Equilibrium of Co planar force systems

1. University of Engineering and Technology
Peshawar, Pakistan
CE-117: Engineering Mechanics
MODULE 7:
Equilibrium of Coplanar force systems
Prof. Dr. Mohammad Javed & Engr. Mudassir Iqbal
mjaved@uetpeshawar.edu.pk mudassiriqbal@uetpeshawar.edu.pk
1

2. Lecture Objectives
To learn determining unknown (reactions or/and applied
forces) in Coplanar force system under Static Equilibrium

3. 3
If a system of forces acting on a body, keeps the body in
a state of rest or in a state of uniform motion along a
straight line, then the system of forces is said to be in
Static Equilibrium (simply referred to as Equilibrium).
stAtic equiLibrium

4. 4
equiLibrium cOnditiOns
The Coplanar force system (in x-y plane) under Static
Equilibrium shall follow the follow eqns. of equilibrium:
ΣFx =0 , ΣFy =0 and ΣMo = 0
The summations in above given Eqs. must include all the
forces that act on the body—both the applied forces and
the reactions (the forces provided by supports).

5. 5
PrOcedure FOr equiLibrium
AnALYsis OF FOrce sYstems
Step 1: Draw a free-body diagram (FBD) of the body that
shows all the applied and reaction forces and couples that
act on the body.
Step 2: Write the equilibrium equations in terms of the
forces and couples that appear on the free-body diagram.
Step 3: Solve the equilibrium equations for the unknowns

6. GrAPhicAL methOds FOr equiLibrium AnALYsis
Triangle Law: If three forces are in equilibrium, then, they form a
closed triangle when represented in aTip toTail arrangement, as shown
in below Figure
Polygonal Law: If more than three forces are in equilibrium, then,
they form a closed polygon when represented in a Tip to Tail
arrangement, as shown in below Figure
F2
F3
F1
F5
F4
F3 F2
F1
6
F1
F1
F1F5
F4
F3 F2
F1
F2
F3
F1

7. If a system of Three forces is in equilibrium, then, each force
of the system is proportional to sine of the angle between
the other two forces (and constant of proportionality is the
same for all the forces). Thus, with reference to Figure, we
have:
Note: While using Lami’s theorem, all the three forces
should be either directed away or all directed towards the
point of concurrence.
LAmi’s theOrem
α
F1
F3 F2
γβ

8. Equilibrium of Coplanar
ConCurrEnt forCE systEm
8

9. 9
equiLibrium cOnditiOns FOr
cOncurrent FOrce sYstem
 In case of Coplanar Concurrent force system, following
equations can be used more conveniently to determine
unknown reactions
ΣFx =0 , ΣFy =0
 ΣMO = 0 can be used to verify the result.
 Moment centre O shall be any point other than point
of intersection of forces and reactions
xy

10. 10
PrObLem 7.1
The device shown is used to straighten the frames of
wrecked autos. Determine the tension of each segment
of the chain, i.e., AB and BC, if the force which the
hydraulic cylinder DB exerts on point B is 3.50 kN, as
shown.

11. 11
steP1: Fbd
Draw the FBD of a point at which the unknown
forces, FAB and FBC acts
FAB
3.5 kN
FBC
α
β
α = Tan-1 |400/450| = 41.63o
β = Tan-1 |250/450| = 29.06o

12. 12
steP2: deveLOP equiLibrium eqns.
 ΣFx= 0
-FAB+3.5 Sin 41.63o-FBC Sin 29.06o=0
FAB = 2.325 -0.486FBC
 ΣFy= 0
3.5 Cos 41.63o+FBCCos 29.06o=0
FBC= - 3 kN
FBC Sin 29.06o
FBCCos29.06o
3.5 Sin 41.63o
3.5Cos41.63o

13. 13
steP2 . cOndt;
 -ve sign indicate the correct direction of FBC is
apposite to the assumed one. However, instead of
revising the calculations it is more appropriate to
carry on with negative sign in rest of calculations work
 The resulting eqn from equilibrium along x-axis is:
FAB = 2.325 -0.486FBC = 2.325-0.343*(-3)
FAB = 3.78 kN (+ve sign indicate that the assumed
direction of FAB is correct

14. 14
steP3 (OPtiOnAL): veriFY the resuLts
ΣMD = -3.78*450+3.5*0
-(-3 Sin29.06o*450)
-(-3Cos29.06o*400)
or ΣMD = 3.66 kN.mm ≈ 0
This indicate that our
calculations are correct.
Moment centre to
verify the results
3.78kN 3 Sin 29.06o
3Cos29.06o
Note: The small error in ΣMD is due to rounding off the value of
forces and angle. Using values up to 3 decimals (α = 41.634o , β =
29.055o) result in FBC= 2.993 KN & FAB= 3.779 kN. Using these
values in ΣMD reduce the error to 0.0833 kN.mm

15. 15
equiLibrium OF cOncurrent FOrce sYstems
exercise 7.1
Ans: FCD= 359N, FBD= 440 N,
FAB= 622 N, FBC = 228 N
1
2

