Modul VI.
Equilibrium of a Particle.
In the preceding sections, we discussed the methods for determining the resultant of
several forces acting on a particle. Although this has not occurred in any of the problems
considered so far, it is quite possible For the resultant to be zero. In such a case, the net
effect of the given forces is zero, and the particle is said to be in equilibrium. We thus
have the following definition: When the resultant of all the forces acting on a particle As
zero, the particle is in equilibrium.
A particle which is acted upon by two forces will be in equilibrium if the two forces have
the same magnitude, same line of action, and opposite sense. The resultant of the two
forces is then zero. Such a case is shown in Fig. 5.1.
Fig.5.1
Fig.5.2
Fig.5.3
Another case of equilibrium of a particle is represented in Fig. 5.2, where four forces are
shown acting on A. In Fig. 5.3, the resultant of the given forces is determined by the
polygon rule. Starting from point O with F1 and arranging the forces in tip-to-tail fashion,
we find that the tip of F4 coincides with the starting point O. Thus the resultant R of the
given system offerees is zero, and the particle is in equilibrium.
The closed polygon drawn in Fig. 5.3 provides a graphical expression of the equilibrium
of A. To express algebraically the conditions for the equilibrium of a particle, we write
R = ΣF = 0 -
(5.1)
Resolving each force F into rectangular components, we have
 (F i  F
x
y
j )  0......or......( Fx )i  ( Fy ) j  0
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Dr. Ir. Abdul Hamid M.Eng.
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We conclude that the necessary and sufficient conditions for the
equilibrium of a particle are
ΣFx = 0
ΣFy = 0
(5.2)
Returning to the particle shown in Fig. 5.4 we check that the equilibrium conditions are
satisfied.
Fig.5.4
We write
ΣFx= 300 Ib - (200 Ib) sin 30° - (400 Ib) sin 30°
= 300 Ib - 100 Ib - 200 Ib = 0
ΣFy = - 173.2 Ib - (200 Ib) cos 30° + (400 Ib) cos 30°
= - 173,2 Ib - 173.2 Ib + 346.4 Ib = 0
3.2. Newton's First Law of Motion. In the latter pi of the seventeenth century, Sir Isaac
Newton formulated thr fundamental laws upon which the science of mechanics is base
The first of these laws can be stated as follows:
// the resultant force acting on a particle is zero, the partk will remain at rest (if originally
at rest) or will move with co stant speed in a straight line (if originally in motion),
From this law and from the definition of equilibrium give in the above section, it is seen
that a particle in equilibrium either is rest or is moving in a straight line with constant
speed. In tl following section, various problems concerning the equilibriu of a particle will
be considered.
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3.3 Problems
Involving the
Equilibrium of Particle.
Free-Body Diagram.
In
practice, a problem engineering mechanics is derived from an actual physical situ tion.
A sketch showing the physical conditions of the problei is known as a space diagram.
The methods of analysis discussed in the preceding sectioi apply to a system of forces
acting on a particle. A large numbt of problems involving actual structures, however, may
be n duced to problems concerning the equilibrium of a particle. Th is done by choosing
a significant particle and drawing a separal diagram showing this particle and all the
forces acting on i Such a diagram is called a free-body diagram.
