Modul IV
System Gaya Tiga Dimensi
Rx 

F
nx
n 1
 0 ........(2.14)
(ii). Untuk tiga dimensi
Rz 

R  Rx  R y  Rz
F
nz
n 1


F
n 1
Ry 
Rx 
nx


n 1
n 1
  Fny   Fnz
....(2.15)

F
n 1
ny

F
n 1
nx
Gaya adalah suatu vektor yg mempunyai arah dan besaran.
untuk dua dimensi:
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Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
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untuk tiga dimensi:
Dgn menggunakan unit vektor
maka: A 
yg berharga “satu”
I, J, K
Ax I  Ay J  Az K
.....(2.16)
atau dpt ditulis :
rB  Ax I  Ay J  Az K .....(2.17)
A
disini:

rB
A
Ax  Ay  Az
2
2
2
.....(2.18)
Bila ordinat P(0,0,0) dan Q(XQ,YQ,ZQ),maka
 X Q  X P I  YQ  YP J  Z Q  Z P K
rQ
......(2.19)
P
disini:

rQ
P
cos  
X
X
 XP
Q
Q
 XP


2
 YQ  YP

2
 z Q  Z P

2
.....(2.20)
rQ
cos  
P
Y
 YP
Q

rQ
cos  
Z
P
Q
 ZP

rQ
P
Unit Vektor Umum
 X Q  X P I  YQ  YP J  Z Q  Z P 
rQ
P
.........(2.21)
 rQ .eQ
P

eQ
P
P
rQ
P
rQ
P
Contoh soal:
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Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
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4.1.Sebuah peti seberat massa 500 kg digantung dengan kabel yg disambung pada titik
A. Hitunglah tegangan pada setiap kabel.
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Dr. Ir. Abdul Hamid M.Eng.
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Example 4.2. A force of 500 N forms angles of 60, 45, and 120°,respectively, with the x,
y, and z axes. Find the components Fx, Fy, and Fz, of the force.
Substituting F = 500 N, θx, = 60°, θy = 45°, θz = 120° into formula, we write
Fx = (500 N) cos 60° = +250 N
Fy = (500 N) cos 45° = +354 N
Fz = (500 N) cos 120° = -250 N
carrying into Eq. the values obtained for the scalar components of F, we have
F = (250 N)i + (354 N)j - (250 N)k
As in the case of two-dimensional problems, the plus sign indicates that the component
has the same sense as the corresponding axis, and the minus sign that it has the
opposite sense.
The angle a force F forms with an axis should be measured from the positive side of the
axis and will always be comprised between 0 and 180°. An angle θx smaller than 90°
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Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
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(acute) indicates that F (assumed attached at O) is on the same side of the xy plane as
the positive x axis; cos θx and Fx,. will then be positive. An angle θx larger than 90°
(obtuse) would indicate that F is on the other side of the yz plane; cos 9X and Fx. would
then be negative. In Example 1 the angles θx and θy are acute, while θz is obtuse:
consequently, Fx , and Fy are positive, while Fz, is negative.
Substituting into (2.20) the expressions obtained for Fx, Fy, Fz in 2.19), we write
F = F (cos θx i + cos θy j + cos θz k) …..(2.21)
which shows that the force F may be expressed as the product of :the scalar F and of
the vector
λ = cos θx i + cos θy j + cos θzk…… (2.22)
Fig.4.1
Clearly, the vector λ is a vector of magnitude equal to 1 and of the same direction as F
(Fig. 4.1). We shall refer to λ as the unit vector along the line of action of F. It follows
from (2.22) that the omponents of the unit vector λ are respectively equal to the direction
cosines of the line of action of F:
λx = cos θx
λy = cos θy
λz= cos θz
(2.23)
We should observe that the values of the three angles θx , θy , θz are not independent.
Expressing that the sum of the squares of the components of λ is equal to the square of
its magnitude, we write
2x  2y  2z  1
or, substituting for λx, λy , λz from (2.23),
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
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cos2 θx + cos2 θy + cos2 θZ = 1
In Example 1, for instance, once the values θx = 60° and θy =45° have been selected, the
value of θz must be equal to 60° or 120° in order to satisfy identity (2.24).
When the components Fx, Fy, Fz of a force F are given, the magnitude F of the
force is obtained from (2.18). The relations (2.19) may then be solved for the direction
cosines,
cos
Fy
Fx
F
..... cos ....... cos z .
F
F
F
and the angles θx , θy, θz characterizing the direction of F may be found.
Example 4.3. A force F has the components Fx= 20 Ib, Fy = -30 Ib, Fz = 60 Ib. Determine
its magnitude F and the angles θx, θy, θz it forms with the axes of coordinates.
From formula (2.18} we obtain:
F  Fx2  Fy2  Fz2
 (20lb ) 2  (340lb ) 2  (60lb ) 2
 4900lb  70lb
Substituting the values of the components and magnitude of F into Eqs. (2.25), we write
cos
Fy
Fx 20lb
F
 30lb
60lb

