Download - 89573289 2 Equilibrium of Force System
2. EQUILIBRIUM OF FORCE SYSTEMS
Definition:-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 equilibrium. ALTERNATIVELY, if the resultant of the force system is zero, then, the force system is said to be in equilibrium.
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Conditions for Equilibrium :
A coplanar concurrent force system will be in equilibrium if it satisfies the following two conditions:
ι) ∑ Fx = 0; and ii) ∑ Fy = 0
i.e. Algebraic sum of components of all the forces of the system, along two mutually perpendicular directions, is ZERO.
XY
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Graphical conditions for Equilibrium
Triangle Law: If three forces are in equilibrium, then, they form a closed triangle when represented in a Tip to Tail arrangement, as shown in Fig 2.1
Fig 2.1
F 3
F2
F1
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 Fig. 2.2.
Fig 2.2
F2
F3
F1
F5
F 4 F3
F2
F1
F5
F4
F3 F2
F1
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 Fig.2.3, we have,
LAMI’S THEOREM
γβα Sin
F
Sin
F
Sin
F 333 == β
α
γ
Fig. 2.3
F1
F3 F2
Note: While using Lami’s theorem, all the three forces should be either directed away or all directed towards the point of concurrence.
When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction.
Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system
∑ Fx = 0; ∑ Fy = 0 ∑M = 0
These requirements are both necessary and sufficient conditions for equilibrium.
EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM
Space Diagram (SPD) : The sketch showing the physical conditions of the problem, like, the nature of supports provided; size, shape and location of various bodies; forces applied on the bodies, etc., is known as space diagram.
eg, Fig 2.4 is a space diagram
SPACE DIAGRAMS & FREE BODY DIAGRAMS
Weight of sphere = 0.5 kN,Radius = 1m
Cable
P = 2kN
30°
Fig. 2.4 SPDwall
3 m θ
Sphere
Free Body Diagram (FBD) :
It is an isolated diagram of the body being analyzed (called free body), in which, the body is shown freed from all its supports and contacting bodies/surfaces. Instead of the supports and contacting bodies/surfaces, the reactive forces exerted by them on the free body is shown, along with all other applied forces.
1) Tensile Force: It is a force trying to pull or extend the body. It is represented by a vector directed away from the body.
2) Compressive Force: It is force trying to push or contract the body. It is represented by a vector directed towards the body.
3) Reactions at smooth surfaces: The reactions of smooth surfaces, like walls, floors, Inclined planes, etc. will be normal to the surface and pointing towards the body.
4)Forces in Link rods/connecting rods: These forces will be acting along the axis of the rod, either towards or away from the body. (They are either compressive or tensile in nature).
A Few Guidelines for Drawing FBD
5) Forces in Cables (Strings or Chords): These can only be tensile forces. Thus, these forces will be along the cable and directed away from the body.
T = Tension in the cable
Rw = Reaction of the wall
W = self weight of the sphere
P = external load acting on the sphere
Free Body Diagrams of the sphere shown in Fig. 2.4
Fig. 2.5 F B D of Sphere
P = 2kN
30°
θ
Sphere
T
W=0.5kN
Rw
Detach the sphere from all contacts and replace that with forces like:Cable contact is replaced by the force tension = TContact with the smooth wall is replaced by the reaction Rw.
Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium.
There are different types of supports.a) Roller Support b) Hinged or pinned support c) Fixed or built in support
Types of Supports Action on body
(a) Flexible cable ,belt ,chain, rope
BODY
BODY
T
Force exerted by cable is always a tension away from the body in the direction of cable
(b) Smooth surfaces
Contact forces are normal to the surfaces
F F900
900
Supports
(c) Roller support
Contact force is normal to the surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component. (Whose direction is known.)
A
A
Supports
(d) pinned Support / hinged support
This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.
A
RvR
Rh
θ
A
Supports
This type of support not only prevents the translatory movement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.
(e) Fixed or Built-in Support
RAV
MRAH
AA
Supports
TYPES OF BEAMS
A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span (distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam.
(a) Simply supported beam
span
A B
span
A B
TYPES OF BEAMS
B
span
A
(b) Cantilever beam
Rv
MRH A B
If one end or both ends of the beam project beyond the support it is known as overhanging beam.
(c) Overhanging beam (right overhang)
A
A
B
B
TYPES OF BEAMS
Statically determinate beam
Using the equations of equilibrium given in Eq. 2.1,if all the reaction components can be found out, then the beam is a statically determinate beam
∑ Fx = 0; ∑ Fy = 0 ∑M = 0
the equations of equilibrium
FRICTION
Friction is defined as the contact resistance exerted by
one body upon another body when one body moves or tends to move past another body. This force which opposes the movement or tendency of movement is known as frictional resistance or friction. Friction is due to the resistance offered by minute projections at the contact surfaces. Hence friction is the retarding force, always opposite to the direction of motion. Friction has both advantages & disadvantages. Disadvantages ---- Power loss, wear and tear etc. Advantages ---- Brakes, traction for vehicles etc.
