When some external forces (which may be concurrent or parallel) are acting on a stationary body, the body may start moving or may start rotating about any point. But if the body does not start moving and also does not start rotating about any point, then the body** is said to be in equilibrium.

**1.12.1. Principle of Equilibrium**. The principle of equilibrium states that, a stationary body which is subjected to coplanar forces (concurrent or parallel) will be in equilibrium if the algebraic sum of all the external forces is zero and also the algebraic sum of moments of all the external forces about any point in their plane is zero. Mathematically, it is expressed by the equations :

ΣF = 0

ΣM =0

The sign l: is known as sigma which is a Greek letter. This sign represents the algebraic sun of forces or moments.

The equation (1.22) is also known as force law of equilibrium whereas the equation (1.23) is known as moment law of equilibrium.

The forces are generally resolved into horizontal and vertical components. Hence equation (1.22) is written as

Σ F _{x} =0

Σ F _{y} =0

where ΣF = Algebraic sum of all horizontal components

and ΣF = Algebraic sum of all vertical components.

**1.12.2. Equations of Equilibrium for Non-concurrent Force Systems**. A non-concurrent force systems will be in equilibrium if the resultant of all forces and moment is zero.

Hence the equations of equilibrium are

ΣF _{x} = 0, ΣF _{y} = 0 and ΣM = 0

**1.12.3. Equations of Equilibrium for Concurrent Force System**. For the concurrent forces, the lines of action of all forces meet at a point, and hence the moment of those force about that very point will be zero or ΣM = 0 automatically.

Thus for concurrent force system, the condition I.M = 0 becomes redundant and only two conditions, i.e., ΣF_{ x }= 0 and ΣF_{Y} = 0 are required.

**1.12.4. Force Law of Equilibrium**. Force law of equilibrium is given by equation (1.22) or by equations (1.24) and (1.25). Let us apply this law to the following important force system:

(i) Two force system (ii) Three force system

(iii) Four force system.

**1.12.5. Two Force System**. When a body is subjected to two forces, then the body will be in equilibrium if the two forces Fig. 1. 71 are collinear, equal and opposite as shown in Fig. 1.71.

If the two forces acting on a body are equal and opposite but are parallel, as shown in Fig. 1.71(a), then the body will not be in equilibrium. This is due to the fact that the three conditions of equilibrium will not be satisfied. This is proved as given below :

(i) Here Σ F _{y} = 0 as there is no horizontal force acting on the body. Hence first condition of equilibrium is satisfied.

(ii) Also here ΣF_{Y} = 0 as F _{1} = F_{2 }.

Hence second condition of equilibrium is also satisfied.

(iii) ΣM about any point should be ·zero. The resultant moment about point A is given by · Fig. 1.71(a)

M _{A} = – F_{2} x AB (- ve sign is due to clockwise moment)

But M_{A} is not equal to zero. Hence the third condition is not satisfied.

Hence a body will not be in equilibrium under the action of two equal and opposite parallel forces.

Two equal and opposite parallel forces produce a couple and moment of the couple is – F 1 x AB [Fig. 1.71(a)].

**1.12.6. Three Force System**. The three forces acting on a body which is in equilibrium may be either concurrent or parallel. Let us first consider that the body is in equilibrium when three forces, acting on the body, are concurrent. This is shown in Fig. 1.72.

(a) When three forces are concurrent. The three concurrent forces F_{1}‘ F_{2} and F_{3} are acting on a body at point 0 and the body is in equilibrium. The resultant of F _{1} and F_{2} is given by R. If the force F_{3} is collinear, equal and opposite to the resultant R, then the body will be in equilibrium. The forceF_{3} which is equal and opposite to the resultant R is known as equilibrant. Hence for three concurrent forces acting on a body when the body is in equilibrium, the resultant of the two forces should be equal and opposite to the third force.

and F_{3} are acting and the body is in equilibrium. If three forces F_{1} F_{2} and F_{3} are acting in the same direction, then there will be a resultant R =F_{1} +F_{2 }+ F_{3} and body will not be in equilibrium. The three forces are acting in opposite direction and their magnitude is so adjusted that there is no resultant force and body is in equilibrium. Let us suppose that F2 is acting in opposite direction as shown in Fig. 1.73.

Now let us apply the three conditions of equilibrium:

(i) ΣF x = 0 as there is no horizontal force acting on the body

(ii) ΣF y.= 0 i.e., F1 + F3 = F2

(iii) ΣM = 0 about any point .

