equilibrium of force system

Definition:-If a system of forces acting on a body, keeps the body in astate of rest or in a state of uniform motion along a straightline, 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:

i) 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.

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 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

F1F5

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,

LAMIS THEOREM

SinF

SinF

SinF 321

Fig. 2.3

F1

F3 F2

Note: While using Lamis 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 physicalconditions of the problem, like, the nature of supportsprovided; size, shape and location of various bodies; forcesapplied 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 = 2kN30

Fig. 2.4 SPDwall

3 m

Sphere

Free Body Diagram (FBD) :

It is an isolated diagram of the body being analyzed (calledfree body), in which, the body is shown freed from all itssupports and contacting bodies/surfaces. Instead of thesupports and contacting bodies/surfaces, the reactiveforces exerted by them on the free body is shown, alongwith all other applied forces.

1) Tensile Force: It is a force trying to pull or extend the body.

2) Compressive Force: It is force trying to push or contract 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.

Fig. 2.5 F B D of Sphere

P = 2kN

30

Sphere

T

W=0.5kN

Rw

Weight of sphere = 0.5 kN,Radius = 1m

Cable

P = 2kN

30

Fig. 2.4 SPDwall

3 m

Sphere

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

BODYT

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

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. 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

MRAHA A

Supports

Types of support

Hinge and roller support

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

Types of beam

Cantilever, fixed, continuous and overhang beams

Types of load

Statically determinate beam

Using the equations of equilibrium, 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 byone body upon another body when one body moves or tendsto move past another body. This force which opposes themovement or tendency of movement is known as frictionalresistance or friction. Friction is due to the resistance offered byminute projections at the contact surfaces. Hence friction is theretarding 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 wedgingaction between the hills and vales of contact surfaces. Thewedging action is dependent on the normal reaction N.

NF (Friction)

P

W

Hills & Vales Magnified Surface

FRICTION

Frictional resistance has the remarkable property ofadjusting itself to magnitude of force producing or tendingto produce the motion so that the motion is prevented.

When P = 0, F = 0 block is under equilibrium

When P increases, F also increases proportionately tomaintain equilibrium. However there is a limit beyondwhich 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 rapidlyand then remains fairly a constant thereafter. However athigh speeds it tends to decrease. This frictional resistanceexperienced by the body while in motion is known asDynamic friction OR Kinetic Friction.

FRICTION

Rolling friction friction experienced bya body when it rolls over a surface.

Dynamic FrictionSliding friction friction experienced when a body slides over another surface.

FRICTION

Where Fmax = Limiting Friction

N= Normal Reaction between the contact surfaces

=Coefficient of friction

Note : Static friction varies from zero to a maximum value. Dynamicfriction is fairly a constant.

FmaxN

Fmax

P

W

N R

F N

Fmax = N

=

FRICTION

Angle of Friction

The angle between N & R dependson the value of F.

This angle , between the resultantR and the normal reaction N istermed as angle of friction.

i.e. tan = (Fmax )/N =

Angle is known as Angle of limiting Friction.

Fmax

P

W

N RAs F increases, also increases and willreach to a maximum value of when F isFmax (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 theinclination of the surface. Beyond that angle, the grains startrolling 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 ratioto the normal reaction between the two surfaces.(Experimentally proved)

2. The force of friction is independent of the area of contactbetween the two surfaces.

3. For low velocities the total amount of friction that canbe developed is practically independent of velocity.It is less than the frictional force correspondingto impending motion.

FRICTION

An uniform ladder 3m in length and weighing 180N is placedagainst a wall with its end A at the floor and the other end Bon the wall, ladder AB making 60 with the floor. Coefficientof friction between the wall and ladder is 0.25 and betweenfloor and ladder is 0.35. In addition to the self weight, theladder has to support a person weighing 900N at its top B.To prevent slipping, a force P is applied horizontally at A atthe level of the floor. Find the minimum force P required forthis condition. Find also the minimum angle at which theabove ladder with the person at the top should be placed toprevent slipping without the horizontal force P.

(Q 2.23)FRICTION

FB

FA

180 N

3cos

3sin

900N

NB

NA

A

B

X +ve

Y +ve

P

FBD of Ladder

(Q 2.23)FRICTION

Q1. A 10kN roller rests on a smooth horizontal floor and isheld by the bar AC as shown in Fig(1). Determine themagnitude and nature of the force in the bar AC and reactionfrom the floor under the action of the forces applied on theroller.

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 infigure. Determine the magnitude and direction of the least force Prequired to pull the rope, so that, the weight is shifted horizontallyby 0.5m. Also, determine, tension in the rope in its new position.

2m

10kN

P

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

7545 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, = 54.780]

Loads are in kNW

22.520

B

C

DA

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 ; C=-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, B=64.6)

2.5m200N

2.5m

A

B60

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: =18.44)

0.5L 2W

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 planeb) to prevent the block moving down the plane.

Take = 0.20

P

25

100N


[Ans.:(a) Pmin = 59.2N

(b) Pmin = 23.7N(b) = 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 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 to cause the motion of500N block to impend down the plane, if for all contactsurfaces is 0.30.

500N

200N

= ?


[Ans.: = 28.4]

Q12. A horizontal bar 10m long and of negligible weightrests on rough inclines as shown in fig. If angle of frictionis 15o, how close to B may the 200N force be appliedbefore the motion impends.

100N 200N

2 m X = ?A B

30 60


[Ans.: x = 3.5m]

Q13. Determine the vertical force P required to drive thewedge B downwards in the arrangements shown in fig.Angle of friction for all contact surfaces is 12o.Weight ofblock A= 1600 N.

AB

P

20


[Ans.: P = 328.42N]

Q14. Determine the force P which is necessary to startthe wedge to raise the block A weighing 1000N. Selfweight of the wedge may be ignored. Take angle of friction, = 15o for all contact surfaces.

20

A

Pwedge


[Ans.: P = 1192N]

Q15. A ladder of weight 200N, 6m long is supported as shownin fig. If between the floor and the ladder is 0.5 & betweenthe wall and the ladder is 0.25 and it supports a vertical load of1000N, determinea) the least value of at which the ladder may be placedwithout slippingb) the reactions at A & B

1000N

5m

A

B


[Ans.: (a) = 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. 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 45oagainst a vertical wall. between the ladder and the wallis 0.20 & between ladder and ground is 0.50. If a manweighing 600N ascends the ladder, how high will he bewhen the ladder just slips. If a boy now stands on thebottom rung of the ladder, what must be his least weightso that the man can go to the top of the ladder.[Ans.: (a) x = 2.92m (b) Wboy = 458N]


equilibrium of force system

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