Frictional resistance is dependent on the amount of wedging action between the hills and vales of contact surfaces. The wedging action is dependent on the normal reaction N.
NF (Friction)
P
W
Hills & Vales
Magnified Surface
FRICTION
Frictional resistance has the remarkable property of adjusting itself in magnitude of force producing or tending to produce the motion so that the motion is prevented.
When P = 0, F = 0 block under equilibrium
When P increases, F also increases proportionately to maintain equilibrium. However there is a limit beyond which the magnitude of this friction cannot increase.
FRICTION
When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest and such frictional resistance is called the static friction.
Further if P is increased, the value of F decreases rapidly and then remains fairly a constant thereafter. However at high speeds it tends to decrease. This frictional resistance experienced by the body while in motion is known as Dynamic friction OR Kinetic Friction.
FRICTION
Rolling friction friction experienced by a body when it rolls over a surface.
Dynamic Friction
Sliding friction friction experienced when a body slides over another surface.
FRICTION
Where Fmax = Limiting Friction
N= Normal Reaction between the contact surfaces
µ =Coefficient of friction
Note : Static friction varies from zero to a maximum value. Dynamic friction is fairly a constant.
Fmax
N
Fmax
P
W
φ
N R
F α N
Fmax = µN
µ =
FRICTION
Angle of Friction
The angle between N & R depends on the value of F.
This angle θ, between the resultant R and the normal reaction N is termed as angle of friction.
i.e. tanφ = (Fmax )/N = µ
Angle φ is known as Angle of limiting Friction.
Fmax
P
W
φ
NRAs F increases, θ also increases and will
reach to a maximum value of φ when F is Fmax (limiting friction)
FRICTION
Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending.
Angle of reposeWhen granular material is heaped, there exists a limit for the inclination of the surface. Beyond that angle, the grains start rolling down. This limiting angle upto which the grains repose (sleep) is called the angle of repose of the granular material.
FRICTION
Significance of Angle of repose:
The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose.
Angle of repose is numerically equal to Angle of limiting friction
FRICTION
Laws of dry friction
1. The magnitude of limiting friction bears a constant ratio
to the normal reaction between the two surfaces.
(Experimentally proved)
2. The force of friction is independent of the area of contact
between the two surfaces.
3. For low velocities the total amount of friction that can
be developed is practically independent of velocity.
It is less than the frictional force corresponding
to impending motion.
FRICTION
A sphere of 100N weight is tied to a wall by a string as shown in Fig.Q2.1. Find the tension in the string and the reaction of the wall.
Fig Q2.1
Numerical Problems & Solutions(Q 2.1)
150
Using Lami’s theorem,
FBD of sphere
NSinSin
R
NSinSin
T
Sin
R
Sin
W
Sin
T
3.3333333333
3.11133333333
33333333
=×=
=×=
==
W
R
T
16590
105
15
W
R
T
(Q 2.1)
Determine the magnitude and nature of the forces in the bars AB and AC shown in Fig.Q2.2. Neglect size and weight of the pulley.
40 kN
A
B
C
30
60
D
Fig. Q2.2
(Q 2.2) Numerical Problems & Solutions
(Q 2.2)
40 kN
A
B
C
30
60
D
FBD of 40kN weight
T
40 kN
T
If the pulley is frictionless then tension in the rope on either side of it is same
T
T
T
(Q 2.2)
40 kN
A
B
C
30
60
D T
The AB and AC may be subjected to either tension or compression
T
T
T
Hence initially assume one direction
FBD of joint A
FAC
FAB
300
600
T
T
Angle between FAB and F AC = 90 0
The joint A is under equilibrium and hence sum of all forces acting at joint A is zero.
kNCosCosF
TWF
FFrom
AC
XXAC
X
333.3333333333
3
,3
−=−−=∴=−−−
=+ ∑
FAC is –ve , FAC is towards ‘A’, So it is Compressive.
Taking FAC as X-axis and FAB as Y– axis
FAB is +ve. FAB is towards ‘A’, So it is Tensile.
(Q 2.2)
Two cylinders A & B of weight 400N and 200N respectively, rest on smooth planes as shown in Fig.Q2.3. Find the force ‘P’ required for equilibrium.