Taking the moments of F_{1}F_{2} and F_{3} about point A,

ΣM_{A} = -F_{2} x AB +F_{3} x AC

(Moment of F3 is anti-clockwise whereas moment of F2 is clockwise)

For equilibrium, Σ.MA should be zero

i.e., -F_{2 }x A_{B} + F_{3} x AC = 0

If the distances AB and AC are such that the above equation is satisfied, then the body will be in equilibrium under the action of three parallel forces.

**1.12.7. Four Force System**. The body will be in equilibrium if the resultant force in horizontal direction is zero (i.e., Σ F x = 0), resultant force in vertical direction is zero (i.e., r. F. = 0) and moment of all forces about any point in the plane of forces is zero (i.e., r. M = 0). >

**Problem 1.32.** Two forces F1 and F2 are acting on a body and the body is in equilibrium. If the magnitude of the force F1 is 100 N and its acting at 0 long x-axis as shown in Fig. 1. 74, then determine the magnitude and direction of force F2.

**Sol.** Given:

Force, F1 = 100 N

The body is in equilibrium under the action of two forcesF1 andF2.

When two forces are acting on a body and the body is in equilibrium, then the two forces should be collinear, equal and opposite.

F _{2} = F_{1} = 100 N

The force F_{2} should pass through 0, and would be acting in the opposite direction of F_{1}

**Problem 1.33**. Three forces F_{1}, F_{2} and F_{3} are acting on a body as shown in Fig. 1. 75 and the body is in equilibrium. If the magnitude of force F_{3} is 400 N, find the magnitudes of force F_{1} and F_{2}.

**Sol.** Given:

Force, F _{3} = 400 N

As the body is in equilibrium, the resultant force in

x-direction should be zero and also the resultant force in

y-direction. should be zero.

(i) for Σ F _{x} = 0

F_{1} cos 30^{o} + F_{2} cos 30^{o} = 0

F_{1} – F_{2 }= 0

ΣF_{Y} = 0 , we get

(ii) for Σ F _{y} = 0 , we have

F_{1} sin 30^{o} + F_{2} sin 30^{o }– 400 = 0

F_{1} x 0.5 + f_{2} x 0.5 = 400

or f_{1} x 0.5 + f_{2} x 0.5 = 400

or f_{1} = 400 N

also f_{2} = f1 = 400 N.

** **

** **

**2nd Method**

If three forces are acting on a body at a point and the body is in equilibrium, Lame’s Theorem can be applied.

Using Lame’s theorem,

F_{1} / sin 120^{o} = f_{2} / sin 120^{ o} = 400 / sin 120^{ o}

F_{1} = F_{2} = 400 N.

**Problem ****1.39.***Three parallel forces F _{1} F_{2} and F_{3} are acting on a body as shown in*1.76

Fig.

*and the body is in equilibrium. Ifforce F*=

_{1}*250*

*Nand F*=

_{3}*1000*

*N and the distance*

between F=

between F

_{1}and F_{2}*1.0*

*m, then determine the magnitude offorce F*

force F

_{2}and the distance ofF_{2}fromforce F

_{3}.**Sol. **Given:

Force, F_{1 }*=*250 N

Force, *F _{3}=*1000 N

Distance *AB =*1. 0 m

The body is in equilibrium.

Find *F *2 and distance *BC. *

For the equilibrium of the body, the resultant force in the vertical direction should be zero (here there

is no force in horizontal direction).

For*Ʃ**F** _{y =}*0, we get

* F _{1}*

*+ F*0

_{3}– F2 =or 250 + 1000 *-F _{2}=*0

*or F _{2}*= 250 + 1000 =

**1250 N. Ans.**

For the equilibrium of the body, the moment of all forces about any point must be zero:

Taking moments of all forces about point *A *and considering distance *BC *= *x, *we get

or (AC = AB + BC = 1 + x)

or 1250 – 1000 – 1000x = 0

or 250 = 1000x

or X=

**Problem 1.40.***The **five forces F _{1}, F_{2}, F_{3}, F_{4} and F_{5}are acting at a point on a body as*1.77

shown in Fig.

*and the body is in equilibrium. IfF*= 18

_{1}*N, F*= 22.5

_{2}*N, F*= 15

_{3}*Nand F*

_{4}*=*

*30*

*N, find the force F*

_{5}in magnitude and direction.* *

**Sol. **Given:

Forces, *F _{1}*= 18 N,

*F*=2.25N,

_{2}*F*= 15 N and

_{3}*F*= 30 N.

_{4}

The body is in equilibrium. Find force F_{5}in magnitude and direction. This problem can

be solved analytically and graphically.

