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SWK122 Summary

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This document is a summary of the SWK122 module which I made from using Jeff Hanson as main inspiration and uses some of his techniques in solving certain questions. It has a couple of example questions which is not filled in.

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  • July 26, 2023
  • 47
  • 2022/2023
  • Summary

1  review

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By: sannelievanzyl • 1 year ago

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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]
=

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