Fig Q2.3
60
60
45
15
B
A P
(Q 2.3)
Fig Q2.360
60
45
15
45
60 B
A P
RA
RB
60
45B
P
RB
15
FAB
Weight
15A
300FAB
Weight
(Q 2.3)
Fig Q2.3a FBD OF ‘A’
135
6015
45
60
120 105
15
Fig Q2.3b FBD OF ‘B’
RB
YWB=200N
XRA
FBA
FAB
P
WA=400N
(Q 2.3)
Considering FBD of ‘A’ and Using Lami’s theorem,
)33.333111111333
(
3.111111333333
333333333
NSinSin
R
NSinSin
F
Sin
F
Sin
W
Sin
R
A
BA
BAAA
=×=
=×=∴
==
(Q 2.3)
)33.333(
33.111
33.1111)3333(
333.3333333
3
,3
333.3333333333
3
,,3
NR
NP
SinCosP
CosSinRPSin
RPF
FFrom
SinCosRPCos
FWRP
FFrom
B
B
XBXABX
X
B
ABYBBYY
Y
==∴
+=+
=+∴=−−+
→=
−−=−∴=−−+−
+↑=
+∑
∑Considering FBD of ‘B’, We have,
-------Eqn(1)
-----------------Eqn(2)
Adding Eqn(1) and Eqn(2), We get,
(Q 2.3)
Determine, the tension in the strings AB, BC, CD and inclination of the segment CD to the vertical, in the system shown in Fig Q2.4.
(Q 2.4)
50
30
θD
C
B
A
20 kN
30 kN
Fig Q2.4 SPD
50
30
θX
Y
20 kN30 kN
Fig Q2.4a FBD of Joint ‘B’
50
150 +VE
+VE
160
Fig Q2.4b FBD of Joint ‘C’
TCDTCB
TBC
TBA
(Q 2.4)
kNT
CosCosT
FFrom
SinSinT
FFrom
CD
o
CD
Y
CD
X
33.3333.3333.3333.33
tan
3333.1133
,3
3333.33
,3
=∴=
=
−=
↑+=
=
→=
∑
∑ +
θ
θ
θ
θ
Considering FBD of Joint ‘C’, We have,
------Eqn (1)
------Eqn (2)
Dividing Eqn(1) by (2), we get,(NOTE: For this FBD, if we use Lami’s Theorem,we have to expand Sin(50+θ) and solve for θ, which can take more time.)
(Q 2.4)
A wire is fixed at two points A and D as shown in Fig Q 2.5. Determine inclination of the segment BC to the vertical and the tension in all the segments.
60
30
θ
D
C
B
A
20kN
25 kNFig Q2.5 SPD
(Q 2.5)
30
θ20 kN
30 kN
Fig Q2.5a FBD at Joint ‘B’
60
150
(210- θ)
Fig Q2.5 FBD at Joint ‘C’
TCDTCB
TBC
TBA
θ
120
(60+θ)
(180- θ)
(Q 2.5)
Considering FBD of Joint ‘B’ and Using Lami’s theorem,
θθ
θ
θθ
SinSin
T
SinSin
T
Sin
T
SinSin
T
BA
BC
BCBA
×−
=
×−
=
=−
=
)333(
33
111)333(
33333)333(
33
---------Eqn(1)
------------Eqn(2)
(Q 2.5)
Considering FBD of Joint ‘C’ and Using Lami’s theorem,
θθ
θ
θθ
SinSin
T
SinSin
TT
Sin
T
SinSin
T
CD
BCCB
CBCD
×+
=
×+
==
=+
=−
)33(
33
333)33(
33333)33(
33)111(
---------Eqn(3)
---------Eqn(4)
(Q 2.5)
Equating R.H.S. of Eqns (1) and (3), we get,
θθ
θθ
θθ
θθ
CosSin
SinCosCosSin
SinCosCosSin
SinSin
SinSin
33.3333.11
)333333(33.33
)3333(33
333)33(
33333
)333(
33
=
−=
+
×+
=×−
(Continued in next slide)
(Q 2.5)
kNT
kNT
kNTTCos
Sin
CosSin
CD
BA
CBBC
o
33.33
33.33
33.33
33.1133.3333.33
tan
33.3333.33
==
==
=∴==
=
θθθθ
θθ
(Continuation)
(Q 2.5)
A beam AB of span 12m shown in the figure is hinged at A and is on rollers at B. Determine the reactions at A and B for the loading shown.
AB
20kN 25kN 30kN
30° 45°
4m 3m 3m 2m
(Q 2.6)
∑ Fx = 0 HA – 25cos 30 – 30cos45 = 0
∑ Fy = 0 VA – 20 – 25 sin30 – 30sin45 +VB = 0
∑MA = 0 -20×4 - 25 sin30×7 - 30 sin 45×10+ VB ×12=0
Solution
HA B
20kN
25kN 30kN
30° 45°
4m 3m 3m 2mVA VB
FBD of Beam ABFBD of Beam AB
(Q 2.6)
Solution
∑MA = 0
0 = -20×4 - 25 sin30×7 - 30 sin 45×10+ VB ×12
HA B
20kN
25kN 30kN
30° 45°
4m 3m 3m 2m
VA VB
FBD of Beam ABFBD of Beam AB
25 sin 30 30 sin 45
A
(Q 2.6)
RA= 48.21 kN
Solution(contd.)