1. **Analytical method **

Let θ= Angle made by force F_{5}with horizontal axis *O-X’. *

As the body is in equilibrium, the resultant forcein x-direction and y-direction should be zero.

*(i) *For *Ʃ**F** _{x}*= 0, we get

*F _{1}*+

*F*cos 45

_{2},^{0}

*– F*cos 30

_{4}^{0}

*-F*cas θ= 0

_{5}or 18 + 22.5 x 0.707 – 30 x 0.866

*-F*cos θ= 0

_{5}or 18 + 15.9 – 25.98

*-F*cos θ= 0

_{5}*or F _{5}*cos θ= 18 + 15.9 – 25.98

*F*case = 7.92 ……(i)

_{5}*(ii) *For *Ʃ**F*= 0, we get

F_{2}sin 45^{0}+ *F _{3}*–

*F*sin 30

_{4}^{0}–

*F*sin θ= 0

_{5}or 22.5 x 0.707 + 15 – 30 x 0.5 *-F _{5}*sin θ= 0

or 15.9 + 15 – 15 – *F _{5}*sin θ= 0

*or F _{5}*sin θ= 15.9

Dividing equation (ii) byequation (i), we get =or tan θ= 2.0075

θ= tan^{-1} 2.0075 = **63.52°. ****Ans.**

Substituting the value of e in equation (i), we get

*F _{5}*cos 63.52° = 7.92

* F5 *= = **17.76 N. Ans. **

** **

2. **Graphical method **

(*i*) First draw a space diagram with given four forces F_{1}, F_{2}F_{3}and F_{4}at correct angles as shown in Fig. *1.78(a). *

*(ii) *Now choose a suitable scale, say 1 cm= 5 N for drawing a force diagram. Take any point O in the force diagram as shown in Fig. *1.78(b). *

*(iii) *Draw line *Oa*parallel to force *F _{1}*and cut

*Oa*=

*F*= 18 N to the same scale.

_{l}

*(iv) *From*a, *draw the line *ab*parallel to *F _{2}*and cut

*ab*=

*F*= 22.5 N

_{2}*(v) *From *b, *draw the line *be *parallel to *F _{3}*and cut

*be*=

*F*= 15 N

_{3}* *

*(vi) *From c, draw the line *cd *parallel to *F _{4}*and cut

*cd*=

*F*= 30 N

_{4}* *

*(vii) *Now join *d *to O. Then the closing side *dO*represents the force *F _{5}*in magnitude and direction. Now measure the length

*dO.*

By measurement, length *dO*= 3.55 em.

Force *F _{5}*= Length

*dO*x Scale = 3.55 x 5 =

**17.75 N. Ans.**

The direction is obtained in the space diagram by drawing the force *F _{5}*parallel to line

*dO.*

Measure the angle θ, which is equal to 63.5°. Or the force *F _{5}*is making an angle of

180 + 63.5 = 243.5° with the force

*F*

_{1}.**Problem ****1.41.***Fig. *1.*78(c) shows the coplanar system of forces acting on a flat plate. *

*Determine: **(i) **the resultant and **(ii) x **and y intercepts of the resultant. *

**Sol. **Given:

Force at *A *= 2240 N

Angle with x-axis = 63.43°

Force at *B =* 1805 N

Angle with x-axis = 33.67°

Force at C = 1500 N

Angle with x-axis = 60°

Lengths *OA =* 4 m

*DB = 3m*

*DC = 2m*

*OD = 3m. *

and

Each force is resolved into *X *and *Y *components as shown in Fig. 1. *78(d).*

*(i) *Force *at A*= 2240 N

Its X-component = 2240 × cos 63.43° = 1001.9 N

Its *Y-component =* 2240 × sin 63.43° = 2003.4 N

*(ii) *Force at *B =* 1805 N.

Its X-component = 1805 × cos 33.67° = 1502.2 N

Its *Y-component *= 1805 × sin 33.67° = 1000.7 N

* (iii) *Force at C = 1500 N.