..... cos

....... cos z 
.
F 70lb
F
70lb
F 70lb
Calculating successively each quotient and its arc cosine, we obtain
θx = 73.4°
Θy = 115.4°
θz = 31.0°
The computations indicated may easily be carried out with a calculator.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
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SAMPLE PROBLEM 4.4.
A tower guy wire is anchored by means of a bolt at A. The tension in the wire is 2500 N.
Determine (a) the components Fx, Fy, Fz of the force acting on the bolt, (b) the angles 0x,
θy, θz defining the direction of the force.
Solution
a. Components of the Force.
The line of action of the force acting on the bolt passes
through A and B, and the force is directed from A to B. The components of the vector
AB, which has the same direction as the force, are
dx = -40 rn
dy = +80 m
dz = +30 m
The total distance from A to B is
AB = d =
d x2  d y2  d z2 = 94.3 m
Denoting by i, j, k the unit vectors along the coordinate axes, we have
AB = -(40 m)i + (80 m)j + (30 m)k
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Dr. Ir. Abdul Hamid M.Eng.
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
AB
Introducing the unit vector  
, we write
AB
F = Fλ = F
AB 2500 N

AB
AB 94.3m
Substituting the expression found for AB, we obtain
F
2500 N
 (40m)i  (80m) j  (30m)k 
94.3m
F = -(1060N)i + (2120 N)j + (795 N)k
The components of F, therefore, are
Fx = -1060N
Fy = + 2120N
Fz = +795N
b. Direction of the Force. Using Eqs. (2.25), we write
Fy  2120 N
Fx  1060 N

..... cos  y 

.......
F
2500 N
F
2500 N
F
 795 N
cos  z  z 
.
F
2500 N
cos  x 
Calculating successively each quotient and its arc cosine, we obtain
θx = 115.1°
θy = 32.0°
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θz, - 71.5°
Dr. Ir. Abdul Hamid M.Eng.
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(Note. This result could have been obtained by using the components and magnitude of
the vector AB rather than those of the force F).
SAMPLE PROBLEM 4.5
In order to move a wrecked truck, two cables are attached to the truck at A and pulled by
winches B and C as shown. Determine the resultant of the forces exerted on the truck by
the two cables, knowing that the tension is 2000 Ib in cable AB and 1500 Ib in cable AC.
Solution. The force exerted by each cable on the truck will be resolved into x,y, and z
components. We first determine the components and magnitudes of the vectors AB and
AC, measuring them from the truck toward each winch. Denoting by i, j, k the unit
vectors along the coordinate axes, we write
AB = -(52 ft)i + (50 ft)j + (40 ft)k
AB = 82.5 ft
AC = -(52 ft)i + (62 ft)j - (50 ft)k
AC = 95.1 ft
Denoting by  AB the unit vector along AB, we have
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Dr. Ir. Abdul Hamid M.Eng.
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T AB  T AB  AB  T AB
AB 2000lb

AB
AB 82.5 ft
Substituting the expression found for AB, we obtain
2000 ft
 (52 ft)i  (50 ft) j  (40 ft)k 
82.5 ft
  (1260lb )i  (1212lb ) j  (970lb )k 
T AB 
T AB
Denoting by λAC the unit vector along AC, we obtain in a similar way
TAC  TAC  AC  TAB
AC 1500lb

AC
AC 95.1lb
TAC   (820lb)i  (978lb) j  (788lb)k 
The resultant R of the forces exerted by the two cables is
R = TAB+ TAC = -(2080 lb)i + (2190 lb)j + (182 lb)k
The magnitude and direction of the resultant are now determined
R=
Rx2  R y2  Rz2 =
 2080
2

 2190  182  3030lb
2
2
From Eqs.2.33 we obtain:
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
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R y  2190 N
R x  2080 N

..... cos  y 

.......
R
3030 N
R
3030 N
R
 182 N
cos  z  z 
.
R
3030 N
cos  x 
Calculating successively each quotient and its arc cosine, we have
θx = 133.4°
θy = 43.7°
θz = 86.6° -.
Soal2 Latihan
2.1.Uraikanlah gaya2 F dalam term2 komponen2nya terhadap dua sistem koordinat XYZ
dan xyz,pada Gambar 2.2 diatas.
2.2.Tentukanlah vector berlokasi disebuah titik P yang mempunyai koordinat Cartesian
(400,400,-200) mm dari titik asal O.Berapakah sudut arah vector ini?.
2.3.Seseorang berada ditik A sedang mengamati keatas puncak gedung setinggi 200m
dititik B. Hitunglah komponen2 posisi titik arah cosinus
posisi vector ini terhadap arah selatan dan timur. (lihat gambar diatas).
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
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2.4.Dua kawat kabel diikatkan pada tiang vertical AB.Sebuah gaya keatas sebesar 2 kN
dikerjakan pada tiang dengan maksud resultante dari tiga gaya yang bekerja pada titik A
// sumbu x. Hitunglah tegangan didua kabel tsb dan besar gaya resultantenya.
2.5 Determine (a) the x, y, and z components of the 250-N force, (b) the angles θx, θy,
and θz, that the force forms with the coordinate axes.
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Dr. Ir. Abdul Hamid M.Eng.
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2.6 Determine (a) the x, y, and z components of the 300-N force, (b) the angles θX, θy,
and θz that the force forms with the coordinate axes.
2.7 The angle between the guy wire AB and the mast is 20°. Knowing that the tension in
AB is 300 Ib, determine (d) the x, y, and z components of the force exerted on the boat
at B, (b) the angles θX, θy, and θz defining the direction of the force exerted at B.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng.
STATIKA STRUKTUR
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