VA
HA
RA
α( )33AAA VHR +=
HA=42.86kN, VA=22.07kN, VB=31.64kN
α = 27.25°
= −
A
A
H
V3tanα
(Q 2.6)
Find the Support reactions for the given beam loaded as shown in the figure.
(Q 2.7)
60°
2m
40kN/m
A
60kN0.5m
5m
1m
B
[Ans: RB=140kN VA=10
HA=61.24 RA= 62.05kN
α = 9.3°]
60°2m
40kN/m
A60kN
1m
B
RBH=RBCos30
RB
RBv = RBCos60
C
HA
VA
∑ Fx = 0 HA + 60 – RB Cos30 = 0
∑ Fy = 0 VA + RB Cos60 – 40 x 2 = 0
∑ MA = 0 -30 - 40×2×4 + RB Cos60×5 = 0
30kNm
αHA
VA
Solution
FBD
2m
RA
(Q 2.7)
Find the Support reactions for the given beam loaded as shown in the figure.
80kN/m
100kN
3m1m
30kN 0.5m
2m
AB
(Q 2.8)
Solutions
80kN/m
100kN
3m1m
30kN
2m
A
B
VA
HA
VB
15kNm
FBD
(Q 2.8)
[ Ans: VB= 112.5kN VA =37.5kN HA= – 100kN
RA= 106.8kN α = 20.56°]
1m
120kN
A
6 m
C B
15kNm
30kN
2m
RA
α
HA
VA
ΣFx = 0 HA + 100 = 0
∑ Fy = 0 VA + VB – 30 –120 = 0
∑ MA = 0 - 30×2 - 15 - (120)x5 + VBx6 = 0
HA
VA
VB
100kN
FBD(Q 2.8)
Find the Support reactions for the beam loaded as shown in the figure.
3m 2m 2m
20kN23kN30kN
15kN/m
(Q 2.9)
∑ Fx = 0 HA = 0
∑ Fy = 0 VA –45 –30 –23 –20 = 0
∑ MA = 0 MA –45x1.5 –30x3 –23x5 –20x7=0
[ Ans: VA = 118kN MA =412.5kNm]
Solution
2m 2m
20kN23kN30KN45kN
MA
VA
HA
1.5m 1.5mFBD
A
(Q 2.9)
2m 3m 1m 2m
AC B
D
10KN/m20KN/m
Find reactions at A,B,C and D
(Q 2.10)
Solution
2m 3m 1m 2m
AC B
D
10kN/m 10kN/m
10kN/m
Rc
(Q 2.10)
Solution
2m 3m
VA
C
Rc
40kN
20kN
1.33mVD
VB
2.0m
FBD of top beam
FBD of bottom beam
C D
A B
.67m2m
(Q 2.10)
Solution
For top Beam :
∑ Fy = 0; Rc – 40 – 20+VD = 0
∑ MD = 0; - Rc × 6 + 40 × 4 + 20 × 3.33 = 0
Solving the above eqns
RC=37.77kN; VD=22.23kN
20kN
2m
0.6
7
RC VD
3.33m
40kN
(Q 2.10)
For bottom beam :
∑ Fy = 0 VA –37.77–VB=0
∑ MB = 0 -VA× 5 +37.77 ×3=0
Solving the above eqns
VA=22.66kN; VB=15.10kN
2m 3m
RC=37.77kN
VA VB
(Q 2.10)
A ladder of length 5m has a weight of 200N. The foot of the ladder rests on the floor and the top of it leans against the vertical wall. Both the wall and floor are smooth. The ladder is inclined at 60° with the floor. A weight of 300N is suspended at the top of the ladder. Find the value of the horizontal force to be applied at the foot of the ladder to keep it in equilibrium.
(Q 2.11)X
FBD OF LADDER
300N
HB
VA
HA
200N2.
5m
2.5m
600
Solution
(Q 2.11)
∑Fy = 0 VA – 200 – 300=0 ::VA=500N
∑ MA = 0
HB x 5 sin60 – 200 ×2.5 cos 60 – 300 ×5cos60=0
:: HB=230.94N
∑Fx = 0 , HA –HB=0
HA=230.94N(Ans.)
