Summary
Summary FULL APPLIED MATH STATISTICS AND MECHANICS A* NOTES
- Module
- M3 - Mechanics 3
- Institution
- WJEC
Full A* revision notes of Applied Math (4th exam)
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,Mechanlcs, Forces
* 2nd
Newton's law : F =
ma
mass-
* Normal reaction (R) : Resolving forces
gravity)
E .
g) Y 19 6N
↑
.
- finding
f = ma
e : ⑯
-
e frictional force :
frictional
force Emotion In table & - 8N
x 4 - = 2x2
opposite = Verticle= IUsin 20
t
2gN
↳
↑ between surfaces
z
rough Adjacent = verticle= 10cos20
=
-
2 * force can Split
be
2 mS
not
movingreally
Pulling force Y(2x9 8) = my of and
* Thrust :
Pushing force
* Tension : .
into verticle
Thrust
s honizontal
- Tensioni
>
- (2x9 8) 19
y=
2g 61
= =
. .
%
!
C
↑
Gravity = 9 8 ms-2
. force JC
in direction = 12 COS23
force in direction = 12 sin 23
y
PNPs
11
in 30 Resolve f= ma
30 horizontally :
PCOS30-4COSYS = 0 - PCOS3O = 4cOS4S
usinus
·
us 4 coS4S
p = Tos30
=
3 27N
.
Resolve F = Ma
Vertically
:
T0
=
4 Sin4S + Psin30= Q = 4 Sin4S + 3 27 Sin 30
.
Q Sin4S 3x9 8
Q=
Vertically
=
:
4 46N
.
.
Q = = 41 6N .
↑
Q Sin 4S 3x9 . S
-
P = Qcos4S
Horizontally
:
Q Cos 45
p
p
=
=
41 bcs4
.
29 4 N .
f = ma
Tsin30 + T =
mg
↓ T = 9 24.
9 24 Sin 30 +9 24
mg
=
30ITS
. .
in so
13 86
my weight
= =
.
PCOS40 QCoSSS
Horizontally : =
usings" Ipsinno S
1-2 = 0
: PSin 40 + QSings +
vertically
-
PCOS40 Marange
: Do horizontally
:
TzCoSGo = T, COS30
QCSS
Tz =
cosso
d COS60
Sub into second equation
↓
SSS Sin40 + Qings- Sub in
T, Sin30 + TeSin60
= 10x9 S
Lxcosuo
:
Vertically
.
site
Tisin3o
QoSS)Sin40 +
QSinss(cos40) -
Cos 40 = 0
too
= costo
(TiSin30
+
(530) Sing a s
↳ T Sin30cos60 +T
, CoszoSin6o = 98
Q(COSSSsin40 +
SinSSCOS40) = COS4D T
, (Sin30coS60 CoS30Sin60)
+ = as
GS
=49
COS40
S
Q=
-
CoSSSsin40 + SinScos40
= 0 769N .
Sin30CoS60 + CoS3Usin60
ANSWERS
P
=
169 coss = 0 576
.
Tz = #cos30 =
84 87 N .
COS40 CoS68
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