1
MEM202 Engineering Mechanics - Statics MEM
2.5 Resolution of A Force into Components(Parallelograms and Laws of Sines and Cosines)
Ar
Br
Ar
Br
DC
BARrr
rrr
+=
+=
Cr
Dr
Rr
αβ γ φ
α
Rr
βαφ += φπβαπγ −=−−=
γβα sinsinsin :Sines ofLaw RBA
==
RAγα
sinsin
= RBγβ
sinsin
=
2
MEM202 Engineering Mechanics - Statics MEM
2.5 Resolution of A Force into Components(Parallelograms and Laws of Sines and Cosines - Examples)
Resolve 900 N into u- and v-components as shownooo 110sin
90025sin45sin
== vu FF
N 405110sin
25sin900
N 677110sin
45sin900
==
==
o
o
o
o
v
u
F
F
Determine F2 and α so that the resultant of F1 and F2 is a 1,000 lb force in the x-direction
lb 81838cos2 122
12 =−+= oRFRFF
oo
o85.1038sin
818250sin
38sin818
sin250 1 =⎟
⎠⎞
⎜⎝⎛=⇒= −α
α
Homework: Problems 2-31, 2-33, 2-37, 2-40
3
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceTwo Dimensions
x
yFr
iFF xx
rr=
jFF yy
rr=
θ
ir
jr
direction- inr unit vecto A
direction- inr unit vecto A
yj
xi
=
=r
r
FyθFF
FxθFF
y
xr
r
ofcomponent scalar :sin
ofcomponent scalar :cos
=
=
( ) ( ) jFiFjFiFF yx
rrrrrθθ sincos +=+=
x
y
FF1tan−=θ
4
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceTwo Dimensions - Example
lb 4.16669.33sin300
sinlb 25069.33cos300
cos
111
111
==
===
=
o
o
θ
θ
FF
FF
y
x
o69.3332tan 1
1 == −θ
( )
( ) lb 4.13725sin325
sinlb 29525cos325
cos
222
222
−=−=
==−=
=
o
o
θ
θ
FF
FF
y
x
o252 −=θ
jiFrrr
4.1662501 += jiFrrr
4.1372952 −=
Determine the xand y scalar components for the two forces shown.
5
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceTwo Dimensions
x
yFr
iFF xx
rr=
jFF yy
rr=
θ xθ
yθ Fe
F FFFF
F
yxrr
rr
of direction theinr unit vecto A :
of Magnitude:22 +==
θθπθ
θθ
sin2
coscos
coscos
FFFF
FFF
yy
xx
=⎟⎠⎞
⎜⎝⎛ −==
==Fyx eFjFiFF rrrr
=+=
ir
jr
Fer
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎠⎞
⎜⎝⎛=
+=+=
−−
FF
θFFθ
jθiθjFF
iFFe
yy
xx
yxyx
F
11 coscos
coscosrrrrr
jFiF
jFiF
eFjFiFF
yx
Fyx
rr
rr
vrrr
θθ
θθ
coscos
sincos
+=
+=
=+=
6
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceTwo Dimensions - Example
lb 4.166coslb 250cos
111
111
====
yy
xx
FFFF
θθ
oo 31.5669.33 11 == yx θθ
lb 4.137coslb 295cos
222
222
−====
yy
xx
FFFF
θθ
( ) oooo 1152590 25 22 =−−=−= yx θθ
jieFF F
rrrr4.166250
111 +== jieFF F
rrrr4.137295
222 −==
ji
jie yxFrr
Determine the xand y scalar components for the two forces shown. rrr
555.0832.0
coscos 111
+=
+= θθ
ji
jie yxFrr
rrr
423.0906.0
coscos 222
−=
+= θθ
7
MEM202 Engineering Mechanics - Statics MEM
x
y
z
Fr
xF
yF
zF
xθyθ
zθ
ir j
rkr
Fer
2.6 Rectangular Components of A ForceThree Dimensions
Fzyx eFkFjFiFF rrrrr=++=
⎟⎠⎞
⎜⎝⎛==
⎟⎟⎠
⎞⎜⎜⎝
⎛==
⎟⎠⎞
⎜⎝⎛==
−
−
−
FFFF
FF
FF
FFFF
zzzz
yyyy
xxxx
1
1
1
coscos
coscos
coscos
θθ
θθ
θθ
222zyx FFFF ++=
axes coordinate positive and between Angles:,,
alongr unit vecto A:
componentsScalar :,, of Magnitude:
F
Fe
FFFFF
zyx
F
zyx
r
rr
r
θθθkji
kFFj
FF
iFFe
zyx
zyxF
rrr
rrrr
θθθ coscoscos ++=
++=
1coscoscos :NOTE 222 =++ zyx θθθ
8
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceThree Dimensions - Example
lb 6340.65cos500,1cos
lb 278,16.31cos500,1cos
lb 4640.72cos500,1cos
===
===
===
o
o
o
zz
yy
xx
FF
FF
FF
θ
θ
θ
lb 500,1222 =++= zyx FFFF
kji
kjie zyxFrrr
rrrr
42268517.03090.0
coscoscos
++=
++= θθθ
lb 634278,1464 kjieFkFjFiFF Fzyx
rrrrrrrr++==++=
9
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceThree Dimensions - Example
( ) ( ) ( ) m 123.4223 222 =+−+−=d
o
o
o
0.61123.42coscos
0.119123.42coscos
7.136123.43coscos
11
11
11
===
=−
==
=−
==
−−
−−
−−
dzdydx
z
y
x
θ
θ
θ
kN 3.240.61cos50cos
kN 3.240.119cos50cos
kN 4.367.136cos50cos
===
−===
−===
o
o
o
zz
yy
xx
FF
FF
FF
θ
θ
θ
lb 3.243.244.36 kjikFjFiFF zyx
rrrrrrr+−−=++=
10
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceAlternative Method: Azimuth and Elevation Angles
x
y
z
Fr
xF
yF
zF
θ
φ
Angle Elevation:Angle Azimuth :
φθ
θφθ
θφθφ
φ
sincossincoscoscos
sincos
FFFFFF
FFFF
xyy
xyx
z
xy
==
===
=
xyF
11
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceAlternative Method: Azimuth and Elevation Angles – Example
oo 47 56 == φθ
lb 42456sin512sin lb 28656cos512cos
lb 54947sin750sin lb 51247cos750cos
======
======oo
oo
θFFθFF
FFFF
xyyxyx
zxy φφ
lb 549424286 kjikFjFiFF zyx
rrrrrrr++=++=
12
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceA Force Defined by Two Points, A(xA, yA, zA) and B(xB, yB, zB)
x
y
z
Fr
AB
AxAy
Az
Bx
By
Bz
( ) ( ) ( )222ABABAB zzyyxxF −+−+−=
kjie zyxF
rrrr θθθ coscoscos ++=
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )222
222
222
cos
cos
cos
ABABAB
ABz
ABABAB
ABy
ABABAB
ABx
zzyyxx
zzzzyyxx
yyzzyyxx
xx
−+−+−
−=
−+−+−
−=
−+−+−
−=
θ
θ
θ
ABzz
AByy
ABxx
zzFFyyFFxxFF
−==
−==−==
θ
θθ
coscoscos
FeFF rr=
13
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceA Force Defined by Two Points, A(xA, yA, zA) and B(xB, yB, zB) - Example
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )2233.0cos
8930.0cos
3907.0cos
222
222
222
=−+−+−
−=
=−+−+−
−=
=−+−+−
−=
ABABAB
ABz
ABABAB
ABy
ABABAB
ABx
zzyyxx
zzzzyyxx
yyzzyyxx
xx
θ
θ
θ
N 223cosN 893cosN 391cos
==
====
zz
yy
xx
FFFFFF
θ
θθ
mm 500mm 750mm 475mm 400mm 350mm 250
======
BBB
AAA
zyxzyx
lb 223893391 kjikFjFiFF zyx
rrrrrrr++=++=
14
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceA Force Defined by Two Points, A(xA, yA, zA) and B(xB, yB, zB)
If A is at origin, xA = yA = zA = 0
Bzz
Byy
Bxx
zFFyFFxFF
==
====
θ
θθ
coscoscos
x
y
z
Fr
A
B
Bx
By
Bz
222BBB zyxF ++=
kjie zyxF
rrrr θθθ coscoscos ++=
222
222
222
cos
cos
cos
BBB
Bz
BBB
By
BBB
Bx
zyxz
zyxy
zyxx
++=
++=
++=
θ
θ
θ
FeFF rr=
15