300N
HB
VA
HA
200N
2.5m
2.5m
600
(Q 2.11)
Find the reactions at the supports A and C of the bent 2
0 kN
/m
B C
3m
2m
A
(Q 2.12)
Solution 20
kN
/m
B C
3m
2mVA
HA
FBD
X
Y
VC
(Q 2.12)
Solution (contd.)
B C
3m
2mVA
HA
FBD
60kN∑Fx = 0 60 –HA=0
∑Fy = 0 VA+VC=0
∑MA = 0 VCx2-60 ×1.5=0
VC
(Q 2.12)
Solving the above
Ans: VA = - 45kN
VC = 45 kN
HA = 60kN
3m
FBD after finding reactns
RA=75 kN
36.90
B
C
2m
VAHA
60kN VC
- ve sign for VA indicates, reaction is downwards and not upwards as assumed initially.
(Q 2.12)
A roller (B) of weight 2000N rests as shown in the fig. on beam CD of weight 500N.Determine the reactions at C and D. Neglect the weight of beam AB.
C
D
BA
30°4m
1m
(Q 2.13)
Solution: 2000N
RAB
R BCD
VD
FBD of Roller
D30°
2.5m
1m
300
500N
Hc
Vc
1.5m
FBD of beam CD
C
(Q 2.13)
R BCD
Solution: 2000N
RAB
300
R BCD
FBD of Roller
FBD of Roller :
∑ Fy = 0 RBCD cos 300 –2000=0
∑ Fx = 0 RAB – RBCD sin 300 =0
Solving above eqns : RBCD=2309.4N;
RAB=1154.7N
(Q 2.13)
For bottom beam :
∑ Fy = 0 VD –500+Vc –2309.4cos30=0
∑ MC = 0
-VD × 5cos30 + 500 × 2.5 × cos30-2309.4 × 1=0
Solving the above eqns: VD=783.33N; VC=1716.67N
∑ Fx = 0
2309.4 sin 30 –HC =0 Hc=1154.7 N
FBD of beam CD D
30°
2.5m
1m
300
500N
Hc
Vc
1.5m
2309.4N
VD
(Q 2.13)
Compute the reactions for the bent beam shown in the figure at A and F.
B C D
3m 3m4m
50 N/m
45°AF
300Nm
4m
(Q 2.14)X
Solution ∑ MF = 0 – VA × 14 +200 × 5 – 300=0
VA=50N
∑FX=0 HF=0
∑FY=0 VA +VF = 200;
VF = 200 – 50 =150N
45°
A
B C D
F
3m 3m4m 4mVA
FBD
200 N
2m
HF
VF
300Nm
(Q 2.14)
Determine the support reactions for the shown truss
(Q 2.15) X
4m 4m 4m
A 3KN
G 3KN
F 3KN
B C D E
4m
Solution
4m 4m 4m
A3KN
G 3KN
F 3KN
B
C D EHB
HA
VA
FBD4m
(Q 2.15)
∑ MA = 0 HB × 4 – 3 × 4 – 3 × 8 – 3 × 12=0
HB=18kN
∑FX=0 : –HA+HB=0
HA=18kN
∑FY=0 VA –3 –3 –3=0;
VA=9kN
18kN
9kN26.570
RA=20.12kN
4m 4m 4m
A 3KN3KN
3KN
HB
HA
VA4m
(Q 2.15)
If coefficient of friction is 0.20 between the contact surfacesa) Find the force P just to cause motion to impend up the
planeb) Find the force P just to prevent motion down the plane c) Determine the magnitude and direction of the friction if P = 80N.
P
200N
30°
(Q 2.16)FRICTION
ΣFy = 0
N1 – 200 cos30 = 0 ∴ N1= 173.2 N
F1 = 0.20N1 = 0.20 1173.20 = 34.64N
ΣFx = 0
P – 200sin30 – F1 = 0 ∴ P = 134.64 N
a)
200cos30
Imp. motion
200sin30
P
N1
F 1 = µN 1
= 0.20N 1
P
200N
30°
X + veY +ve
300
(Q 2.16)FRICTION
ΣFx = 0
P – 200 sin30 + 0.20 3173.2 = 0 ∴ P = 65.36 N
(Q 2.16)
b)
200cos30
Imp. motion
200sin30
P
N1
F 1 = µN 1
= 0.20N 1
X + veY +ve
FRICTION
ΣFx = 0
80 – 200Cos60 + F1 = 0 ∴ F1 = 20 N
C) Block will be under rest for the value of P between 134.64 & 65.36N..
