Definition, Formulas and Example Problems

1. Equilibrium
Equilibrium – the state in which there is no tendency in
motion of an object to change.

2. 2 Kinds of Equilibrium
1. Static Equilibrium – exist when an
object is at rest.
2. Dynamic Equilibrium – exist when
an object is in uniform motion.

3. General Conditions
First Condition – “The vector sum of all forces
acting on a body must be zero.
In symbol: F = 0 FX = 0
=
Fy= 0
=

4. Second Condition
“The algebraic sum of all moments about any axis
(within or outside) must be zero.
Mclockwise = Mcounterclockwiseclockwise
=

5. • WEIGHT
• The weight of an object, (Fw )
or (W), is essentially the force
with which gravity pulls
downward upon the object.
m
W = mg
Magnitude of the weight: W = mg
where: m = mass, kg
g = 9.8 m/s2
Direction of the weight: downward
Point of application: center of gravity

6. m
T1
W
TENSILE FORCE
The tensile force, (FT
) or T, acting on a
string or cable or chain is the applied force
tending to stretch it. The scalar magnitude of
the tensile force is the tension.

7. How to solve problems dealing with
current in the equilibrium:
1. Draw the sketch of the forces that act on the
object (free-body diagram)
2. Resolve the force into component along x and y
axis.
3. Write the equation of the first condition for
equilibrium.
4. Solve the equation.

8. Exercise:
An object, weighing 100 N and suspended by rope
A, is pulled aside by a horizontal force exerted by
rope B and held so that rope A makes an angle of
30o
with the vertical. Find the tension in the two
ropes.

9. rope A
30o
100 lb
rope B
rope B
rope B
F1
60o
30o
F2
W = 100 lb

10. Force components: Fx
= F cos  and Fy
= F sin 
Fx
= 0 Fy
= 0
– F1
(cos 60o
) + F2
+ 0 = 0 F1
(sin 60o
) + 0 – 100
lb = 0
F2
= F1
(cos 60o
) F1
(sin 60o
) = 100 lb
F2
= F1
(0.5) ------ (1)
60
sin
lb
100
F o
1 
F1
= 115.47 lb ---- subst. to (1)
then: F2
= F1
(0.5)
F2
= (115.47 lb)
(0.5)
F2
= 57.735 lb

11. 2. Two strings support a small 8 kg body. Find the
tension in the strings if both make 30o
angles with
the vertical.
30o
30o
m = 8 kg
F1
F2
F2
F2
60o
60o
m = 8 kg
m = 8 kg
m = 8 kg
m = 8 kg
m = 8 kg

12. W = mg
W = 8 kg (9.8 m/s2
)
W = 78.4 kg-m/sec2
= 78.4 N
Fx = 0 Fy=0
– F1 (cos 60o
) + F2 (cos 60o
) + 0 = 0 F1 (sin 60o
) + F2 (sin 60o
) – 78.4 N = 0
F2 (cos 60o
) = F1 (cos 60o
) F1 (sin 60o
) + F1 (sin 60o
) = 78.4 N
2 F1 (sin 60o)
= 78.4 N
F2 = F1
F1 = 45.264 N
 F1 = F2 = 45.264 N
o
o
1
2
60
cos
)
60
(cos
F
F  o
o
1
2
60
cos
)
60
(cos
F
F  o
o
1
2
60
cos
)
60
(cos
F
F 
o
o
1
2
60
cos
)
60
(cos
F
F 
)
60
(sin
2
78.4
F o
1 