
↳M
L
7
angle
54.52: b)Find the of Er projected
<Fz
magnitude
210N
=
Oz =?
2m -
Ea &= Onto rector E..
L 7 Y
r
120
↑
sch
W cos", +cosOy coz +
1
=
2i 4k cos120'+cos 50'+cos28z=1
3j
- +
-
12+3 +4 ...fz
54.52,
= + a
5.385
=
or 1 25.5
Tr =
-0.37141 -
0.55715 0.7428k
+
F, FUr,= = -
46.385 -
69.635 92.88k
+
E F cosfx
= 210cos 120 = -
=
105 i
F 122.0k
135.0y
=
-105i
+ +
21010s50
Fy FcosOy 134.995
=
=
=
F,.Fz 4870
=
-
9400 11322
+
6792
=
Fz Fcos8z
= 210
=
cos 54.5
=
121.95/
F 122.0k
135.0y
=
-105i
+ +
F,.Fz 6792
=
=cos 8
(125)(210) 26250
cos8 0.259
=
Fine 38.96-75.2
= 90.6
+
54.36
=
N
...f
75]
=
Jeff Hanson
, to
nting
F =
Fx
F 2500
=
X AB
, ~T
2500N
=
A)
-
(B
Mason,Filien
-
Bios 140 602
+
so,
8021
+
see
=
40 60 80k
-
- +
187.7
F FY=
F
2500( 0.5575 0.143)
0.3712 0.557 0.743
=
0.3712
= - + -
-
+ -
-
Fa ( = -
927.52 1392.5
+ -
1857.54)N
e
, Resultantof Vector forces:
① Blue
Z Triangle
is
-
② Directional
it
Cosines
③ U -
I method
Goal:Find of
the resultant vectors Ben:F x FsinOzcos4 =
Fy F sin
=
Ozs in
-y Fz=Fcos Oz
F, Fz, Fs Find the directional
Cosine the resultant.
angles
of
Directional Cosines:
F 80N
For F,: Blue
Using
=
Triangle 0 600 Fe Fcos- Fy Fcosty Fz FcosOz
= =
=
=
-
48.96 80cosfy
80cosfx -51.42= 80cos
Oz
Oz 1350
..-
..28.28
=
=
=
:. Ox 69.30,
=
:.fy=127.7, =.0z 130.0,
=
... F Fsinz
= cosd Fy Fsin
=
Oz sin
=
(80) sin 135°cos(-60) 80sin
=
135:sin (60) Fz Fcosz
=
=28. 28N =-
48.99N 88 cos 135
=
51.42A
Calculate Fresultant:
= -
48.995-51.42K)
minee
-
N
F, (28.28i 48.985 56.57)N
=
- -
For Fe: Directional ( +
42.43y
30.00k)N
Using
+
cosines. F2 =
-
30.00
Fz (20.86i 26.085 10.434)N
+ +
=
Fx Fcos c
=
Fy FcosOy
=
Fz Fcos 8z
=
68
=
cos 128 =
60 cos 45 -60 cos 60 Fres=(19.14 2
19.53y-16.141) N
+
30N 42.43 N 30A IFres)=31.75N
=-
=
-
30i) 3scosty cost is
"435
+
N coste =
-
:.x 52.93°:fy
=
52.04°:.0z 120.55°
=
=
For Fs:X/U-method
4A
(0.5963: 0.1454 0.29813)
=
+ +
AB (4i i
5j 2i)
=
=
+ +
1AB) =
6.708 N
E 1F =
(35)(0.5963i 0.74525 0.2981k)
=
+ +
i) N
ne i
10.43
26.095
+
+
, Dot Product
of vectors:
I main
equations:
0.
coso-l
for 'f'
finding angle between 2 rectors
⑳I Finel Fo =
line
for
finding the
magnitude
ofthe
projection
of a force on
any
line.
F, 125N
=
E
a rA
...in
a) Find the between E, and Fr
↳M
L
7
angle
54.52: b)Find the of Er projected
<Fz
magnitude
210N
=
Oz =?
2m -
Ea &= Onto rector E..
L 7 Y
r
120
↑
sch
W cos", +cosOy coz +
1
=
2i 4k cos120'+cos 50'+cos28z=1
3j
- +
-
12+3 +4 ...fz
54.52,
= + a
5.385
=
or 1 25.5
Tr =
-0.37141 -
0.55715 0.7428k
+
F, FUr,= = -
46.385 -
69.635 92.88k
+
E F cosfx
= 210cos 120 = -
=
105 i
F 122.0k
135.0y
=
-105i
+ +
21010s50
Fy FcosOy 134.995
=
=
=
F,.Fz 4870
=
-
9400 11322
+
6792
=
Fz Fcos8z
= 210
=
cos 54.5
=
121.95/
F 122.0k
135.0y
=
-105i
+ +
F,.Fz 6792
=
=cos 8
(125)(210) 26250
cos8 0.259
=
Fine 38.96-75.2
= 90.6
+
54.36
=
N
...f
75]
=