As an example, consider the 75-kg crate shown in the spac diagram of Fig. 5.5a. This
crate was lying between two build ings, and it is now being lifted onto a truck, which will
remov it. The crate is supported by a vertical cable, which is joine at A to two ropes
which pass over pulleys attached to the build ings at B and C. It is desired to determine
the tension in cad of the ropes AB and AC.
In order to solve this problem, a free-body diagram must be drawn, showing a particle in
equilibrium. Since we are interested in the rope tensions, the free-body diagram should
include at least one of these tensions and, if possible, both tensions. Point A is seen to
be a good free body for this problem. The free-body diagram of point A is shown in Fig.
5.5b.
Fig.5.5a.
Fig.5.5b.
It shows point A and the forces exerted on A by the vertical cable and the two ropes. The
force exerted by the cable is directed downward and is equal to the weight W of the
crate. Recalling Eq. (5.1), we write
W = mg = (75 kg)(9.81 m/s2) = 736 N
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and indicate this value in the free-body diagram. The forces exerted by the two ropes are
not known. Since they are respectively equal in magnitude to the tension in rope AB and
rope AC, we denote them by TAB and TAC and draw them away from A in the directions
shown in the space diagram. No other detail is included in the free-body diagram.
Since point A is in equilibrium, the three forces acting on it must form a closed triangle
when drawn in tip-to-tail fashion. This force triangle has been drawn in Fig. 2.29c. The
values TAB and TAC of the tension in the ropes may be found graphically if the triangle is
drawn to scale, or they may be found by trigonometry. If the latter method of solution is
chosen, we use the law of sines and write
When a particle is in equilibrium under three forces, the problem may always be solved
by drawing a force triangle. When a particle is in equilibrium under more than three
forces, the problem may be solved graphically by drawing a force polygon. If an analytic
solution is desired, the equations of equilibrium given in preceding section should be
solved:
ΣFx =0
ΣFy = 0
(5.3)
These equations may be solved for no more than two unknowns; similarly, the force
triangle used in the case of equilibrium under three forces may be solved for two
unknowns.
The more common types of problems are those where the two unknowns represent (1)
the two components (or the magnitude and direction) of a single force, (2) the magnitude
of two forces each of known direction. Problems involving the determination of the
maximum or minimum value of the magnitude of a force are also encountered .
SAMPLE PROBLEM 5.1
In a ship-unloading operation, a 3500-lb automobile is supported i a cable, A rope is tied
to the cable at A and pulled in order to cenle the automobile over its intended position.
The angle between th cable and the vertical is 2°, while the angle between the rope am
the horizontal is 30°. What is the tension in the rope?
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Solution. Point A is chosen as a free body, and the complete free-body diagram is
drawn. TAB is the tension in the cable AB, an., TAC is the tension in the rope.
Equilibrium Condition. Since only three forces act on the free bodi. we draw a force
triangle to express that it is in equilibrium. Using thi law of sines, we write
T
TAB
3500lb
 AC  

