Chap.

3
Equilibrium of a Particle
 Chapter Outline
„ Condition for the Equilibrium of a Particle
„ The Free-Body Diagram
„ Coplanar Systems
„ Three-Dimensional Force Systems

2
Condition for the Equilibrium of a
Particle

„ ΣF = 0
(充要條件)
i.e. ma = 0 ⇒ a = 0
(靜者恆靜,動者呈等速直線運動)

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 Free Body Diagram (FBD)
„ 自由體圖

“The diagram of the particle (body) which

represents it as being isolated or free from
its surroundings ”, it is necessary to show
all the forces that act on the particle.

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Free Body Diagram (FBD)

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Free Body Diagram (FBD):
Procedure
„ Draw or sketch the outlined shape of the
particle (body), and imagine the particle is
isolated or cut free from its surroundings
„ Indicate all the forces that act on the particle,
including active and reactive forces
„ Forces that are known should be labeled with
their proper magnitudes and directions

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 Free Body Diagram (FBD)
„ Springs
(spring constant)

Fs

unstretched position

彈簧力 Fs = -ks
使彈簧變形的外力 F = ks

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 Free Body Diagram (FBD)
„ Cables and Pulleys

C
magnitude is constant
for the whole cable
tension pulling force

往外為tension T T

往內為compression C C

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Ex. 3-1,
draw the FBD of the sphere, cord CE and
the knot at C

FBD of sphere FBD of Cord CE

FBD of knot at C
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Coplanar Force Systems 共面力

平衡的條件：
∑F = ∑F i+∑F
x y j=0

⇒ ∑ Fx = 0
∑F y =0

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Ex. 3-4,
Determine the required length of cord AC so that the 8-kg lamp is
suspended in the position shown. The undeformed length of spring
is 0.4 m and has a stiffness of kAB = 300 N/m.

ΣFx = 0：TAB －TAC cos30° = 0

ΣFy = 0：TAC sin30°－78.48 = 0
⇒ TAC = 156.96 N , TAB = 135.93 N
Spring on cable

SAB = 135.93/300 = 0.4531 m

Deformation,cable on spring
8 kg = 78.48 N ∴ spring length = 0.4531 + 0.4 = 0.8531 m
total horizontal length：
2 = 0.8531 + AC‧cos30°
⇒ AC = 1.324 m
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p.95, Problem 3-4
If cables BD and BC can withstand a maximum tensile force of 20-
kN, determine the maximum mass of the girder that can be
suspended from cable AB so that neither cable will fail. The center
of mass of the girder is located at point G.

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significant number 13
p.97, Problem 3-20
Determine the tension developed in each wire used to support the
50-kg chandelier.

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p.99, Problem 3-32
Determine the magnitude and direction θ of the equilibrium force
FAB exerted along link AB by the tractive apparatus shown. The
suspended mass is 10-kg. Neglect the size of the pulley at A.

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Three-Dimensional Force
Systems

Σ F = ΣFx i + ΣFy j + ΣFz k = 0

ΣFx = 0
Σ F = ΣFx i + ΣFy j + ΣFz k = 0
⇒ ΣFy = 0
ΣFz = 0

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Example 3.7
Determine the force developed in each cable
used to support the 40-kN crate.

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A(0, 0, 0), B(-3, -4, 8), C(-3, 4, 8)

FBD at Point A
To expose all three unknown forces in the cables.
Equations of Equilibrium
Expressing each forces in Cartesian vectors,
FB = FB(rB / rB)
= -0.318FBi – 0.424FBj + 0.848FBk
FC = FC (rC / rC)
= -0.318FCi + 0.424FCj + 0.848FCk
FD = FDi
W = -40k

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For equilibrium,
∑F = 0; FB + FC + FD + W = 0
-0.318FBi – 0.424FBj + 0.848FBk - 0.318FCi
+ 0.424FCj + 0.848FCk + FDi - 40k = 0

∑Fx = 0; -0.318FB - 0.318FC + FD = 0

∑Fy = 0; -0.424FB + 0.424FC = 0
∑Fz = 0; 0.848FB + 0.848FC - 40 = 0

Solving,
FB = FC = 23.6 kN
FD = 15.0 kN
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Ex. 3-8,
Determine the tension in each cord and the stretch of
the spring.

AD = (−1,2,2) AD = ( −1) 2 + 2 2 + 2 2

1
u AD = ( −1,2,2)
3
FD
FD = (−1,2,2)
3
F B = FB i
F C = FC (cos120°, cos135°, cos 60°)

W = − 981 kN k
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 ΣF = 0 :
F B + FC + F D +W = 0
⎧ 1
Σ
⎪ x F = 0 : FB + FC cos 120 ° − FD = 0
3
⎪
⎪ 2
⎨ΣF y = 0 : FC cos 135 ° + FD = 0
⎪ 3
⎪ 2
Σ
⎪⎩ ZF = 0 : F C cos 60 ° + FD − 981 = 0
3
∴ FB = 694 N , FC = 813N , FD = 862 N 又 FB = ks
⇒ 693 .7 = 1500 s ⇒ s = 0.462 m
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p.110, Problem 3-56
The ends of the three cables are attached to a ring at A and to the
edge of a uniform 150-kg plate. Determine the tension in each of
the cables for equilibrium.
A(0, 0, 12), B(4, -6, 0), C(-6, -4, 0), D(-4, 6, 0)

AB = (4,−6,−12), AB = 14
AC = (−6,−4,−12), AC = 14
AD = (−4,6,−12), AD = 14

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p.112, Problem 3-68
The three outer blocks each have a mass of 2 kg, and the central
block E has a mass of 3 kg. Determine the sag s for equilibrium of
the system.

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p.115, 3-76.
The ring of negligible size is subjected to a vertical force of
1000-N. Determine the longest length l of cord AC such that
the tension acting in AC is 800-N. Also, what is the force
acting in cord AB? Hint: Use the equilibrium condition to
determine the required angle θ for attachment, then
determine l using trigonometry applied to ΔABC.

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(-sin 40°)×[1] + cos 40°)×[2]
⇒ 800 sin(40°+θ ) = 1000 cos 40°

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