Given, P = 80N
Assume direction of friction
(Q 2.16)
200cos30
200sin30
P
N1F 1
X + veY +ve
FRICTION
Compute the magnitude of P that will cause the motion to impend up the plane. Coefficient of friction, μ = 0.20
200Sin30
200Cos30
PF 1
= 0.20N 1
P
200N
30°R
N1
φ
(Q 2.17)FRICTION
200Sin30
200Cos30
P F 1 = 0.20N 1
R
N1
φ
(Q 2.17)
300
NP
PNFy
NPFx
3.333
33cos33333sin3
33cos3333.333cos3
3
3
=−−∑ ==
−∑ −+==
FRICTION
P
R
N1
φ
μN 1OR
tan φ = µ = 0.20
∴ φ = 11.3°
11.30
78.70
200(Q 2.17)
600
33sin3.333sin3.333sin
333 RP ==
P = 175.7
FRICTION
Block A weighing 1000N rests over block B of weight 2000N as shown in fig. Block A is tied to the wall with a horizontal string. If coefficient of friction between A & B is 0.25 and between B and the floor is 0.33, what should be the value of P just to move the block B ?
P
A
B
(Q 2.17) XFRICTION
T
F1 N1
A
RELATIVE MOTION
FBD of Block A
Block A: ∑Fy = 0 ∴ N1 - 1000 = 0 N1 = 1000 N F1 = µ1 N1
= 0.251 1000 = 250 N ∑Fx = 0 F1 – T = 0 ∴ T= 250 N
1000N
F1N1
B
Imp. motionFBD of Block B
P
F2
N2
2000
=1000
X +ve
Y +ve
(Q 2.17)FRICTION
F1
N1
B
Imp. motionFBD of Block B
Block A: ∑Fy = 0 N2 - N1 - 2000 = 0 N2 - 1000 -2000 = 0 N2 = 3000 N ∑Fx = 0 P - F1 -F2 = 0 P- 250 - 0.33 1N2 = 0 P - 250 -0.33 3 3000 = 0 ∴ P = 1250 N
P
F2
N2
2000
=1000
X +ve
Y +ve
(Q 2.17)FRICTION
The bodies shown in the following figure are separated by an uniform strut weighing 100N which is attached to the bodies with frictionless pins. Coefficient of friction under each body is 0.30. Determine the force P that will just start the system rightward. Weight of block A= 400 N, B= 200N
PA
B
30°
45°
(Q 2.18)FRICTION
PA
B
30°
45°
FBD of the Strut
100 N
T100/2 = 50 N
50 N
T
tan φ = 0.30
∴ φ = 16.7°
(Q 2.18)FRICTION
F2
R
200+50
16.70
FBD of B
45°N2
250
R
T
30°
60°61.7°
45+16.70
61.7°
60°58.3°
250
R
T
250/Sin58.3° = T/Sin61.70°
∴ T = 258.72N
T
30
(Q 2.18)FRICTION
FBD of A
30°T = 258.72
N1
F1 = 0.30N1
P
400+50
∑Fy = 0 N1 - 450 - 258.72Sin30 = 0 N1 = 579.36N∑Fx = 0 P - F1- 258.72Cos30 = 0 P-0.30 3579.36-258.72cos30=0 P- 173.81- 224.06=0 ∴ P = 397.87 N
X +ve
Y +ve
(Q 2.18)FRICTION
What horizontal force P is required on the wedges B and C just to raise the weight 1000N resting on A. Angle of limiting friction between all contact surfaces is 10o.
A
B15°
CP P 15°
(Q 2.19) XFRICTION
FBD of A1000N
15°10° 15°
10°
N2
R2
F2F1
R1
N1
1000N
R1
R2
10+15
=25°25°
1000N
25°
25°
130°
R1
R2
1000/Sin130 = R1/ Sin25 = R2/ Sin25 ∴ R1 = R2 = 551.69N
(Q 2.19)FRICTION
FBD of B
10°N3
R3
F3
R1=551.69
R3
10°
65°
35°
80°
R3
P/Sin35 = 551.69/ Sin80
∴ P = 321.32N
P
10°15°
N1
R1 = 551.69
F1 25°
65°
80°
P
R 1=5
51.6
9
P
Note: FBD of Block (C) can also be considered. No need to consider the FBD of both the blocks (B) & (C).
(Q 2.19)FRICTION
Determine the force P required just to start the wedge A shown in the figure. Angle of limiting friction between all contact surfaces is 15°.