sin 120
sin 2
sin 58
With a calculator, we first compute and store the value of the quotient. Multiplying this
value successively by sin 120° and sin 2° we obtain
TAB = 3570 Ib TAC. = 144 Ib
SAMPLE PROBLEM 5.2
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Determine the magnitude and direction of the smallest force F whidi will maintain the
package shown in equilibrium. Note that the fore; exerted by the rollers on the package
is perpendicular to the incline.
Solution. We choose the package as a free body, assuming that it may be treated as a
particle. We draw the corresponding free-bod y, diagram.
Equilibrium Condition. Since only three forces act on the fret body, we draw a force
triangle to express that it is in equilibrium. Line 1-1 represents the known direction of P.
In order to obtain the minimum value of the force F, we choose the direction of F
perpendicular to that of P. From the geometry of the triangle obtained, we find
F = (294 N) sin 15° = 76.1 N
α= 15°
F = 76.1 N
SAMPLE PROBLEM 5.3
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Two forces P and Q of magnitude P=- 1000 Ib and Q = 1200 Ib are applied to the aircraft
connection shown. Knowing that the connection is in equilibrium, determine the tensions
T1 and T2.
Solution. The connection is considered a particle and taken as a free body. It is acted
upon by four forces directed as shown. Each force is resolved into its x and y
components.
P = -(10001b)j
Q= -(1200 Ib) cos 15°i + (1200 Ib) sin 15°j
= -(1159 lb)i + (311 lb)j
T1 = T1 i
T2 = T2 cos 60°i 4- T2 sin 60°j
= 0.500 T2i + 0.866 T2j
Equilibrium Condition. Since the connection is in equilibrium, the resultant of the forces
must be zero. Thus
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R=P+Q+ T1 + T2=0
Substituting for P, Q, T1 , and T2 the expressions obtained above, and factoring the unit
vectors i and j, we have
(-1159 Ib + T1 + 0.500 T2)i+(- 1000 Ib + 311 Ib + 0.866 T2)j = 0
This equation will be satisfied if, and only if, the coefficients of i and j are equal to zero.
We thus obtain the following two equilibrium equations, which express, respectively, that
the sum of the x components and the sum of the y components of the given forces must
be zero.
[Σ Fx]
-1159 Ib + T1 + 0.500 T2 = 0
[Σ Fy ]
-1000 Ib + 311 Ib + 0.866 T2 = 0
Solving these equations, we find
T1 = 761 Ib
T2 = 796 Ib
In drawing the free-body diagram, we assumed a sense for each unknown force. A
positive sign in the answer indicates that the assumed sense is correct. The complete
force polygon may be drawn to check the results.
Basic Persamaan:
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Hukum Newton I telah menyatakan dengan jelas bahwa kondisi untuk keseimbangan
static dari sebuah partikel yang menyatakan bahwa sebuah partikel akan tetap diam
hanya jika resultanta semua gaya yang bekerja pada benda tsb=nol.
Thus,persamaan untuk keseimbangan static:
R   Fx I   FY J   FZ K  0 …..(5.4)
Karena suatu vector nol hanya mempunyai komponen nol,maka persamaan tiga scalar
untuk keseimbangan static adalah:
F
F
F
x
0
Y
0
Z
0
…….(5.5)
Contoh2 Soal.
1.Sebuah peluru diletakkan dipita karet diantara ujung2 A dan B, lihat Gambar dibawah
.Pita karet tsb mengalami deformasi dan menghasilkan gaya tegang yg sebanding
terhadap elonggasinya,yaitu F=kΔ,disini Δ=elonggasi pita karet dan k=konstanta pegas.
Hitunglah gaya P yg dibutuhkan untuk menjaga peluru tsb dalam keseimbangan static
sebagai fungsi dari jarak x.
Gambar.
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Dr. Ir. Abdul Hamid M.Eng.
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Jawab:
Berat peluru tidak dinyatakan sehingga berat peluru dapat diabaikan.
Sudut θ untuk gaya pegas selalu tergantung pada harga x (dlm meter ) berdasarkan :
  sin 1
0.05
0.05  x
2
2
x
 cos 1
0.05 2  x 2
....(i).
Untuk keseimbangan static ,didpt:
F
F
x
y
 P  Fk
1  Fk 2 cos   0.....(ii ).
 Fk 2  Fk 1 sin   0......(iii )
dari (iii) didapat (Fk)1=(Fk)2.
Gaya tegang (Fk)1=kΔ,disini Δ =pertambahan panjang karet.
Panjang pita karet semula,sebelum ada beban peluru=2 x 0.05m=0.1m.
Panjang pita karet setelah ada beban peluru=2 x ((0.05)2+x2)1/2 .Maka:
Fk 1
 k  k 
2

