Summary FULL APPLIED MATH STATISTICS AND MECHANICS A* NOTES

,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