P
2000N
A
B500N
θ = 15°
θ
(Q 2.20)FRICTION
FBD of B
500 N
N1
F1 = 0.27N1
N2
2000N
From (1) & (2)500 + 0.27N1 = 3.70N1 – 7407.413.43N1 = 7907.41∴ N1 = 2305.37 N
F2 = 0.27N2
φ = 15°tan φ = µ ∴ µ = 0.27∑Fx = 0 N2 - 500- 0.27N1 = 0 N2 = 500 + 0.27N1 ---------(1)
∑Fy = 0 N1- 2000 - 0.27N2 = 0 0.27N2 = N1 -2000N2= 3.70N1 -7407.41 ------(2)
∴N2=1122.45N
X +ve
Y +ve
(Q 2.20)FRICTION
FBD of Wedge A
15°
75°
F2=0.27N2
=303.06
R2= √ 1122.452 + 303.062 = 1162.64
N2=1122.45
N3
R3
P
R2=1162.64
R2=1162.64
30°
60°
60°
75° 45°
R3
P/Sin45
= 1162.64/ Sin60
∴ P = 949.29N
P
15°φ =15°
P15°
R3
R2=1162.64
F3
(Q 2.20)FRICTION
Determine the minimum value of P to prevent the blocks from slipping. Neglect the weights of the link rods. Co-efficient of friction for all contact surfaces is 0.25.Find the frictional force under the block B and comment on the result.
A
B
Pin Joint
P60°30°
WA= WB=2000N
C
Pin Joints
(Q 2.21)XFRICTION
30°
F1=0.25 N1
N1
2000
FBD of A
T1
∑Fx = 0 N1 - T1 Cos30= 0 N1 = T1 Cos30 = 0.866T1-------(1)
∑Fy = 0 - 2000 + F1 + T1Sin30 = 0 -2000 +0.25N1 + 0.5T1= 0------(2)
From (1) & (2)-2000 + 0.25(0.866T1 )+ 0.5T1 = 0∴ T1 = 2791.32 NX +ve
Y +ve
(Q 2.21)FRICTION
P30°
60°
T2
P/Sin90 = 2791.32/Sin60= T2/Sin30∴ P=3223.14 NT2 = 1611.57N
60°
P
T2
Joint (C)
T1 = 2791.32T1
90°30°
(Q 2.21)FRICTION
60°
T2=1611.57
F2
2000
FBD of B ∑Fy = 0 N2 - 2000 – 1611.57Sin60 = 0 ∴ N2 =3395.60
∑Fx = 0 F2 – 1611.57Cos60 = 0 F2 = 805.79 N (Friction Developed under block B)
Limiting friction = µN2 = 0.25 x 3395.60 = 848.92N
Limiting friction is greater than Friction developed. Hence the block B is at rest.
N2
X +vet
Y +ve
(Q 2.21)FRICTION
An uniform ladder of length 7m rests against a vertical wall with which it makes an angle of 45o. Coefficient of friction between the ladder and the wall is 1/3 and between ladder and the floor is 1/2. If a person whose weight is half that of the ladder ascends it, how high will he be when the ladder just slips?
FB
NB
NA
FA
W
0.5W
45
7ma
7cos45
7sin45
Σ Fx=0
FA-NB=0 ∴0.5NA-NB=0
NB=0.5NA--------(1)ΣFy=0
NA-W-0.5W+FB=0
NA+0.33NB=1.5W------(2)
B
A
3.5m
X +ve
Y +ve
(Q 2.22)FRICTION
From (1) & (2)
NA+0.33(0.5NA)=1.5W
∴NA=1.29W
NB=0.64W
ΣMB=0
(FA× 7sin45)-(NA × 7cos45)+(W × 3.5cos45)+(0.5W × acos45)=0
a = 2m from the top
+ ve moment
(Q 2.22)FRICTION
An uniform ladder 3m in length and weighing 180N is placed against a wall with its end A at the floor and the other end B on the wall, ladder AB making 60° with the floor. Coefficient of friction between the wall and ladder is 0.25 and between floor and ladder is 0.35. In addition to the self weight, the ladder has to support a person weighing 900N at its top B. To prevent slipping, a force P is applied horizontally at A at the level of the floor. Find the minimum force P required for this condition. Find also the minimum angle α at which the above ladder with the person at the top should be placed to prevent slipping without the horizontal force P.
(Q 2.23)FRICTION
FB
FA
180 N
3cosα
3sinα
3m
900N
NB
α
NA
A
B1.
5m
X +ve
Y +ve
P
FBD of Ladder
(Q 2.23)FRICTION
FB
FA
180 N
3cosα
3sinα
3m
900N
NB
α
NA
A
B
1.5m
a) When α = 60°
ΣFx = 0, FA+P-NB=0 0.35NA+P-NB=0 NB=P+0.35NA---------(1)
ΣFy=0, NA-180-900+FB=0 NA+0.25NB=1080----(2)
X +ve
Y +ve
P
(Q 2.23)FRICTION
FB
FA
180 N
3cosα
3sinα
3m
900N
NB
α
NA
A
B
1.5m
X +ve
Y +ve
P
From(1),(2)&(3) NB=499.16N NA=955.21N ∴P=164.80N
ΣMB=0
= 0.35NA 1 3sin60 + P 3 3sin60
+ 180 x 1.5cos60
-NA3 3cos60
= 0 --------(3)
(Q 2.23)FRICTION
(b) Force P is removed, α=?