Dari (ii) dan (iv) →
0.052
 x 2  0.10 
.....(iv )

P  2Fk 1 cos  
2k  2

0.052

 4kx1 

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 x 2  0.10 

x
0.052
 x2



0.0025  x 
0.05
2
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2.Sebuah peti seberat massa 500 kg digantung dengan kabel yg disambung pada titik A.
Hitunglah tegangan pada setiap kabel.
Jawab.
Dapat dihitung bahwa
FAB  FAB e B / A  FAB
rB / A
rB / A
 FAB
 6I  9 J  2K
62  92  22
9
2
 6

 FAB  
I
J 
K
11
11 
 11
FAC  FAC eC / A  FAC
rC / A
rC / A
 FAC
 1.5 J  2 K
1.5 2  2 2
4 
 3
 FAC   J  K 
5 
 5
FAD  FAD e D / A  FAD
rD / A
rD / A
 FAD
6 I  1.5 J  2 K
6 2  1.5 2  2 2
3
4 
 12
 FAD  I 
J
K
13
13 
 13
FAE  FAE e E / A  FAE
rE / A
rE / A
 FAE
2K
22
 FAE K
Untuk kondisi keseimbangann static ,jumlah komponen gaya dititik A=0,maka:
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6
12
FAB  FAD  0.....(i )
11
13
9
3
3
 FY  11 FAB  5 FAC  13 FAD  0....(ii )
2
4
4
 FZ   11 FAB  5 FAC  13 FAD  FAE ....(iii )
F
X

Tegangan kabel langsung dengan peti
F
Z
 mg  FAE  0
500  9.806  FAE
FAE  4903 N .....(iv )
Dari (i) didapat: FAD 
13 6

FAB  0.5909 FAB
12 11
5 9
3

FAB 
FAD 

3  11
13

Dari (ii) didapat:  1.66670.8182 FAB  0.2308(0.5909) FAB 
FAC 
 1.1364 FAB
2
4
4
FAB  (1.1364) FAB  (0.5909) FAB  4903N
11
5
13
 3852 N
FAE 
Dari (iii) didapat: FAB
FAD  0.5909(3852)  2276 N
FAC  1.1364(3852)  4377 N
Soal2 Latihan
5.1. Sebuah sistem gaya seperti terlihat pada Gamb.
,disamping.Diketahui W=20 N
0
dan   75 sedangkan ukuran
dan berat pulley dpt diabaikan.Hitunglah tegangan T2 dan sudut  sehingga tercapai
keseimbangan statik.
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Dr. Ir. Abdul Hamid M.Eng.
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Gambar soal 1.
Gambar soal 2.
5.2.Lihat Gambar 1 diatas .Bila W 1=10 N,tentukan harga W 2 dan W 3 supaya tercapai
kesimbangan statik.
5.3.Sebuah mobil seberat 3000 lb diam ditanjakan bersalju. Sebuah kabel penarik diikat
pd mobil tsb dan truk penggerek.Gesekan pd kerekan dpt diabaikan ,juga massa
kerekan dpt diabaikan.Bila tanjakan sangat licin , hitunglah gaya dlm kabel dimana mobil
dan truk penggerek dlm keadaan diam.
Gamb.
Jawab:
Steps: (i). Gambarkan FBD lengkap dgn sumbu2 x dan
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y
(ii). Uraikan gaya2 tsb pd sumbu2 x dan y
(iii).Applikasikan prinsip MekTek,dlm hal ini ka
rena sistem gaya dlm keadaan diam maka:
Fx  0, dan..Fy  0 ..........(1.12)
FBD I:
Dari pers.diatas didapat:
F
F
x
 F  3000 sin 20 0  0.........i
y
 N  3000 cos 20 0  0.......ii
Dari (i) didpt F=3000lb sin200=1026 lb.
Dari (ii) dpt dihitung N=3000cos200=2819 lb.
FBD II:
F
x
 2T  F  0
2T  F  1026lb
T  513lb.
Problems.
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5.4 through 5.6
Two cables are tied together at C and as shown. Determine the
tension in AC and BC.
5.7 A 600-lb block is supported by the two cables AC i (a) For what value of (a)For what
value of α is the tension in cable AC minimum? (b) What are the corresponding values of
the tension in cables AC and BC.
5.8 A 600-lb block is supported by the two cables AC and BC. Determine (a) the value of
α for which the larger of the cable tensions is as small as possible, (b) the corresponding
values of the tension in cables AC and BC.
5.9 Two ropes are tied together at C. If the maximum permissible tension in each rope is
2.5 kN, what is the maximum force F that may be applied? In what direction must this
maximum force act?
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5.10 The force P is applied to a small wheel which rolls on the cable ACB. Knowing that
the tension in both parts of the cable is 750 N, determine the magnitude and direction of
P.
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