Its X-component = 1500 × cos 60° = 750 N

Its *Y-component *= 1500 × sin 60° = 1299 N

The net force along X-axis,

*R _{x} *= Ʃ

*LF*= 1001.9 – 1502.2 – 750 = – 1250.3 N

_{x }The net force along Y-axis,

*R _{y}*=

*ƩLF*= – 2003.4 – 1000.7 + 1299 = – 1705.1 N

_{y}* *

*(i) The resultant force is given by,*

R= √(R_(x^2 )+R_(y^2 ) )= √((-1250.3)^2+(-1750.1)^2 )

= √(1563250+2907366)=2114.4 N. Ans

The angle made by the resultant with x-axis is given by

The net moment* about point O,

*M _{o} *= 2003.4 × 4 + 1000.7 × 3 -1299 × 2 -1502.2 × 3 – 750 × 3

= 8012.16 + 3002.1- 2598 – 4506.6 – 2250

= 11014.26 – 9354.6 = 1659.55 Nm

As the net moment about O is clockwise, hence the resultant must act towards right of origin O, making an angle = 53.7° with x-axis as shown in Fig. 1. *78(e).*The components R_{x} and R_{y }are also negative. Hence this condition is also satisfied.

* *

*(ii) Intercepts of resultant on x-axis and y-axis *[Refer to Fig. *1.78(e)]. *

Let *x= *Intercept of resultant along x-axis.

*y *= Intercept of resultant along y-axis.

The moment of a force about a point is equal to the sum of the moments of the components of the force about the same point. Resolving the resultant *(R) *into its component *R _{x} *and

*R*at

_{y}*F.*

Moment of *R *about 0 = Sum of moments of *R _{x}* and R

_{y}at O

*Considering clockwise moment positive and anti-clockwise moment as negative. At *A, *the X component of 1001.9 N passes through O and hence has no moment.

But moment of *R *about O = 1659.66 (M_{o} = 1659.66)

1659.66 = *Rx *× O + *R _{y}* ×

*x*

(as *R _{x} *at

*F*passes through O hence it has no moment)

1659.66 = 1705.1 × *x*

To find y-intercept, resolve the resultant *R *at G into its component *R _{x} *and

*R*

_{y}.Moment of *R *about O = Sum of moments of *R _{x} *and

*R*at O

_{y}or 1659.66 = R_{x}* × y *+ *R _{y} *× O.

(At G, *R _{y} *passes through O and hence has no moment)

1659.66 = 1250.3 × *y *

**Problem 1.42**. *A lamp weighing *5 *N is suspended from the ceiling by a chain. It is
pulled aside by a horizontal cord until the chain makes an angle of 60° with the ceiling as
shown in Fig. *1.79.

*Find the tensions in the chain and the cord by applying Lame’s theorem and*

also by graphical method.

also by graphical method.

**Sol.** Given:

Weight of lamp= 5 N

Angle made by chain with ceiling = 60°

Cord is horizontal as shown in Fig. 1.79.

*(i) ***By ****Lame’s theorem **

Let *T _{1} *= Tension (or pull) in the cord

*T*= Tension (or pull) in the chain.

_{2}Now from the geometry, it is obvious that angles between *T _{1}* and lamp will be 90°, between lamp and

*T*150° and between

_{2}*T*and

_{2}*T*

_{1 }*120°.*[Refer to Fig.

*1.79(b)*]

*.*

Applying Lame’s theorem, we get

* *

* (ii) ***By Graphical method **

(1) First draw the space diagram at correct angles as shown in Fig. *1.79(b). *Now choose

a suitable scale say 1 cm= 1 N for drawing a force diagram as shown in Fig. 1.79(c). Take any

point O in the force diagram.

(2) From O, draw the line *Oa *vertically downward to represent the weight of the lamp. Cut Oa = 5N.

(3) From *a, *draw the line *ab *parallel to *T _{2.} *The magnitude

*of T*is unknown. Now from

_{2}0, draw the line

*Ob*horizontally

*(i.e.,*parallel to

*T*

_{1}*)*cutting the line

*ab*at point

*b.*

(4) Now measure the lengths *ab *and *bO). *

Then *ab *represents *T _{2} *and

*bO*represents

*T*By measurements,

_{1}.*ab*= 5.77 cm and

*bO*

= 2.9 cm.

Pull in the cord = *bO*= 2.9 cm × scale = 2.9 × 1

= **2.9 N. Ans. **

Pull in the chain *= **ab*= 5.77 cm × scale = 5.77 × 1

= **5.77 N. Ans. **

**Problem 1.43.***On a horizontal line PQRS *12 c*m long, where PQ *= *QR *= *RS *= 4 c*m,
forces of *

*1000*

*N,*

*1500*

*N,*

*1000*

*N and*

*500*

*N are acting at P, Q, Rand*S

*respectively, all*

downwards, their lines of action making angles of

downwards, their lines of action making angles of

*90°, 60°,*45°

*and*

*30°*

*respectively with PS.*

Obtain the resultant of the system completely in magnitude, direction and position graphically

and check the answer analytically.