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceAlong an arbitrary direction n
x
y
z
Fr
Ferα
FeFF rr= nen of Direction:r
( )nFnFnn eeFeeFeFF rrrrrr⋅=⋅=⋅=
ner( ) ( ) nnFnnnnn eeeFeeFeFF rrrrrrrr
⋅=⋅==
( )( )
FF
FeF
FFeF
nn
nn
11 coscos
cos1
−− =⋅
=⇒
==⋅rr
rr
α
α
16
MEM202 Engineering Mechanics - Statics MEM
2.6 Rectangular Components of A ForceAlong an arbitrary direction n - Example
Determine the rectangular scalar component Fn of F along line OA
( ) ( )( )( ) ( )( ) ( )( ) lb 5313245.08738112.04364867.0218
3245.08112.04867.0873436218=++−=
++−⋅++=
⋅=
kjikji
eFF nnrrrrrr
rr
( ) ( ) ( )kji
kjien
rrr
rrrr
3245.08112.04867.0
253
253222
++−=
++−
++−=
( )( ) orr9.57
000,1531coscoscos1 11 ===⇒==⋅ −−
FFFFeF n
nn αα
lb 873436218 kjieFF F
rrrrr++==
( ) ( ) ( ) lb 000,1873436218 222 =++=F
17
MEM202 Engineering Mechanics - Statics MEM
2.7 Resultants by Rectangular ComponentsExample
( ) ( ) ( )kji
kjie
rrr
rrrr
6674.04767.05721.0
5.35.23
5.35.232221
+−=
+−+
+−=
( ) ( ) ( )kji
kjie
rrr
rrrr
5433.06985.04657.0
5.35.43
5.35.432222
+−−=
+−+−
+−−=
( ) ( ) ( )kji
kjie
rrr
rrrr
5657.07071.04243.0
453
4532223
++=
++
++=
kip 628.22284.28972.16
kip 299.16955.20971.13
kip 348.13534.9442.11
333333
222222
111111
kFjFiFkjieFF
kFjFiFkjieFF
kFjFiFkjieFF
zyx
zyx
zyx
rrrrrrrr
rrrrrrrr
rrrrrrrr
++=++==
++=+−−==
++=+−==
18
MEM202 Engineering Mechanics - Statics MEM
2.7 Resultants by Rectangular ComponentsExample
kip 628.22284.28972.16
kip 299.16955.20971.13
kip 348.13534.9442.11
333
222
111
kjieFF
kjieFF
kjieFF
rrrrr
rrrrr
rrrrr
++==
+−−==
+−==
( ) ( ) ( )kip 275.52205.2443.14 kji
kFjFiF
kRjRiRR
iziyix
zyx
rrr
rrr
rrrr
+−=
++=
++=
∑∑∑
kip 3.54222 =++= zyx RRRR
R
ooo 6.15cos 3.92cos 6.74cos 111 ====== −−−
RR
RR
RR z
zy
yx
x θθθ
19
MEM202 Engineering Mechanics - Statics MEM
Concurrent Force Systems – A Summary (2-D)
xF4
r
yF1
r
yF4
r
043214123
321312123
2112
=+++=+=
++=+=
+=
FFFFFRR
FFFFRR
FFR
rrrrrrr
rrrrrr
rrr
1Fr
2Fr
3Fr
4Fr
04321 =+++== ∑ FFFFFRrrrrrr
Parallelogram and Laws of Sines and Cosines
Rectangular Components
1F Vector Additionr
2Fr
3Fr
4Fr
12Rr
123Rr
00
0
4321
4321
=+++==+++=
=+=
yyyyy
xxxxx
yx
FFFFRFFFFR
jRiRRrrr
1Fr
2Fr
4Fr
xF1
rxF2
r
yF2
r
xFF 33
rr=
04321 =+++== ∑ FFFFFRrrrrrr
1Fr
2Fr
3Fr
4Fr
xF4
r
yF1
r
yF4
r xF1
r xF2
r
yF2
r
20
MEM202 Engineering Mechanics - Statics MEM
kFjFiFF
kFjFiFFkFjFiFF
nznynxn
zyx
zyx
vrrrM
vrrr
vrrr
++=
++=++=
2222
1111
∑∑∑
=+++=
=+++=
=+++=
++=
znzzzz
ynyyyy
xnxxxx
xxx
FFFFR
FFFFR
FFFFRiRiRiRR
L
L
L
rrrr
21
21
21
222zyx RRRR ++=
RF
RR
RF
RR
RF
RR
zzz
yyy
xxx
∑
∑
∑
−−
−−
−−
==
==
==
11
11
11
coscos
coscos
coscos
θ
θ
θ
ReRR rr =
kjie zyxR
rrrr θθθ coscoscos ++=
Concurrent Force Systems – A Summary (3-D)Rectangular Components
Forces in components
Resultant in components
Magnitude and direction of the resultant