16. 16
Determine the values of α and θ so that the forces
shown in Figure will be in equilibrium.
Ans: α = 46.57o,θ =28.96o
Ans: FBA=80.7 lb, FCD= 65.9 lb,
FBC= 57.1 lb, θ = 2.95o
equiLibrium OF cOncurrent FOrce sYstems
exercise 7.1
3
4

17. Cords are loop around a small spacer
separating two cylinders each weighing 400 lb
and pass, as shown in Figure over a frictionless
pulleys to weights of 200 lb and 400 lb .
Determine the angle θ and the normal pressure
N between the cylinders and the smooth
horizontal surface.
Ans: N= 453.6 lb, θ =60o
17
equiLibrium OF cOncurrent FOrce sYstems
exercise 7.1
5
6

18. hOme AssiGnment 7.1
 Section B: 2 and 4
 Section D: 3 and 5
18

19. Equilibrium of Coplanar
parallEl forCE systEm
19

20. 20
equiLibrium cOnditiOns FOr
PArALLeL FOrce sYstem
 Assume that all the forces lying in the xy-plane are parallel
to the y-axis. Thus equation Σ Fx = 0 automatically
vanishes, and the number of independent equilibrium
equations are reduced from three to two.
 A more convenient approach is to use 1 force and 1
moment equilibrium eqn. (to find the unknown forces
or/and reactions) i.e.,
ΣFy =0 , ΣMO =0 (O is any moment centre)
 Applying ΣM = 0 at a point other than O can be used to
verify the results.

21. The cantilever beam shown in Figure is built into a
wall 2 ft thick so that it rests against points A and
B. The beam is 12 ft long and weighs 100 lb per ft.
21
PrObLem 7.2
A

22. 22
100*12 = 1200 lb
12/2= 6 ft
RA
RB
By substituting value of RB in eqn. I, RA = -3200+15600 = 12400 lb
+ve answers of RA and RB indicate that our assumed directions are correct
drAw Fbd & deveOP equibrium eqns.

23. 23
100*12 = 1200 lb
12/2= 6 ft
RA =12400 lb
RB =15600 lb
veriFY resuLts (OPtiOnAL)
Select a moment centre
other than A

24. For the system of pulleys shown
in figure determine the force P
required to lift weight, W= 900 lb.
Neglect axle friction and the
weights of the pulleys.
24
PrOb. 7.3
A
C
B
D
E1
2 3
4
5
6
7
8
P

25. 25
C
A
B
T2
T3 = 3P
PrOb. 7.3 cOntd.
ΣFy =0 (Pulley C)
-T3+3P = 0
T3= 3P
A
1 2
2
2 3
34
5
T1

26. 26
C
D
E
T4
PrOb. 7.3 cOntd.
W
ΣFy =0 (Pulley D)
-W+9P = 0
P= W/9 =900/9= 100 lb
675
8
67

27. 27
equiLibrium OF PArALLeL FOrce sYstems
excercise7.2
1
2
Determine the reactions R1 and R2 of
the beam loaded with a concentrated
load of 1600 lb and a load varying from
zero to an intensity of 400 lb per ft.
Ans: R1= 1900 lb, R2= 2100 lb
The wheel loads on a jeep are given
in Figure. Determine the distance x so
that the reaction of the beam at A is
twice as great as the reaction at B.
Ans: x= 4 ft

28. 28
equiLibrium OF PArALLeL FOrce sYstems
exercise 7.2
3 The roof truss in Figure is supported by a roller at A and a
hinge at B. Find the values of the reactions.
Ans: RA= 96.67 kN, RB= 93.33 kN

29. 29
equiLibrium OF PArALLeL FOrce sYstems
hOme AssiGnment 7.2
4
Determine the force required
to hold the 1000-lb weight in
equilibrium
Ans: 125 lb
5

30. hOme AssiGnment 7.2
 Section B: 1 ,2 and 5
 Section D: 3, 4 and 5
30

31. Equilibrium of non parallEl,
non ConCurrEnt forCE systEm
31

32. 32
equiLibrium cOnditiOns FOr nOn-
PArALLeL, nOn-cOncurrent FOrce sYstem
 One approach is to use 2 force and 1 moment
equilibrium eqns. (to find the unknown forces or/and
reactions) i.e.,
ΣFx =0 , ΣFy =0 , ΣMO =0 (O is any moment centre)
Applying ΣM = 0 at a point other than O can be used to
verify the results.
 Another approach is to use 1 force and 2 moment
equilibrium eqns. (to find the unknown forces or/and
reactions)
Remaining force equilibrium eqn can be used to verify the
results.