Obtain the resultant of the system completely in magnitude, direction and position graphically

and check the answer analytically.

** **

**Sol. **Given:

*PQ *= *QR *= *RS *= 4 cm

Force at *P *= 1000 N. Angle with *PS *= 90°

Force at *Q *= 1500 N. Angle with *QS *= 60°

Force at *R *= 1000 N. Angle with *RS *= 45°

Force at S = 500 N. Angle with *PS *= 30°

**Graphical method **

Draw the space diagram of the forces as shown in Fig. *1.79(d). *The procedure is as follows:

*(i)* Draw a horizontal line *PQRS *= 12 cm in which take *PQ *= *QR *= *RS *= 4 cm.

*(ii) *Draw the line of action *of forces P, **Q, **R, S *of magnitude 1000 N, 1500 N, 1000 N and

500 N respectively at an angle of 90°, 60°, 45° and 30° respectively with line *PS *as shown in

Fig.1.79.

** **

**Magnitude and Direction of Resultant Force (R*) **

To find the magnitude and direction of the resultant force, the force diagram is drawn

as shown in Fig. 1. *79(e) *as given above:

*(i) *Draw the vector *ab *to represent the force 1000 N to a scale of 1 cm= 500 N. The

vector *ab *is parallel to the line of action of force *P. *

* *

*(ii) *From point *b, *draw vector *bc *= 1500 N and parallel to the line of action of force *Q. *Similarly the vectors, *cd *= 1000 N and parallel to line of action of force *R *and *de *= 500 N and

parallel to the line of action of force *S, *are drawn.

*(iii) *Join *ae *which gives the magnitude of the resultant. Measuring *ae, *the resultant force is equal to 3770 N.

*(iv) *To get the line of action of the resultant, choose any point O on force diagram (called

the pole) and join *Oa, Ob, Oc, Od *and *Oe. *

*(v) *Now choose any point *X; *on the line of action of force *P *and draw a line parallel to Oa.

*(vi) *Also from the point X_{1} draw another line parallel to *Ob, *which cuts the line of action

of force *Q *at *X _{2}. *Similarly from point

*X*draw a line parallel to Oc to cut the line of action of

_{2},force

*R*at X

_{3}. From point X

_{3}, draw a line parallel to

*Od*to cut the line of action of force

*S*atX

_{4}.

*(vii) *From point X_{4}, draw a line parallel *Oe. *

*(viii) *Produce the first line *(i.e., *the line from X_{l} and parallel to *Oa) *and the last line *(i.e.,
*the line from X

_{4}and parallel to

*Oe)*to interest at.X, Then the resultant must pass through this

point.

*(ix) *From point X, draw a line parallel *to ae *which determines the line of action of resultant force. Measure *PX. *By measurements:

Resultant force, *R* *= 3770 N

Point of action, *PX *= 4.20 cm

Direction, θ = 60° 30′ with *PS.*

**Analytical method **

In analytical method, all the forces acting can be resolved horizontally and vertically. Resultant of all vertical and horizontal forces can be calculated separately and then the final

resultant can be obtained.

Resolving all forces and considering the system for vertical forces only.

Vertical force at *P *= 1000 N

Vertical force at *Q *= 1500 sin 60° = 1299 N

Vertical force at *R *= 1000 sin 45° = 707 N

Vertical force at *S *= 500 sin 30° = 250 N

Let *Rv**= the resultant of all vertical forces and acting at a distance *x cm *from *P. *

= 1000 + 1299 + 707 + 250 = 3256 N

Taking moments of all vertical forces about point *P, *

*Rv* *× *x *= 1299 × 4 + 707 × 8 + 250 × 12 = 13852

Now consider the system for horizontal forces only, horizontal force *at P*= 0

Horizontal force at *Q *= 1500 × cos 60° = 750 N

Horizontal force at *R *= 1000 × cos 45° = 707 N

Horizontal force at *S *= 500 × cos 30° = 433 N

Resultant of all horizontal forces will be,

R_{H}* = 0 + 750 + 707 + 433 = 1890 n

The resultant R* of R_{V}* and R_{H}* will also pass through point X which is at a distance of 4.25 cm from P.

The resultant will make an angle θ with *PS *and is given by

θ = tan^{-1} 1.723 = 59.9°

Thus the resultant of 3764 N makes an angle 59.9° with *PS *and passing through point

X which is at a distance of 4.25 cm from point *P.*

This result confirms closely with the values obtained by graphical method.