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5. static equilibrium 1

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02/03/2015
EQUILIBRIUM OF PARTICLES & RIGID BODIES
 A particle or body is said to be in equilibrium if the resultant force and moment acting
on it is zero.
 This is necessary condition for Newton’s first law.
The equilibrant of a system of forces acting on a particle can easily be obtained applying
the already discussed techniques for finding the resultant of forces on a particle.
Problems on equilibrium often require equations of equilibrium to obtained and solved.
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
# 136
EQUILIBRIUM OF PARTICLES & RIGID BODIES
Solving problems on equilibrium involve three main steps;
Draw a free body diagram for the problem
Obtain the equations of equilibrium for the problem
Solve the equations and interpret your results
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
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02/03/2015
EQUILIBRIUM OF PARTICLES & RIGID BODIES
Static Equilibrium of a Particle
 For particles,

R   F 0
  Fx  0
F
y
0
F
z
0
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DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
EQUILIBRIUM OF PARTICLES & RIGID BODIES
Static Equilibrium of a Particle
 Example
Determine if the particle P is in equilibrium under the influence of the forces shown in the
diagram.
120 N
60o
P
150 N
31o
70 N
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
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02/03/2015
EQUILIBRIUM OF PARTICLES & RIGID BODIES
Static Equilibrium of a Particle
 Solution
Equations of Equilibrium

  Fx  120 sin 60o  70 sin 31o  150  11.02 N
y
   Fy  120 cos 60o  70 cos 31o  0.0017 N  0.00 N
120 N
60o
P
150 N
x
31o
For equilibrium,  Fx  0   Fy
But  Fx  0
Hence, P is not in equilibrium
70 N
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
# 140
EQUILIBRIUM OF PARTICLES & RIGID BODIES
Static Equilibrium of a Particle
 Example
Determine the required tensions in cables BC and BA so they can sustain the 60kg weight.
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
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02/03/2015
EQUILIBRIUM OF PARTICLES & RIGID BODIES
Static Equilibrium of a Particle
 Solution
Equations of Equilibrium
FBD

  Fh  0 : TBC sin 45o  TBA sin 53.1o  0
v
TBA
TBC
53.1o
- - - (1)
   Fv  0 : TBC cos 45  TBA cos 53.1  TBD  0
o
o
 TBC cos 45o  TBA cos 53.1o  588.6 N
45o
- - - (2)
h
TBD  m.g  60  9.81
Solving (1) and (2) simultaneously,
TBC  475.41 N
TBA  420.43 N
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
# 142
EQUILIBRIUM OF PARTICLES AND BODIES
Static Equilibrium of a Particle
Example
Three cables are used to tether a balloon as shown. Knowing that the balloon exerts an 800-N
vertical force at A, determine the tension in each cable assuming equilibrium.
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
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02/03/2015
EQUILIBRIUM OF PARTICLES AND BODIES
Static Equilibrium of a Particle
Solution
FBD
800 N
Equations of Equilibriu m

 4.2i  5.6 j
TAB  TAB
 0.6TABi  0.8TAB j
4.2 2  5.6 2

2.4i  5.6 j  4.2k
TAC  TAC
 0.37TAC i  0.87TAC j  0.65TAC k
2.4 2  5.6 2  4.2 2

 5.6 j  3.3k
TAD  TAD
 0.86TAD j  0.51TAD k
5.6 2  3.32
FAy  800 Nj
F
F
F
x
 0.6TAB  0.37TAC  0
- - - (1)
y
 0.8TAB  0.87TAC  0.86TAD  800  0
- - - (2)
z
 0.65TAC  0.51TAD  0
- - - (3)
Solving (1), (2) and (3) simultaneously,
TAB  200.6 N
TAC  325.3 N
DDEK/2015/ ME 164 - STATICS of SOLID MECHANICS
TAD  414.6 N
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