33. Determine the reactions at the contact points of
the rod due to 10 lb force. Assume that bar has a
negligible thickness and smooth points of contact at
A, B, and C
33
PrObLem 7.4

34. 34
RA
RB
RC
PrObLem 7.4, sOLutiOn
RAx
RAy
RBy
RBx
B
C
ΣFx =0
RBx –RAx +10 = 0
or RBCos 30o –RACos 30o +10=0
or RB= RA -11.55 -------------(I)
ΣFy =0
RC+ RBy –RAy =0
or RC +RBSin 30o –RASin 30o =0
or RC +(RA -11.55 )Sin 30o –RASin 30o = 0
or RC = 5.775 lb

35. 35
RA
RB
RC
PrObLem 7.4, sOLutiOn (cOntd.)
RAx
RAy
RBy
RBx
B
C
ΣMB =0
Rc* BC+RA *AB-10 Cos 30o*DB
+ 10 Sin 30o*0 = 0
CE/BC= Cos 30o or BC= 3/Cos 30o = 3.464 in.,
AB= 5- BE= 5-3.464*Sin30o = 3.268 in.,
DB= (8+5)-3.464* Sin30o = 11.268 in. D
5.775* 3.464+RA *3.268-10 Cos 30o*11.268= 0
RA = 23.74 lb, Substituting RA in eqn I result in RB = 12.19 lb
What does +ve sign of all reactions indicate ?
E

36. 36
RA
RB
RC
PrObLem 7.4, sOLutiOn (cOntd.)
RAx
RAy
RBy
RBx
B
C
In order to verify the result, apply
the moment equilibrium eqn at
any point other than B
ΣMC = -RB *BE+ RA *AE -
10 Cos 30o*DE+ 10 Sin 30o*CE
D
E
BE= 3.464 Sin 30o = 1.732 in., AE= 5 in.,
DE= 13 in. , CE= 3 in.
ΣMC = -12.19 *1.732+ 23.74 *5- 10 Cos 30o*13 + 10 Sin 30o*3
= 0.0036 lb.in ≈ 0
What does ΣMC = 0 indicate ?

37. Compute the tension in the cable BD when the 165
lb man stands 5 ft off the ground, as shown. The
weight of the stepladder and friction may be
neglected.
Ans: 36.7 lb
37
PrObLem 7.5

38. 38
ΣME =0
RA*4-165*(4-AB ʹ) =0
ABʹ=2/7.5*5 = 1.333
RA= 165*(4-1.333)/4
 RA= 110 lb
ΣMA =0
RB*4-165*1.333=0
RB= 55 lb
Check: ΣFy =110+55-165= 0
165 lb
RA
RB
PrObLem 7.5, sOLutiOn
B ʹʹ
B
B ʹ

39. 39
ΣMc =0
110*2-T*3-165*(2-1.333)=0
T = 36.64 lb
Check ΣMA =0 or not
Cx=T =36.64 lb and Cy = 165-110=
55 lb by equilibrium
ΣMA =- 36.64*7.5 - 55*2
+ 165*1.333 +36.64*4.5 = 0.0745≈ 0
Results are correct.
PrObLem 7.5, sOLutiOn (cOntd)
T
Cx
Cy
110 lb
C
B ʹʹ
B ʹ

40. 1. The beam shown in Figure is supported by a hinge at A and a
roller on a 1 to 2 slope at B. Determine the resultant reactions at
A and B.
Ans: Ax = 15 kN, Ay = 10 kN
RB= 33.54 kN
40
exercise 7.3
2. The uniform rod in Figure weighs 420 lb
and has its center of gravity at G. Determine
the tension in the cable and the reactions at
the smooth surfaces at A and B

41. 41
4. The winch cable on a tow truck is subjected to a force of
T= 6kN when the cable is directed at θ =60o . Determine the
magnitudes of the total brake frictional force F for the rear set of
wheels B and the total normal forces at both front wheels A and
both rear wheels B for equilibrium. The truck has a total mass of 4
Mg and mass center at G
exercise 7.3
3
Ans: F=5.2 kN, Ay=17.3kN,
By=24.9kN

42. 42
6. Determine the mass of the heaviest
uniform bar that can be supported
in the position shown in Figure if the
breaking strength of the horizontal
cable attached at C is 15 kN.
Neglect friction.
Ans: m= 2850 kg
5. Determine the horizontal and vertical
components of reaction at the pin A
and the tension developed in cable BC
used to support the steel frame.
Ans:T= 34.62 kN, Ax= 20.8 kN Ay= 87.7kN
exercise 7.3

43. 7. Determine the horizontal and vertical
components of reaction at A and the tension
in cable BC on the boom in given crane.
Weight of the boom (acting at G) is 650 lb
and weight of crate suspended at B is 1250 lb
Ans: Ax = 10.2 kip, Ay=6.15 kip, TBC=11.1 kip.
8. The articulated crane boom has a weight
of 125 lb and center of gravity at G. If it
supports a load of 600 lb, determine the
force acting at the pin A and the force in the
hydraulic cylinder BC when the boom is in
the position shown
43
exercise 7.3
Ans: Ax = 3.21 kip, Ay=1.97 kip, FB=4.19 kip

44. 9. The uniform 15-m pole has a mass of
150 kg and is supported by its smooth
ends against the vertical walls and by the
tension T in the vertical cable.
Compute the reactions at A and B.
Ans. A = B = 327 N
44
10. Determine the tension in the cable
at B, given that the uniform cylinder in
figure weighs weighs 350 lb. Neglect
friction and the weight of bar AB
Ans: 200 lb
exercise 7.3

45. hOme AssiGnment 7.3
 Section B: 3 ,7 and 9
 Section D: 4, 8 and 10
45