ΣFx=0
FA-NB=0 0.35NA-NB=0 0.35NA=NB---(1)
ΣFy=0
NA-180-900+FB=0 NA+0.25NB=1080-----(2)
ΣMB=0
0.35NA 33sinα+180 31.5cosα-NA 33cosα=0----(3)
From(1), (2) &(3)
α=68.95°
X +ve
Y +ve
(Q 2.23)FRICTION
Q1. A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kN(T),R=14.98 kN]
C
7kN
5kN
Fig(1)
A
450
300
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q1. A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kN(T),R=14.98 kN]
C
7kN
5kN
Fig(1)
A
450
300
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q2. A 10 kN weight is suspended from a rope as shown in figure. Determine the magnitude and direction of the least force P required to pull the rope, so that, the weight is shifted horizontally by 0.5m. Also, determine, tension in the rope in its new position. [Ans: P= 2.43 kN, θ = 14.480 ; T= 9.7kN.]
2m
10kN
Pθ
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q3. Determine the value of P and the nature of the forces in the bars for equilibrium of the system shown in figure.[Ans: P = 3.04 kN, Forces in bars are Compressive.]
60
75
45 45
P2kN
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q4. A cable fixed as shown in Fig. supports three loads. Determine the value of the load W and the inclination of the segment BC. [Ans: W=25kN, θ = 54.780]
Loads are in kNW
22.5 20
B
C
DA
θ 60
30
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q5. Find the reactions at A,B,C and D for the beam loaded as shown in the figure. (Ans.RA=RB =34kN;RC=28.84kN;
MC=-140kNm ; θC=-33.69 ˚ )
4kN/m
12kN/m12kN/m
4kN/m
20 kN
30kN
1m 2m 1m 1m 2m 1m 1m 2m
A BC
34
40kNm
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q6. A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B. (Ans. T=529.12N;RB=807.15N, θB=64.6˚)
2.5m 2.5m200N
2.5m
A
B
60˚
string
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q7. Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal..
(Ans.x=2m.)
2.0m 1.4m1.0m 3.0m0.6
15kN18kN/m
10kN/m
x
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q8. A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical (Ans:θ =18.44˚)
0.5L 2W
θ
B
L m
WC
A
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q9. For the block shown in fig., determine the smallest
force P required
a) to start the block up the plane
b) to prevent the block moving down the plane.
Take μ = 0.20 P
25°
θ100N
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.:
(a)P
min = 59.2N (b)
Pmin = 23.7N
(b) θ = 11.3o]
Q10. A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If μ between the block and the plane is 0.35, determine the unknown force P for impending motion
(a) to the right
(b) to the left
2000N P800N30°
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.: (a) P = 132.8N (b) P = 1252N]
Q11. Determine value of angle θ to cause the motion of 500N block to impend down the plane, if μ for all contact surfaces is 0.30.
500N
200N
θ = ?
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.: θ = 28.4°]
Q12. A horizontal bar 10m long and of negligible weight rests on rough inclines as shown in fig. If angle of friction is 15o, how close to B may the 200N force be applied before the motion impends.
100N 200N
2 mX = ?
A B
30° 60°
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.: x = 3.5m]
Q13. Determine the vertical force P required to drive the wedge B downwards in the arrangements shown in fig. Angle of friction for all contact surfaces is 12o.Weight of block A= 1600 N.
A
B
P
20°
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.: P = 328.42N]
Q14. Determine the force P which is necessary to start the wedge to raise the block A weighing 1000N. Self weight of the wedge may be ignored. Take angle of friction, φ = 15o for all contact surfaces.
20°
A
Pwedge
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.: P = 1192N]
Q15. A ladder of weight 200N, 6m long is supported as shown in fig. If μ between the floor and the ladder is 0.5 & between the wall and the ladder is 0.25 and it supports a vertical load of 1000N, determine a) the least value of α at which the ladder may be placed without slippingb) the reactions at A & B
α
1000N
5m
A
B
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]
Q16. An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. μ between the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor?
Smooth wall
12m
5mA
B
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
[Ans.: FA = 52 N]
Q17. A ladder of length 5m weighing 500N is placed at 45o against a vertical wall. μ between the ladder and the wall is 0.20 & between ladder and ground is 0.50. If a man weighing 600N ascends the ladder, how high will he be when the ladder just slips. If a boy now stands on the bottom rung of the ladder, what must be his least weight so that the man can go to the top of the ladder.
[Ans.: (a) x = 2.92m (b) Wboy = 458N]
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS