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
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 ΣFY = 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 ΣFY = 0 as F 1 = F2 .
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 = – F2 x AB (- ve sign is due to clockwise moment)
But MA 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 F1‘ F2 and F3 are acting on a body at point 0 and the body is in equilibrium. The resultant of F 1 and F2 is given by R. If the force F3 is collinear, equal and opposite to the resultant R, then the body will be in equilibrium. The forceF3 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 F3 are acting and the body is in equilibrium. If three forces F1 F2 and F3 are acting in the same direction, then there will be a resultant R =F1 +F2 + F3 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 F1F2 and F3 about point A,
ΣMA = -F2 x AB +F3 x AC
(Moment of F3 is anti-clockwise whereas moment of F2 is clockwise)
For equilibrium, Σ.MA should be zero
i.e., -F2 x AB + F3 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.
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 = F1 = 100 N
The force F2 should pass through 0, and would be acting in the opposite direction of F1
Problem 1.33. Three forces F1, F2 and F3 are acting on a body as shown in Fig. 1. 75 and the body is in equilibrium. If the magnitude of force F3 is 400 N, find the magnitudes of force F1 and F2.
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
F1 cos 30o + F2 cos 30o = 0
F1 – F2 = 0
ΣFY = 0 , we get
(ii) for Σ F y = 0 , we have
F1 sin 30o + F2 sin 30o - 400 = 0
F1 x 0.5 + f2 x 0.5 = 400
or f1 x 0.5 + f2 x 0.5 = 400
or f1 = 400 N
also f2 = f1 = 400 N.
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,
F1 / sin 120o = f2 / sin 120 o = 400 / sin 120 o
F1 = F2 = 400 N.
Problem 1.39.Three parallel forces F1 F2 and F3 are acting on a body as shown in
Fig. 1.76 and the body is in equilibrium. Ifforce F1= 250 Nand F3= 1000 N and the distance
between F1 and F2= 1.0 m, then determine the magnitude offorce F2 and the distance ofF2 from
Force, F1 =250 N
Force, F3=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).
ForƩFy =0, we get
F1+ F3- F2 = 0
or 250 + 1000 -F2=0
or F2= 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
Problem 1.40.The five forces F1, F2, F3, F4 and F5are acting at a point on a body as
shown in Fig. 1.77 and the body is in equilibrium. IfF1= 18 N, F2= 22.5 N, F3= 15 Nand F4= 30 N, find the force F5 in magnitude and direction.
Forces, F1= 18 N, F2=2.25N,
F3= 15 N andF4= 30 N.
The body is in equilibrium. Find force F5in magnitude and direction. This problem can
be solved analytically and graphically.
1. Analytical method
Let θ= Angle made by force F5with horizontal axis O-X’.
As the body is in equilibrium, the resultant forcein x-direction and y-direction should be zero.
(i) For ƩFx= 0, we get
F1+ F2, cos 450- F4cos 300-F5cas θ= 0
or 18 + 22.5 x 0.707 – 30 x 0.866 -F5cos θ= 0
or 18 + 15.9 – 25.98 -F5cos θ= 0
or F5cos θ= 18 + 15.9 – 25.98
F5case = 7.92 ……(i)
(ii) For ƩF= 0, we get
F2sin 450+ F3- F4sin 300 – F5sin θ= 0
or 22.5 x 0.707 + 15 – 30 x 0.5 -F5sin θ= 0
or 15.9 + 15 – 15 – F5sin θ= 0
or F5sin 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
F5cos 63.52° = 7.92
F5 = = 17.76 N. Ans.
2. Graphical method
(i) First draw a space diagram with given four forces F1, F2F3and F4at 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 Oaparallel to force F1and cut Oa= Fl= 18 N to the same scale.
(iv) Froma, draw the line abparallel to F2and cut ab= F2= 22.5 N
(v) From b, draw the line be parallel to F3and cut be = F3= 15 N
(vi) From c, draw the line cd parallel to F4and cut cd = F4= 30 N
(vii) Now join d to O. Then the closing side dOrepresents the force F5in magnitude and direction. Now measure the length dO.
By measurement, length dO= 3.55 em.
Force F5= Length dOx Scale = 3.55 x 5 = 17.75 N. Ans.
The direction is obtained in the space diagram by drawing the force F5parallel to line
Measure the angle θ, which is equal to 63.5°. Or the force F5is making an angle of
180 + 63.5 = 243.5° with the force F1.
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.
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.
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,
Rx = ƩLFx = 1001.9 – 1502.2 – 750 = – 1250.3 N
The net force along Y-axis,
Ry= ƩLFy= – 2003.4 – 1000.7 + 1299 = – 1705.1 N
(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,
Mo = 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 Rx and Ry 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 Rx and Ry at F.
Moment of R about 0 = Sum of moments of Rx and Ry 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 (Mo = 1659.66)
1659.66 = Rx × O + Ry × x
(as Rx 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 Rx and Ry.
Moment of R about O = Sum of moments of Rx and Ry at O
or 1659.66 = Rx × y + Ry × O.
(At G, Ry 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.
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 T1 = Tension (or pull) in the cord
T2 = Tension (or pull) in the chain.
Now from the geometry, it is obvious that angles between T1 and lamp will be 90°, between lamp and T2 150° and between T2 and T1 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 T2. The magnitude of T2 is unknown. Now from
0, draw the line Ob horizontally (i.e., parallel to T1) cutting the line ab at point b.
(4) Now measure the lengths ab and bO).
Then ab represents T2 and bO represents T1. By measurements, 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 cm long, where PQ = QR = RS = 4 cm,
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 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.
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°
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
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 X1 draw another line parallel to Ob, which cuts the line of action
of force Q at X2. Similarly from point X2, draw a line parallel to Oc to cut the line of action of
force R at X3. From point X3, draw a line parallel to Od to cut the line of action of force S atX4.
(vii) From point X4, draw a line parallel Oe.
(viii) Produce the first line (i.e., the line from Xl and parallel to Oa) and the last line (i.e.,
the line from X4 and parallel to Oe) to interest at.X, Then the resultant must pass through this
(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.
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,
RH* = 0 + 750 + 707 + 433 = 1890 n
The resultant R* of RV* and RH* 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.