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

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


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


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


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


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


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


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


[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°


[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 = ?


[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°


[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°


[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


[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


[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


[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]


89573289 2 equilibrium of force system

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