Notes in Statics of Rigid Bodies
Module 3 Title: Equilibrium of Force Systems
Module Introduction/Rationale:
In this module, we will discuss the three equations for static equilibrium, the analysis of the
problem using the three equations and how to solve the reactions in every type of supports.
Module Outcomes:
CO1: Understand the principles of equilibrium of particles.
CO2: Determine forces of 2D and 3D structures.
Lesson 1 Title: Equilibrium of Force System
Lesson Outcomes:
At the end of this topic the learner should be able to:
1. define equilibrium
2. solve the reactions at the supports
3. solve the unknowns using the three equations of static equilibrium
EQUILIBRIUM
The subject matter of statics, as its name implies, deals essentially with the action of forces on
bodies which are at rest. Such bodies are said to be in equilibrium. Specifically, equilibrium is the term
used to designate the condition where the resultant of a system of forces is zero. Basically, when we say
equilibrium, the body is in a balance condition.
Free-Body Diagram
An isolated view of a body which shows only the external forces exerted on the body is
called a free-body diagram.
Three Equations for Static Equilibrium
1. ΣFx = 0
Right = Left
2. ΣFy = 0
Up = Down
3. ΣM = 0
Clockwise = Counter Clockwise
Three Types of Supports
1. Roller Support
2. Hinge or Pin Support
3. Fixed-End Support
M
RH
Ɵ
RN
RN
RH
RH
M
RV
RV
RV
RN = Reaction Normal to the Plane
RV = Reaction along the vertical axis
RH = Reaction along the horizontal axis
M = Fixed-end moment (moment at the support)
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Notes in Statics of Rigid Bodies
Sample Pictures
Cable
Rocker
Pinned or Hinged
Fixed
Modelling
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Notes in Statics of Rigid Bodies
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Notes in Statics of Rigid Bodies
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Notes in Statics of Rigid Bodies
Sample Problem:
1. The two cables as shown in the figure support a load of 80 KN. Determine the force along member AB
and AC that maintain equilibrium of the system.
Free-Body Diagram (FBD):
C
A
AB
400
ABX
BCX
400
600
600
ABY
B
80 KN
BC
BCY
B
80 KN
Solution:
ΣFx = 0
Right = Left
BCX = ABX
0
0
BC sin 60 = AB sin 40
BC =
AB sin 40 o
sin 60 o
ΣFy = 0
= 0.74 AB ----------- equation 1
Up = Down
ABY + BCY = 80 KN
0
0
AB cos 40 + BC cos 60 = 80 -------- equation 2
Substitute equation 1 in equation 2
0
0
AB cos 40 + (0.74 AB) cos 60 = 80
0.77 AB + 0.37 AB = 80
1.14 AB = 80
AB = 70.16 KN answer
Substitute the value of AB in equation 1
BC = 0.74 AB
BC = 0.74 (70.16 KN)
BC = 51.92 KN answer
Tensile Force in the Cable AB = 70.16 KN
Tensile Force in the Cable BC = 51.92 KN
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Notes in Statics of Rigid Bodies
2. The cable and the boom support a load of 400 N. Determine the tensile force T in the cable and the
compressive force C in the boom.
Free-Body Diagram (FBD):
A
ABX
AB
300
B
ABY
300
B
500
500
BCX
400 N
400 N
BC
BCY
C
Solution:
ΣFx = 0
Right = Left
BCX = ABX
0
0
BC cos 50 = AB cos 30
AB cos 30 o
BC =
= 1.35 AB ----------- equation 1
cos 50 o
ΣFy = 0
Up = Down
ABY + BCY = 400 N
0
0
AB sin 30 + BC sin 50 = 400 -------- equation 2
Substitute equation 1 in equation 2
0
0
AB sin 30 + (1.35 AB) sin 50 = 400
0.50 AB + 1.03 AB = 400
1.53 AB = 400
AB = 261. 44 N answers
Substitute the value of AB in equation 1
BC = 1.35 AB
BC = 1.35 (261.44)
BC = 352.94 N answer
Tensile Force in the Cable = AB = 261.44 N
Compressive Force in the Boom = BC = 352.94 N
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Notes in Statics of Rigid Bodies
3. A cylinder weighing 600 lbs is held against a smooth incline by means of the weightless rod AB.
Determine the force P and N exerted on the cylinder by the rod and the inclined plane respectively.
W = 600 lbs
B
B
250
AB
25
A
0
550
250
A
N
Y
Right = Left
ABX = NX
0
0
AB cos 25 = N sin 55
Nsin 55o
AB =
= 0.90 N ----------- equation 1
cos 25o
ΣFy = 0
550
Free-Body Diagram (FBD):
Solution:
ΣFx = 0
550
W = 600 lbs
ABX
X
250
AB
NX
550
Up = Down
ABY
ABY + NY = 600 lbs
0
0
AB sin 25 + N cos 55 = 600 -------- equation 2
N
NY
Substitute equation 1 in equation 2
0
0
(0.90 N) sin 25 + N cos 55 = 600
0.38 N + 0.57 N = 600
0.95 N = 600
N = 631. 58 lbs answer
Substitute the value of N in equation 1
AB = 0.90 N
AB = 0.90 (631.58)
AB = 568.42 lbs answer
Compressive Force in the Rod = AB = 568.42 lbs
Normal Force exerted by the inclined plane = N = 631.58 lbs
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Notes in Statics of Rigid Bodies
4. The 700-N force and the 400-N force are to be held in equilibrium by the third force F acting at an
unknown angle Ɵ with the horizontal. Determine the value of F and Ɵ.
Free-Body Diagram (FBD):
Y
Y
Q = 400 N
QX
300
P = 700 N
QY
X
300
P = 700 N
X
Q = 400 N
Ɵ
Ɵ
F
FX
FY
F
Solution:
ΣFx = 0
Right = Left
Q X + FX = P
0
400 cos 30 + F cos Ɵ = 700
F cos Ɵ = 700 – 346.41
F cos Ɵ = 353.59
F=
ΣFy = 0
353.59
----------- equation 1
cos θ
Up = Down
Q Y = FY
0
400 sin 30 = F sin Ɵ
200 = F sin Ɵ
F=
200
------------ equation 2
sin θ
Equate 1 = 2
353.59 200
=
cos θ
sin θ
sin θ
200
=
cos θ 353.59
200
tan Ɵ =
353.59
0
Ɵ = 29.49
(cross multiply 353.59 and sin θ)
but tan Ɵ =
sin θ
cos θ
answer
Substitute θ in equation 1
353.59
F=
cos 29.49 o
F = 406.22 N
answer
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Notes in Statics of Rigid Bodies
5. Determine the reactions at the supports of the loaded beam shown in the figure.
20 KN
10 KN/m
20 KN
15 KN
10 KN/m
B
A
5.0 m
2.0 m
A
1
L
2
4.0 m
15 KN
B
1
L
2
5.0 m
RA
RBH
2.0 m
4.0 m
RBV
Solution:
Start at the support with two unknowns
Take a moment at point B
ΣMB = 0
0 = RA (11 m) – (15 KN) (4 m) – (20 KN) (6 m) – (10 KN/m) (5) (2.5 m + 6 m)
0 = RA (11 m) – 60 – 120 – 425
RA = 55 KN
answer
Take a moment at point A
ΣMA = 0
0 = (10 KN/m) (5 m) (2.5 m) + (20 KN) (5 m) + (15 KN) (7 m) – RBV (11 m)
0 = 125 + 100 + 105 – RBV (11 m)
RBV = 30 KN
answer
ΣFx = 0
Right = Left
0 = RHB
RHB = 0
Check the answers
Use ΣFy = 0
Up = Down
RA + RBV = (10 KN/m) (5 m) + 20 KN + 15 KN
55 KN + 30 KN = 85 KN
85 KN = 85 KN  OK the system is in equilibrium
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Notes in Statics of Rigid Bodies
6. Determine the reactions at the supports of the loaded beam shown in the figure.
15 KN/m
20 KN
15 KN/m
20 KN
B
A
3.0 m
RAH
2.0 m
6.0 m
2
L
3
A
2.0 m
3.0 m
RAV
B
1
L
3
6.0 m
RB
Solution:
Start at the support with two unknowns
Take a moment at point A
ΣMA = 0
0 = ½ (15 KN/m) (6 m) (4 m + 2 m) – (20 KN) (3 m) – RB (8 m)
0 = 270 – 60 – RB (8 m)
RB = 26.25 KN
answer
Take a moment at point B
ΣMB = 0
0 = RAV (8 m) – ½ (15 KN/m) (6 m)(2 m) – (20 KN) (3 m + 2 m + 6 m)
0 = RAV (8 m) – 90 – 220
RAV = 38.75 KN answer
ΣFx = 0
Right = Left
RAH = 0
Check the answers
Use ΣFy = 0
Up = Down
RAV + RB = 20 KN + ½ (15KN/m) (6 m)
38.75 KN + 26.25 KN = 65 KN
65 KN = 65 KN  OK the system is in equilibrium
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Notes in Statics of Rigid Bodies
7. Determine the reactions at the supports of the loaded beam shown in the figure.
30 KN
16 KN/m
B
A
30 KN
16 KN/m
B
A
RBH
3.0 m
5.0 m
2.0 m 2.0 m
3.0 m
5.0 m
RA
2.0 m 2.0 m
RBV
Solution:
Start at the support with two unknowns
Take a moment at point B
ΣMB = 0
0 = RA (9 m) – (30 KN) (2 m) – (16 KN/m) (8 m) (4 m + 2 m + 2 m)
0 = RA (9 m) – 60 – 1024
RA = 120.44 KN answer
Take a moment at point A
ΣMA = 0
0 = (16 KN/m) (5 m) (2.5 m) + (30 KN) (7 m) – (16 KNm) (3 m) (1.5 m) – RBV (9 m)
0 = 200 + 210 – 72 – RBV (9 m)
RBV = 37.56 KN answer
ΣFx = 0
Right = Left
RBH = 0
Check the answers
Use ΣFy = 0
Up = Down
RA + RBV = 30 KN + (16 KN/m) (8 m)
120.44 KN + 37.56 KN = 158 KN
158 KN = 158 KN  OK the system is in equilibrium
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Notes in Statics of Rigid Bodies
8. Determine the tension T in the cable and the horizontal and vertical components of the reaction at A.
Free-Body Diagram (FBD):
C
C
TBC
7 ft
7 ft
85
7
TY
6
B
TX
A
A
300 lbs
3 ft
3 ft
B
100 lbs RAH
2 ft
300 lbs
RAV
3 ft
3 ft
100 lbs
2 ft
Solution:
Start at the support with two unknowns
Take a moment at point A
ΣMA = 0
0 = (300 lbs) (3 ft) + (100 lbs) (8 ft) – TY (6 ft)
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0 = (300 lbs) (3 ft) + (100 lbs) (8 ft) – TBC (
) (6 ft)
85
42
0 = 900 + 800 + TBC (
)
85
TBC = 373.17 lbs answer
Take a moment at point B
ΣMB = 0
0 = RAV (6 ft) – (300 lbs) (3 ft) + (100 lbs) (2 ft)
0 = RAV (6 ft) – 900 + 200
RAV = 116.67 lbs answer
ΣFx = 0
Right = Left
RAH = TX
RAH = T (
6
85
) = (373.17) (
6
85
) = 242.86 lbs
answer
Check the answers
Use ΣFy = 0
Up = Down
7
RAV + T (
) = 300 lbs + 100 lbs
85
7
116.67 lbs + (373.17 lbs) (
) = 400 lbs
85
400 lbs = 400 lbs  OK the system is in equilibrium
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Notes in Statics of Rigid Bodies
9. Determine the reactions at the supports of the loaded beam shown in the figure.
Free-Body Diagram (FBD):
50 KN
50 KN
B
B
A
250
2.0 m
RAH
RBX
A
2.0 m
5.0 m
RAV
5.0 m
250 RB
RBY
Solution:
Start at the support with two unknowns
Take a moment at point A
ΣMA = 0
0 = (50 KN) (2 m) – RBY (7 m)
0
0 = 100 – RB cos 25 (7 m)
RB = 15.76 KN
answer
Take a moment at point B
ΣMB = 0
0 = RAV (7 m) – (50 KN) (5 m)
0 = RAV (7 m) – 250
RAV = 35.71 KN answer
ΣFx = 0
Right = Left
RAH = RBX
0
0
RAH = RB sin 25 = (15.76) sin 25 = 6.66 KN
answer
Check the answers
Use ΣFy = 0
Up = Down
RAV + RBY= 50 KN
0
35.71 + 15.76 cos 25 = 50 KN
50 KN = 50 KN  OK the system is in equilibrium
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Notes in Statics of Rigid Bodies
10. Compute the reactions at the supports of the loaded truss shown in the figure.
20 KN
20 KN
15 KN
40 KN
15 KN
40 KN
5m
B
A
10 KN
30 KN
4m
5m
4m
4m
A
10 KN
30 KN
RAV
4m
B
RAH
4m
RB
4m
Solution:
Start at the support with two unknowns
Take a moment at point A
ΣMA = 0
0 = (30 KN) (4 m) + (20 KN) (4 m) + (40 KN) (5 m) – (15 KN) (5 m) – (10 KN) (4 m) – RB (8 m)
0 = 120 + 80 + 200 – 75 – 40 – RB (8 m)
RB = 35.625 KN answer
Take a moment at point B
ΣMB = 0
0 = RAV (8 m) – (30 KN) (4 m) – (20 KN) (4 m) – (10 KN) (12 m) – (15 KN) (5 m) + (40 KN) (5 m)
0 = RAV (8 m) – 120 – 80 – 120 – 75 + 200
RAV = 24.375 KN answer
ΣFx = 0
Right = Left
40 KN = 15 KN + RAH
RAH = 25 KN
answer
Check the answers
Use ΣFy = 0
Up = Down
RAV + RB = 20 KN + 30 KN + 10 KN
24.375 KN + 35.625 KN = 60 KN
60 KN = 60 KN  OK the system is in equilibrium
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Notes in Statics of Rigid Bodies
Problem Set No: 1 (Module 3)
1. Determine the force acting in member AB and BC.
B
100 N
2m
A
300 N
3m
C
3m
3m
2. Determine the force acting in member AB and BC.
B
200 KN
0
60
0
30
C
A
3. Determine the magnitude of P and F in order to maintain equilibrium
Y
F
0
P
60
3
4
X
0
30
400 N
200 N
4. Determine the force acting in member AB and BC.
80 KN
B
3m
6m
A
C
4m
4m
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Notes in Statics of Rigid Bodies
5. Determine the force acting along chord AB and rod BC if the system carries a load of 120 N.
A
0
30
C
0
40
B
120 N
6. A 600 lb-box is held at rest on a smooth inclined plane by force P. Determine the value of P and the
normal force N exerted by the inclined plane.
W= 600 lbs
P
0
45
0
30
7. Determine the reactions at the supports of the loaded beam shown in the figure.
30 KN
20 KN/m
10 KN
A
B
4m
1m 2m
2m
8. Determine the reactions at the supports of the loaded beam shown in the figure.
14 KN
18 KN/m
16 KN/m
A
B
3.0 m 1.0 m
6.0 m
9. Determine the reactions at the supports of the loaded beam shown in the figure.
30 KN/m
15 KN
25 KN/m
B
A
6.0 m
2.0 m
5.0 m
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Notes in Statics of Rigid Bodies
10. Determine the reactions at the supports of the loaded truss shown in the figure.
50 KN
3m
10 KN
5m
A
3m
3m
3m
B
20 KN
11. Determine the reactions at the supports of the loaded beam shown in the figure.
20 KN
30 KN
10 KN/m
3
2
B
A
3m
6m
3m
1m
12. Determine the reactions at the supports of the loaded truss shown in the figure.
40 KN
10 KN
4m
5m
5m
5m
B
4m
20 KN
A
13. Determine the value of the weight W and the reactions at the support. Assume that the pulley at point
B is frictionless.
16 KN
B
12 KN/m
0
30
3m
A
5m
W
3m
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Notes in Statics of Rigid Bodies
14. Determine the reactions at the supports of the loaded truss shown in the figure.
30 KN
50 KN
2m
20 KN
A
2m
B
3m
3m
3m
3m
15. A frictionless pulley 4 ft in diameter supports a load of 300 lbs. Compute the reactions at the hinge
and roller supports.
0
60
B
A
C
300 lbs
6 ft
2 ft
4 ft
16. A 10 m bar AB is supported by a hinge at point A and a smooth vertical surface at B. Determine the
reactions at point A and the normal force at point B.
5m
B
5m
20 KN
A
6m
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