1
UNIT –I Analytical conditions of Equilibrium, Virtual
Work and Catenary
STRUCTURE
1.1 Introduction
1.2 Objectives
1.3 Analytical conditions of equilibrium of Coplanar forces
1.4 Virtual Work
1.5 Catenary
1.6 Unit Summary
1.7 Assignments
1.8 References
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1.1 Introduction:
This unit introduces the basics of Analytical conditions of equilibrium of
Coplanar forces, Virtual Work, Catenary. It also help us to understand the basic concepts
of above topics.
In this unit we shall study the basic ideas of Analytical conditions of equilibrium
of Coplanar forces, Virtual Work, Catenary. It is hopped the unit help students in
studying.
1.2 Objectives:
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At the end of the unit the students would be able to understand the concept of:
Analytical conditions of equilibrium of Coplanar forces
Equivalent Force and Couple
Equilibrium of a Rigid Body acted on by three forces
Trigonometrical Theorems
General Condition of Equilibrium: Analytical Method
Virtual Work
Positions of Equilibrium
Method of Virtual Work
Principal of virtual work
Forces, which may be omitted
Tension and Thrust
Roberval‟s Balance
Catenary
Equation of Catenary
Relation between x, y and 
Cartesian Equation of the catenary
Relation between x and s
Tension at a point
Geometrical Properties of Catenary
Approximations to the common catenary
Sag of a tightly stretched wire
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1.3 Analytical conditions of equilibrium of Coplanar forces
Theorem1:
A system of forces acting in one plane at different points of a rigid body can
always be reduced to a single force through any given point and a couple.
Proof:
To proving this theorem, let us consider the
Y
forces F1, F2, F3, …., act at points (x1, y1), (x2, y2),
(x3, y3), …., the coordinates of the points being given
Y1
with reference to any rectangular axes OX and OY
F1
through the given point O.
Firstly consider the force F1 acting at A1(x1, y1).
A1
Let it be resolved into two forces X1 and Y1
X1
parallel to the coordinate axes. At O introduce two
equal and opposite forces X1, one along OX and another
Y2
along OX. This will have no effect on the body.
F2
Now the forces X1 at A1 and X1 at O along OX form
A2
X2
a couple of moment –X1 (the negative sign is prefixed
|
since the tendency of the couple is to rotate the body
|
X
clockwise). A1M1 or –X1y1 and we are left with the force
O
M1
X1 along OX. Hence the force X1 at A1 is equivalent to
a force X1 at O along OX and a couple of moment –X1y1.
Similarly if we introduce at O equal force Y1 along OY and OY, it is easy to see
that the force Y1 at A1 is equivalent to a force Y1 at O along OY and a couple of moment
Y1x1.
It follows therefore that the force F1 at A1 is equivalent to forces X1 and Y1 at O
along the axes OX and OY respectively and a couple of moment Y1x1 – X1y1.
Proceeding in the same manner with the force F2 (whose components along the
coordinate axes are X2 and Y2, say) at (x2,y2), we will see that the force F2 at (x2,y2) is
equivalent to forces X2 and Y2 along OX and OY at O and a couple of moment Y2x2 –
X2y2.
Applying this process again and again we see that the given system of forces is
equivalent to forces
Rx = X1 + X2 + X3 + ……, i.e. X1 along OX,
Ry = Y1 + Y2 + Y3 + ……, i.e. Y1 along OY
and a couple of moment
G = (Y1x1 – X1y1) + (Y2x2 – X2y2) + …….
= (Y1x1 – X1y1)
.
.
.
(1)
The forces Rx and Ry can be compounded into a single force through O of
magnitude R given by
R2 = (Rx)2 + (Ry)2,
.
.
.
(2)
and acting at an angle to the axis of X, given by
 = tan-1(Ry/Rx)
.
.
.
(3)
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Hence the system of forces can be reduced to a single force R through O and
couple of moment G given by the equation (2) and (1) respectively. It is evident that G
depends upon the position of the origin O while R does not.
Theorem2:
A system of forces acting in one plane at different points of a rigid body can be
reduced to a single force, or a couple.
Proof:
We have just seen that a system of forces
Y
F1, F2, F3, ….. can be reduced to a single force R
and a couple of moment G given by (2) and (1) in
above Th.1.
If R = 0, the forces reduce to a couple. But
R
if R  0, we shall show that the force R and the
couple G can be reduced to a single force R acting
in the direction given by (3) of Th.1 but in a different
O
 
line. To prove this replace the couple G by two equal
X
and opposite forces of intensity R, one along OB in
R
R C
the direction opposite to R and the other along OC,
B
O
where OO is perpendicular to OB, O lies to the right
or left of OB as required by the sign of G and OO.R = G, .
.
(1)
or,
(Y1x1 – X1y1)
OO = --------------------{(Rx)2 + (Ry)2}
The forces at O balance each other and we are left with the force R acting at O
along C.
Since
tan = Ry/Rx,
we have
sin = Ry/R and cos = Rx/R.
(2)
The equation of the line C is therefore
x cos{-(/2 - )} + y sin{-(/2 - )} = OO,
or
x sin – y cos = OO,
or, by (1) and (2),
xRy – yRx = G.
.
.
(3)
Equivalent Force and Couple:
If G = 0, the given forces reduce to the single force R. Hence in every case the
forces can be reduced either to a single force, or a couple.
If the coordinates of any point Q be given by (a, b) and we are required to reduce
the system of forces to a force through the point Q and a couple, the value of the
corresponding couple G may be obtained as follows:
In the notation of Th.1 the moment of F1 about (a, b) is evidently
Y1(x1 – a) – X1(y1 – b),
or
(Y1x1 – X1y1) – Y1a + X1b.
The moment of F2 about (a, b) is
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(Y2x2 – X2y2) – Y2a + X2b,
and similarly for the other forces. Hence the total moment, G., of the forces about the
point is given by the equation
G = G – aY + bX1 = G – aRy + bRx.
If the point Q lies on resultant, then
G – aRy + bRx = 0,
Since the algebraic sum of the moments about any point on the line of action of the
resultant is zero. Change the current coordinate for a, b, we have equation (3) of Th.2 for
the line of action of the resultant.
Ex.1 If a system of forces in one plane reduces to a couple whose moment is G and when
each force is turned round its point of application through a right-angle it reduces to a
couple H; prove that when each force is turned through an angle ; the system is
equivalent to a couple whose moment is G cos + H sin
Sol.
Let the forces be given by Fr acting through the point (xr, yr) inclined at r to axis
of x (where r = 1, 2, 3, ….), then
Rx = Fr cosr = 0, .
.
(1)
Ry = Fr sinr = 0; . .
.
(2)
and
G = (xrsinr – yrcosr) Fr .
(3)
at the origin.
If each force be rotated through a right angle in the positive sense, say, and the
resultant is R, having components Rx and Ry, then
Rx = Fr cos(r + /2) = - Frsinr = 0, by (2)
Ry = Frsin(r + /2) = Frcosr = 0, by (1)
and the resultant moment of the forces in their positions about O is given by
H = {xrsin(r + /2) – yrcos(r + /2)}Fr
= (xrcosr + yr sinr)Fr
.
(4)
Now if the directions are changed by in the same sense and the resultant is R,
with components Rx and Ry, then
Rx = Frcos(r +)
= Fr(cosrcos – sin rsin)
= cosFrcosr – sinFrsinr = 0, .
by (1) and (2)
and
Ry = Frsin(r +)
= Fr(sinrcos + cosrsin)
= cosFrsinr + sinFrcosr = 0, . by (1) and (2)
and the resultant moment of the forces in their final positions is given by
G = {xr sin(r +) – yrcos(r +)}Fr
= {cos(xrsinr – yrcosr) + sin(xrcosr + yrsinr)}Fr
= Gcos + Hsin,
.
.
by (3) and (4)
Ex.2 The algebraic sums of the moments of a system of coplanar forces about points
whose coordinates are (1, 0), (0, 2) and (2, 3) referred to rectangular axes are G1, G2 and
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G3 respectively. Find the tangent of the angle, which the direction of the resultant force
makes with the axis of x.
Sol.
Let R represent the magnitude of the resultant force and let its equation be
y = mx + c,
where c is positive.
Then
(– m – c)R
G1 = -----------------,
(1 + m2)
(2 – c)R
G2 = --------------,
(1 + m2)
(3 – 2m – c)R
G3 = -------------------(1 + m2)
These equations give
G1
m+c
G2
c-2
---- = -------and
---- = ------------ ,
G2
c–2
G3
2m + c - 3
i.e.
G2m + c (G2 – G1) + 2G1 = 0
and
2G2m + c (G2 – G3) – 3G2 + 2G3 = 0
solving we get
- 2G1 (G2 – G3) – (3G2 – 2G3)(G2 – G1)
m = ---------------------------------------------------------G2(G2 – G3) – 2G2(G2 – G1)
and
=
G1G2 – 3G22 + 2G2G3
-----------------------------2G1G2 – G22 – G2G3
G1 – 3G2 + 2G3
=
---------------------- ,
2G1 – G2 – G3
which gives the tangent of the angle which the direction of the single force makes with
the axis of x.
Ex.3 If six forces, of relative magnitudes 1, 2, 3, 4, 5 and 6 act along the sides of a
regular hexagon, taken in order, show that the single equivalent force is of relative
magnitude 6 and that it acts along a line parallel to the force 5 at a distance from the
centre of hexagon 3½ times the distance of the side from the centre.
Sol.
Let ABCDEF be a regular hexagon of side 2a. Suppose that forces of magnitudes
k, 2k, 3k, 4k, 5k and 6k act along the sides AB, BC, CD, DE, EF and FA respectively.
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Let O be the center of the hexagon. Choose through O a set of rectangular axes OMX and
OCY.
Y
It is easy to see that the six forces inclined to
C
OMX at the angles, 3/6, 5/6, 7/6, 9/6, 11/6 and
3k
2k
13/6. Hence the sum of their resolved parts along the
D
B
axes of x and y are
Rx = k[cos/2 + 2cos5/6 + 3cos7/6 + 4cos9/6
4k
O
M X
+ 5cos11/6 + 6cos13/6]
k
= ,
E
A
and
Ry = k[sin/2 + 2sin5/6 + 3sin7/6 + 4sin9/6
5k
6k
+ 5sin11/6 + 6sin13/6] = -3k
F
Therefore the magnitude of the resultant force
= {(k)2 + (- 3k)2} = 6k,
and it makes an angle
tan– 1(-3k/k) = tan– 1(-1/3) = 11/6
with OMX. It is, therefore, parallel to the force 5k acting along EF.
Further since all the sides of the hexagon are at a distance OM from O, the
algebraic sum of the moments of the forces about O
= (k + 2k + 3k + 4k + 5k + 6k)OM = 21k.OM
Hence the distance of the resultant force from the center
= (21k.OM)/6k = 7/2.OM
Ex.4 A system of forces in one plane is equivalent to a couple of moment G. If the line of
action of each force is turned about its point of application in the same direction through
a right angle, prove that the new system is also equivalent to a couple. Also prove that if
the moment of this couple be H and if the lines of action of the original forces be each
turned through 2tan-1[H/G], they would still be equivalent to a couple of moment G.
Sol.
Let P1, P2, …….. be the forces acting at points (x1, y1), (x2, y2),….. and making
angles a1, a2, ….. with OX, then
X = P1cosa1
Y = P1sina1
G = (P1sina1x1 – P1cosa1y1) = P1(x1sina1 – y1cosa1)
…(1)
Since the system is equivalent to a couple so X = 0 i.e.
P1cosa1 = 0
….(2)
Y = 0 i.e.
P1sina1 = 0
….(3)
(i) When each force is turned about its point of application through a right angle in the
+ve direction,
X = P1cos(a1 + /2) = (-P1sina1) = 0 by (3)
Y = P1sin(a1 + /2) = P1cosa1 = 0 by (2)
The system again reduces to a couple. If H be its moment
H = [P1cosa1x1 – (- P1sina1)y1] = P1(x1cosa1 + y1sina1)
(ii) Let each original force be now turned through an angle b
X = P1cos(a1 + b) = P1(cosa1cosb – sina1sinb)
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if
i.e.
or
i.e.
= cosb P1cosa1 – sinb P1sina1 = cosb x 0 – sinb x 0 = 0
Y = P1sin(a1 + b) = P1(sina1cosb + cosa1sinb)
= cosb P1sina1 + sinb P1cosa1 = cosb x 0 + sinb x 0 = 0
The system again reduces to a couple. If H be the moment
H = [P1sin(a1 + b)x1 – P1cos(a1 + b)y1]
= P1[x1(sina1cosb + cosa1sinb) – y1(cosa1cosb – sina1sinb)]
= cosb P1(x1sina1 – y1cosa1) + sinb P1(x1cosa1 + y1sina1)
= cosb.G + sinb.H = G cosb + H sinb
Now H = G
G cosb + H sinb = G
H sinb = G(1- cosb)
H
1 – cosb
--- = ------------- = tanb/2
G
sinb
b = 2tan-1[H/G]
Ex.5 Forces equal to 3P, 7P and 5P act along the sides AB, BC, and CA of an equilateral
triangle ABC; find the magnitude, direction, and line of action of the resultant.
Sol.
Let the side of the triangle be a, and let the
resultant force meet the side BC in Q. Then the sum
of the moments of the forces about Q vanishes.
3P x (QC + a)sin600 = 5P x Qcsin600
QC = 3a/2
A
The sum of the components of the forces
perpendicular to BC
3P
5P
0
0
= 5Psin60 – 3Psin60 = P3
7P
C
Also the sum of the components in the
B
Q
direction BC
= 7P – 5Pcos600 – 3Pcos600 = 3P
Hence the resultant is P12 inclined at an
angle tan-13/3, i.e. 300, to BC and passing through Q
where CQ = (3/2) BC.
Equilibrium of a Rigid Body acted on by three forces:
Theorem:
If three forces, acting in one plane upon a rigid body, keep it in equilibrium, they
must either meet in a point or be parallel.
Proof:
To prove this, suppose that the given forces P, Q and R are not all parallel. Then
at least two of them, say, P and Q must meet in a point O. R must balance the resultant of
P and Q as the three forces P, Q and R are in equilibrium. Since the resultant of P and Q
passes through O, R must also pass through O, i.e. the three forces are concurrent.
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But if two of them are parallel, their resultant is parallel to them, and therefore so
is the force, which is to balance them, i.e. the third force is also parallel to them.
Ex.1 A sphere of given weight rests on two smooth planes inclined to the horizon at
given angles; determine the pressures on the planes.
Sol.
Let the sphere ABC, of weight W, rest on the planes OA and OB, inclined to the
horizon at angles and respectively, A and B being the points of contact of the sphere
with the planes.
Since there are three forces acting on
the sphere, namely the reaction R and S at
A and B and the weight W, therefore, for
C
equilibrium they must be in the same plane
and their lines of action must meet in a point.
The figure represents a vertical
G
section of the sphere and the planes, containing
S
R
the lines of action of the reactions and the
W
weight. The reactions R and S, being
B
perpendicular to the planes, pass through the
centre G of the sphere through which the weight
A
W acts vertically downwards.

Then for the equilibrium of the sphere,
O
since the three forces meet at G, we have,
by Lami‟s theorem
R
S
W .
=
=
sin (S, W)
sin (W, R)
sin (R, S)
or
R
S
W .
=
=
sin ()
sin ()
sin ()
giving
R = W sin and S = W sin
sin ()
sin ()
But the reactions of the planes are equal to the pressures on the planes. Therefore
the pressure on the plane OA whose inclination to the horizon is is
W sin
sin ()
and the pressure on the plane OB whose inclination is is
W sin
sin ()
Ex2. Three uniform rods AB, BC and CD, whose weights are proportional to their
lengths a, b and c are jointed at B and C and are in a horizontal position resting on two
pegs P and Q; find the actions at the joints B and C, and show that distance between the
pegs must be
a2
c2
------- + ------- + b
2a + b 2c + b
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Sol.
Let w be the weight of a rod per unit length,
and let R and S be the vertical actions on BC at
A
the points B and C respectively.
For the equilibrium of BC,
R + S = bw
and symmetry demands that R must be equal to S,
so that
R = S = ½ bw
This gives that the action at each joint is
equal to half the weight of the middle rod. Now
P
B
C
R
Q
D
S
B
C
bw
consider the equilibrium of the rod AB. The
pressure of the peg P vertically upwards
A
B
= aw + R = aw + ½ bw
P
Moments about B of the forces acting on AB give that aw
aw .a/2 = (aw + ½ bw).BP. Therefore
BP = a2/(2a + b)
Also the horizontal action at B is zero. Similarly we can show that
CQ = c2/(2c + b)
Hence
a2
c2
PQ = ------- + --------- + b
2a + b 2c + b
Trigonometrical Theorems:
Theorem: If a straight line CD drawn from the vertex C of a triangle ABC to the
opposite side AB divides it into two segments in the ratio of m: n, then
(m + n) cot = m cot – n cot and (m + n) cot = n cotA – m cotB,
where and are the angles which CD makes with CA and CB and the angle which
CD makes with the base AB.
Proof:
We have
m
AD
AD DC
---- = ------ = ----- ----n
DB
DC DB
sin ACD sin DBC
= ----------- --------(1)
sin DAC sin DCB
sin
sin()
cot + cot
= --------- ---------------- = ---------------sin()
sin
cot – cot
Therefore
(m + n) cot = m cot – n cot
Again from (1) we have also
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m
---n
sin( A) sin B
= ------------- -----------sin A
sin(+ B)
cot A – cot
= ---------------cot B + cot
Hence (m + n) cot = n cot A – m cot B.
Ex.1 A heavy uniform rod, of length 2a, rests partly within and partly without a fixed
smooth hemispherical bowl, of radius r; the rim of the bowl is horizontal, and one point
of the rod is in contact with the rim; if be the inclination of the rod to the horizon, show
that
2r cos2 = a cos.
Sol.
Suppose the given figure represent
that vertical section of the hemisphere, which
passes through the rod. Let AB be the rod, G
D
its centre of gravity, and C the point where the

rod meets the edges of the bowl.
S
B
The reaction at A is along the line to the
O

centre, O, of the bowl; for AO is the only line
R
C
G
through A which is perpendicular to the surface

W
of the bowl at A. Also the reaction at C is

perpendicular to the rod; for this is the only
A
E
direction that is perpendicular to both the rod and the rim of the bowl.
These two reactions meet in a point D, which lies on the geometrical sphere of
which the bowl is a portion. Hence the vertical line through G, the middle point of the
rod, must pass through D.
Through A draw AE horizontal to meet DG in E and join OC.
Then
OAC = OCA = CAE = .
So
a cos = AE = AD cos2 = 2r cos2.
Also, by Lami‟s Theorem, if R and S be the reactions at A and C, we have
R
S
W .
=
=
sin
sin ADG
sin ADC
or
R
S
W .
=
=
sin
cos2
cos
Ex.2 A hemispherical bowl, of radius r, rests on a smooth table and partly inside it rests a
rod of length 2l and of weight equal to that of the bowl. Show that the position of
equilibrium is given by the equations
l sin() = r sin = – 2r cos(),
where is the inclination of the base of the hemisphere to the horizon, and 2 is the
angle subtended at the centre by the part of the rod within the bowl.
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Sol.
Let O be the centre of the base of the bowl and A and B the points of contact of
the rod AH with the bowl. The angle AOB is 2 and OFE is , E being the point of
contact of the bowl with the table. Let G be the middle point of the rod. The centre of
gravity, G, of the bowl lies at the mid-point of the central radius OJ.
The rod is in equilibrium under the action of three forces viz., its weight W acting
vertically through G, the reaction R at A
acting normally to the hemisphere, i.e.,
D
through O and the reaction S at B acting
at right-angles to AB. The lines of action
of these three forces must meet at a point, D,
say, and since ABD is a right-angle, D must
lie on the circle AEB and AD = 2r.
Also AG = l.
O
N
The system consisting of the bowl
and the rod is in equilibrium under the
G
S
action of the reaction at E and the weights W
R
B
H
acting along GK and GL; since the
J
G
A
reaction passes through O, the weights must

M

have equal and opposite moments about E so
K
E
L
F
that KE = EL; (K and L being the points where
the vertical lines through G and G meet the
table). Since further AO = OD, it is easy to see
W
W
that GK passes through A.
We shall now find in three ways the length AM, the perpendicular from A on the
vertical DG.
For this we require the angles ADB, DON and OAB, ON being the perpendicular
from O on the vertical DG.
Since AOB = 2, ADB = ,
DON = 1800 – (AOB + BON) = 1800 – (2),
and
OAB = OBA = 900 – .
Hence AM = AD cos DAM = 2r cos DON
= 2r cos(180 ). = – 2r cos()
Also AM = AG cos GAM = l cos(900 ) = l sin(),
and lastly
AM = 2KE = 2GI, where GI is at right angles to OE,
= 2OG sin = r sin.
It follows therefore that
l sin() = r sin = – 2r cos(),
Ex.3 Equal weight P and P are attached to two strings ACP and BCP passing over a
smooth peg C. AB is a heavy beam, of weight W, whose centre of gravity is a feet from
A and b feet from B; show that AB is inclined to the horizon at an angle
–1
tan– 1 a – b tan sin W
a+b
2P
13
Sol.
Let AB be the rod and G its centre of
C
gravity dividing AB in the ratio a : b. Let C
be the smooth peg. The tension in the strings
BC and AC are each P; since W balances the
P
b
resultant of two equal forces, CA and CB are
B
equally inclined to the vertical. Suppose we
a
G
denote ACG or BCG by .
Resolving vertically, we have
A  
–1
2P cos – W = 0, i.e. = cos (W/2P).
W
Therefore
(a + b) cot BGC = a cot – b cot.
Also BGC = ½ where is the inclination of AB to the horizon.
It follows that
a – b  cot = a – b  tan(½ )
tan =
a+b
a+b
–1
= a – b  tan [½  – cos (W/2P)]
a+b
Hence
a – b tan sin– 1 W
tan– 1
a+b
2P
Ex.4 A square, of side 2a, is placed with its plane vertical between two smooth pegs,
which are in same horizontal line and at a distance c, show that it will be in equilibrium
when the inclination of one of its edges to the horizon is either
450 or ½ sin– 1[(a2 – c2)/c2].
Sol.
Let OABC be the square, its side OA
being inclined at an angle to the horizontal
through O and let DE be the horizontal line
joining the pegs. The reactions at D and E are
perpendicular respectively to OC and OA.
Since the only other force acting on the square
is weight, these three forces must meet in a
point, say, O. The figure OEOD is obviously a
rectangle. We can now solve the problem by
two methods.
Method1:
Let the reactions at D and E are R and
R respectively and W is the weight of the
square. Resolving horizontally, we get
R cos = R sin.
(1)
Taking moments about G, we have
R (a – c cos) = R (a – c sin).
(2)
B
C
G
A
O
R
R
D
E

O N
W
14
Eliminating R and R between (1) and (2), we get
cos (a – c cos) = sin (a – c sin).
or
c (sin2 – cos2) = a (sin – cos).
Hence, either
sin – cos = 0, where = 450,
or
c (sin + cos) = a.
(3)
Squaring either side of (3), we have
c2 (1 + sin2) = a2.
Therefore
 ½ sin– 1[(a2 – c2)/c2].
Method2:
Let ON be the perpendicular from O on the vertical through G. Then we can write
the value of ON in two ways.
ON = OG cos( ¼)
= a2 (cos¼ cos – sin¼ sin) = a (cos – sin).
Also ON = OO cos OON.
But
OO = DE = c,
and
OON = OOE + EON = = 2.
Hence ON = c cos2.
It follows therefore that
a (cos – sin) = c cos2,
or
a (cos – sin) = c (cos2 – sin2), etc.
Ex.5 A beam whose centre of gravity divides it into two portions, a and b, is placed
insides a smooth sphere; show that, if be its inclination in the horizon in the position of
equilibrium and 2a be the angle subtended by the beam at the centre of the sphere, then
tan =
b – a  tan
b+a
Sol.
In this case both the reactions, R and
S, at the ends of the rod pass through the
centre, O, of the sphere. Hence the centre
of gravity, G, of the rod must be vertically
below O. Let OG meet the horizontal line
through A in N. Draw OD perpendicular to AB.
Then AOD = BOD = ,
and
DOG = 900 – DGO = DAN = .
Therefore
(a + b) cot OGB = b cot OAB – acot OBA,
i.e.
(a + b) tan = (b – a) tan.
Also by Lami‟s Theorem,
R
S
W .
=
=
sin BOG
sin AOG
sin AOB
O
R
S
a

b
GD
N W
B
15
or
R
S
=
=
sin()
sin()
giving the reactions.
W .,
sin2
General Condition of Equilibrium: Analytical Method:
Theorem1:
The resultant R and the couple G must separately vanish.
Proof:
We know that the system of forces acting at different points may be reduced to a
force R through an arbitrary point O and a couple G. Since a finite force R cannot balance
a couple G, it is necessary for equilibrium that the resultant force R and the couple G
should separately vanish. The vanishing of R involves the condition that Rx = 0 and Ry =
0.
Note: Now we arrive at the following necessary and sufficient conditions of equilibrium.
A system of forces in a plane will be in equilibrium if the algebraic sums of their
resolved part in any two perpendicular directions vanish, and if the algebraic sum of their
moments about any point also vanishes.
Theorem 2:
A system of forces in a plane will be in equilibrium if the algebraic sum of the
moments of all the forces with respect to each of three non-collinear points is zero.
Proof:
To proving this theorem, consider the origin at one of the three points and let the
coordinates of the other points be (x1, y1) and (x2, y2). If we denotes the algebraic sum of
the moments of the forces about the above points by G, G‟, G‟‟ then by
G = 0,
G‟ = G – x1Ry + y1Rx = 0,
and
G‟‟ = G – x2Ry + y2Rx = 0.
These reduce to
- x1Ry + y1Rx = 0
and
- x2Ry + y2Rx = 0.
Since the three points are not collinear, y1/x1 y2/x2 and the above conditions
reduce to
Rx = 0, Ry = 0 and G = 0
a set of conditions involved in Theorem 1.
Theorem 3:
A system of forces in a plane will be in equilibrium if the algebraic sum of the
moments about each of any two different points is zero and the algebraic sum of the
resolved parts of the forces in any given direction not perpendicular to the line joining the
given points is zero.
16
Proof:
Let the two different points be A and B. As before we take origin at A and x-axis
along the given direction. Let the coordinates of B be (x1, y1). Then the conditions are
G = 0,
G‟ = G – x1Ry + y1Rx = 0,
Rx = 0,
which lead to
Rx = 0, Ry = 0 and G = 0
provided that x1 is not zero, a set of conditions involved in Theorem 1.
Theorem4:
Resultant forces and couple corresponding to any base point O of a system of
coplanar forces.
Proof:
Through O take any pair of rectangular axes Ox, and Oy. At P1, the point (x1, y1),
let there act a force whose components parallel to the axes are X1 and Y1.
Then X1 at P1 is equivalent to a parallel force X1 at O together with a couple
y1X1 . So Y1 at P1 is equivalent to a parallel force Y1 at O together with a couple x1Y1 .
Hence the force at P1 is equivalent to components X1, Y1 along Ox, Oy and a
couple x1Y1 – y1X1 .
So for the other forces at P2, P3, etc.
Hence the system of forces is equivalent to components X, Y along Ox, Oy, and a
couple G about O, such that
y
X = X1 + X2 + X3 + … = X1,
Y1
. P2
.
Y = Y1 + Y2 + Y3 + … = Y1,
P1
P3
and
G = (x1Y1 – y1X1) + (x2Y2 – y2X2) + …
x1
y1
X1
= (x1Y1 – y1X1).
O
x
X and Y compound into a single force R acting at O.
Theorem5:
Equation to the resultant of a system of forces in one plane.
Proof:
We know that the system can be reduced to components X and Y along any two
rectangular axes Ox and Oy, and a couple G about O .
Let Q be any point (h, k), which
lies on the resultant of the given system.
y
The moment of the system about it is
equal to the moment of the resultant about
Y
Q
it and is therefore zero.
G
Now the moment of the system about Q
= G + X.NQ – Y.ON
O
x
= G – hY + kX,
X
N
so that
G – hY + kX = 0.
Hence the locus of (h, k), i.e. the resultant, is the straight line
G – xY + yX = 0.
17
Ex.1 Two equal uniform rods, AB, AC, each of weight W, are freely joined at A and rest
with the extremities B and C on the inside of a smooth circular hoop, whose radius is
greater than the length of either rod, the whole being in a vertical plane, and the middle
points of the rods being joined by a height string; show that, if the string is stretched, its
tension is W(tan – 2 tan), where 2 is the angle between the rods, and the angle
either rod subtends at the centre.
Sol.
Let O be the centre of the circular hoop
O
BFC, AB, AC the equal rods and DE the string
joining their middle points.

If R be the pressure of the hoop on either
rod, then resolving vertically for the system
A
consisting of the two rods

W = R cos
(1)

Taking moments about A for either rod,
D
E
suppose of length 2l, we have
B
C
R.2l sin() = W.l sin + T.l cos,
F
i.e.
2R
sin
()
–
W
sin
T=
cos
= 2W sin () – W sincos
cos
on using (1)
= W (tan2 tan)
Ex.2 Three uniform equal heavy cylinders, each of which touches the other two, are tied
together by a string passing round them and laid with their axes horizontally upon a
horizontal plane. The tension of the string being given, find the pressure between the
cylinders, the string is supported to lie in the vertical plane through the centres of gravity.
Sol.
Consider the section through the string
and A, B, C, the centres of gravity of the three
cylinders. Let W be the weight of each cylinder,
and T the tension of the string. The reaction
between the higher and each of the two lower
T
A
T
cylinders will by symmetry be equal. Let us
denote it by R. Let S be the reaction between
R
R
the two lower cylinders and R the reaction of
T
W
T
the plane XY upon each of the lower cylinders.
R
R
R
R
It is clear that ABC forms an equilateral triangle,
B
C
and the lines of the action of the tension will
S
S
also forms a similar triangle.
Resolving vertically the forces acting on
the upper cylinder alone, we have
X
T
T
Y
0
0
2T cos 30 + W = 2R cos 30 ,
W
W
giving
R = T + W/3
(1)
Again resolving vertically the forces on either of the two lower cylinders, we get
T cos 300 + R = R cos 300 + W,
18
R = ½3(T + W/3) + W – T.½ 3 = (3/2)W
(2)
If we resolve horizontally for either of the two lower cylinders, we have
T cos 600 + T = S + R cos 600,
and therefore
S = (3/2)T – ½ (T + W/3) = (T – W/23)
(3)
The equations (1) and (3) give the required reactions R and S. It should be noted
that the pressure on the plane by both the lower cylinders is, by (2), equal to 3W, i.e. to
the sum of the weights of the three cylinders.
hence
Ex.3 An elliptic lamina is acted upon at the extremities of pairs of conjugate diameters by
forces in its own plane tending outwards and normal to its edge; show that there will be
equilibrium if the force at the end of each diameter is proportional to the conjugate
diameter.
Sol.
Suppose that the forces acting at the
ends P, P and D, D of the pair of conjugate
B
diameters PP and DD be respectively X1,
X2, and Y1 and Y2. Since the normals at P
and D are parallel to the normals at P and
D
P
D respectively, the forces X1 and X2 at P
and P are acting in opposite senses along
A
S
S
A
O
parallel lines and so the forces Y1 and Y2 at
D and D respectively. Obviously the lines
of action of X1 and Y1 are not parallel.
P
D
Moreover the resultant of X1 at P and X2 at P
B
and that of Y1 at D and Y2 at D will not act
in the same straight line even though the
magnitudes of these resultants may be equal. Hence for equilibrium, we must have X1 =
X2 and Y1 = Y2, so that the forces at P and P form a couple, as also the forces at D and
D. Since the tendencies of rotation of these couples are opposite, for equilibrium we
must have the magnitudes of the couples to be equal.
Suppose that the angle POD is equal to . Since tangent at D is parallel to OP, the
normal at D is perpendicular to OP, hence the moment of the couple formed by the equal
and opposite forces Y1 at D and D is equal to Y1.2OD cos(), as the perpendicular
from D on OP is OD cos().
Similarly the moment of the couple formed by the equal and opposite forces X1 at
P and P is equal to X1.2OP cos.
Magnitudes of these moments will be equal if X1/OD = Y1/OP. Therefore
X1/DD = Y1/PP.
Hence there will be equilibrium if the force at the end of each diameter is
proportional to the conjugate diameter.
Ex.4 Three equal uniform rods, each of weight W, are smoothly jointed so as to form an
equilateral triangle. If the system be supported at the middle point of one of the rods,
19
show that the action at the lowest angle is 3W/6, and that at each of the others is
W(13/12).
Sol.
Let ABC be the triangle formed by the rods, and D the middle point of the side
AB at which the system is supported.
Let the action of the hinge at A on the rod AB consist of two components,
respectively equal to Y and X, acting in vertical and horizontal directions; hence the
action of the hinge on AC consists of components equal and opposite to these. Since the
whole system is symmetrical about the vertical line through D, the action at B will consist
of components, also equal to Y and X, as in the figure.
Let the action of the hinge C on CB consist of Y1 vertically upward, and X1
horizontally to the right, so that the action of the same hinge on CA consists of two
components opposite to these, as the figure.
S
Y
S
Y
A
D
B
X
A
W
X
X
B
D
Y
W
Y
X
Y1
W
W
C
W
X1
C
Y1
W
X1
For AB, resolving vertically, we have
S = W + 2Y
(1),
where S is the vertical reaction of the peg at D.
For CB, resolving horizontally and vertically, and taking moments about C, we
have
X + X1 = 0
(2),
W = Y + Y1
(3),
0
0
0
and
W.a cos60 + X.2a sin60 + Y.2a cos60
(4).
For CA, by resolving vertically, we have
W = Y – Y1
(5).
Solving these equations, we have
X1 = – 3W/6, Y1 = 0, Y = W, X = 3W/6 and S = 3W.
Hence the action of the hinge at B consists of a force (X2 + Y2) [i.e. W(13/12)],
acting at an angle tan– 1(Y/X) [i.e. tan– 123], to the horizon; also the action of the hinge
at C consists of a horizontal force equal to 3W/6.
Ex.5A heavy uniform bar, of length 2a and weight w is movable in a vertical plane round
a smooth hinge fixed at one extremity; a heavy smooth sphere of weight W and radius r is
attached to the hinge, by a cord of length l; the two bodies rest in contact. If and be
the inclinations of the bar and the cord to the vertical, show that
sin() = r/(l + r) and W (l + r) sin = wa sin.
Find the pressure between the bar and the sphere.
20
Sol.
Let AB be the bar hinged at A, G its middle point, and D the point at which it is in
contact with the sphere. Let E be the point of the sphere to which the cord AE is attached.
Let C be the centre of the sphere. AB and AE make angle and respectively with the
vertical.
The forces acting on the sphere only are:
N
A
L
(i)its weight W acting vertically downwards
through its centre C,
T
E 
(ii)reaction R of the bar on the sphere along DC,
(iii)tension T of the string along EA.
R

Since forces (i) and (ii) meet at C. So for
equilibrium, the third force must also pass through
C
D
C i.e., AEC must be a straight line.
R
Now from
CAD,
sin.CAD = CD/AC = CD/(EA + CE)
Therefore
sin () = r/(l + r).
G
Now consider the equilibrium of the sphere
and the bar as a whole. In this case, the two equal
and opposite reactions of the sphere on the rod and
W
w
B
of the rod on the sphere will not come in the picture,
and the forces will be
(i)weight W at C acting vertically downwards,
(ii)weight w at G acting vertically downwards,
(iii)tension T along EA,
(iv)reaction of the hinge at A on the bar.
To avoid the reaction of the hinge, we take moment of the forces about A.
Therefore
W.AN = w.AL
where AN, AL are perpendiculars from A on vertical lines drawn through C and G.
W.AC sin = w.AG sin,
W.(l + r) sin = wa sin,
Finally to get R, the reaction of the sphere on the bar, we consider the force on the
bar only. They are
(i)weight w of the bar at G vertically downward,
(ii)reaction R of the sphere along CD,
(iii)reaction of the hinge at A.
Therefore taking moments about A, w.AL = R.AD
R = w.AL /AD
Now AL = AG sin = a sin.
AD = AC cos () = (l + r) cos ()
Therefore
wa sin 
R=
(l + r) cos ()
W (l + r) sin
=
(l + r) cos ()
W sin
=
cos ()
21
1.4 Virtual Work
Positions of Equilibrium:
Now we come to a very powerful method
for attacking problems on equilibrium. Consider a
Y
heavy particle on a smooth curve. If the axis of Y
B
is vertical it is obvious that the particle can rest n
equilibrium at the point A, B or C. These are points
A
of maxima or minima on the curve, i.e. , points for
C
which dy/dx = 0. The physical interpretation of the
positions of equilibrium may be given as follows:
O
X
If a small displacement be given along the curve to the particle from a position of
equilibrium, the work done is zero, for, if W be the weight of the particle, the work done
is Wy, i.e., W(dy/dx)x (to a first approximation). Again to find the position of
equilibrium we have to find those positions on the curve for which Wy = 0, i.e., those
points for which dy/dx = 0.
It should be noted that y is equal to zero to a first approximation only. For, if y =
f(x) be the equation of the curve,
y + y = f(x + x) = f(x) + xf (x) + [(x)2/2!]f (x) + ….
As f (x) = 0 at A, B, or C,
y = [(x)2/2!]f (x) + ….
i.e., y = 0 if squares and higher powers of x are neglected. Hence y = 0 to a first
approximation only. Thus it is necessary that the displacement given be small.
Method of Virtual Work:
In any given problem, we imagine the body to be displaced a little and find out
the work done during the displacement. The condition of equilibrium is obtained by
equating to zero the sum of the work done. Since the body is not actually displaced, the
work done is called virtual work. The virtual work that is calculated is the amount of
work that would have been done if the displacements had actually been made.
We shall now formally enunciate the principle of virtual work and establish the
same. Since the proof is simpler for the case of a particle acted upon by a number of
forces, we consider this first.
Principal of virtual work:
If a system of forces acting on a body be in equilibrium and the body undergo a
slight displacement consistent with the geometrical conditions of the system, the
algebraic sum of the virtual works is zero; and conversely, if this algebraic sum be zero,
the forces are in equilibrium. In other words, if each force P have a virtual displacement
p in the direction of its line of action, then, to the first order of small quantities, (P.p)
= 0; also conversely, if (P.p) be zero, the forces are in equilibrium.
22
Theorem:
The necessary and sufficient condition that a rigid body acted upon by a number
of coplanar forces be in equilibrium is that the algebraic sum of the virtual works done by
the forces in any small displacement consistent with the geometrical conditions of the
system is zero.
Proof:
Let any number of coplanar forces F1,
Y
F2
R
F2, F3, … act at points A1, A2, A3, … of a
F3
F1
rigid body. Through an arbitrary chosen point
P
Q
O, take two fixed rectangular axes OX and OY
A2

in the plane of the forces. Let the coordinates
A3
A1

of the points A1, A2, A3, … be (x1, y1), (x2, y2),
O
X
(x3, y3), .. and suppose that the component of
A4
A5
the forces F1, F2, F3, … along the coordinate
axes are X1 and Y1, X2 and Y2, X3 and Y3, …
F4
F5
respectively.
If the body undergoes a certain displacement parallel to a fixed plane, it may be
brought from its old position to its new position by
(i) a motion of rotation which has the same magnitude and sense for all the points of the
rigid body, and
(ii) a motion of translation which has also the same magnitude and direction for all its
points.
The latter displacement can, in general, be decomposed into a motion of
translation along each of the two perpendicular axes OX and OY and these component
displacements will also be shared by all the points of the body. Let us denote by and
respectively the displacements of rotation and translations along the axes of x and y
corresponding to a slight displacement of the body.
If we assume that the polar coordinates of A1 are (r1, 1), the Cartesian
coordinates of R, the new position of A1, can be written as
r1 cos(1 + ) + and r1 sin(1 + ) + 
i.e.
r1 cos1 - r1 sin1 + 
[ is small, so sin = , cos = 1]
and
r1 sin1 + r1 cos1 + 
neglecting squares and higher powers of .
The relative coordinates of R with respect to A1 are therefore
[(r1 cos1 - r1 sin1 + ) - r1 cos1]
and
[(r1 sin1 + r1 cos1 + ) - r1 sin1],
i.e.,
 - r1 sin1 and  + r1 cos1,
i.e.,
 - y1 and  + x1.
Since the work done by a force is equal to the sum of the works done by its
components, the virtual work of the force F1 is
X1( - y1) + Y1( + x1),
i.e.,
X1 + Y1 + (x1Y1 - y1X1).
 - y1 and  + x1 being the displacements of the point A1 of the application of the
force F1 along the axes of x and y respectively.
Similarly the virtual work of the force F2 acting at the point (x2, y2) is
23
X2 + Y2 + (x2Y2 - y2X2)
and we get similar expression for the other forces.
The algebraic sum of the virtual works is, therefore,
X1 + Y1 + (x1Y1 - y1X1).
Since the body is in equilibrium under the action of the forces,
X1 = 0, Y1 = 0 and (x1Y1 - y1X1) = 0.
This gives that
X1 + Y1 + (x1Y1 - y1X1) = 0.
Hence the condition is necessary; in other words, if a body in equilibrium under
the action of a number of coplanar forces undergo a small displacement, then the
algebraic sum of the virtual works done by different forces is zero.
Now we shall show that the condition is sufficient. It is given that the algebraic
sum of the virtual works of the various forces is zero for all displacements and has to
show that there is equilibrium. Since the equation
X1 + Y1 + (x1Y1 - y1X1) = 0.
(1)
is true for all displacements, and are independent of each other. Suppose then that
and are the translations along OX and OY and rotation respectively for a new
displacement of the rigid body. We still have
X1 + Y1 + (x1Y1 - y1X1) = 0.
(2)
Subtracting (2) from (1), we have
( - ) (x1Y1 - y1X1) = 0.
But since ( - ) is not equal to zero, therefore
(x1Y1 - y1X1) = 0.
Similarly by varying alone, we prove that Y1 = 0 and by making alone vary,
X1 = 0. But X1 = 0, Y1 = 0 and (x1Y1 - y1X1) = 0 are the analytical conditions for
equilibrium and hence the body is in equilibrium.
Forces, which may be omitted:
The equation (1) is known as the equation of virtual work. In a given problem of
equilibrium, we write down this equation corresponding to a suitable displacement. The
displacement should be such as to exclude the forces, which are not required and to
include those, which are required in the final result. There are, however, certain forces,
which may be omitted in forming the equation of virtual work. We shall now enumerate
them.
1.If the distance between two particles of a system is invariable, the work done by the
mutual action and reaction between the two particles is zero.
To show this, suppose that T is the
force acting between the particles A, B
B
which suffer displacements to A, B
respectively such that AB = AB. If be
A
the small angle between AB and AB and
M, N the projections of A, B in AB, then
A
M
T
T B N
the work done by the two forces
= T.AM – T.BN = T(AM + MB) – T(BN + MB)
24
= T(AB – MN)
= T.(AB – AB cos)
= T.AB(½2 – higher powers of )
= 0, to the first order of small quantities.
2.The reaction R of any smooth surface with which the body is in contact does no work.
For, if the surface is smooth, the reaction R on the point A of the body is normal
to the surface. If A moves to a neighbouring point A, then AA is at right angles to the
forces and accordingly the work done by R is zero.
In the case of a rough surface, however, the work done by the frictional force F,
viz., F.(-AA) must come into the equation, since it is not in general zero.
3.If the force between two bodies of a system is a mutual pressure at a point of contact,
then no work is done in any virtual displacement of the system by the action and reaction
if the same points of the bodies remain in contact during the displacement.
For clearly the work done by the action balances that done by the reaction.
4.When a body rolls without sliding on any fixed surface the work done in a small
displacement by the reaction of the surface on the rolling body is zero.
For the point of contact P of the body is momentarily at rest and so its
displacement is zero. Hence the normal reaction at P and the frictional force F there have
zero displacements.
5.If any body is constrained to run round a point or on an axis fixed in space, the virtual
work of the reaction at the point or on the axis is zero.
For the displacement of the point of application of the force is zero.
Tension and Thrust:
In many cases we are required to find the tension of a string or the thrust of a rod.
The geometrical conditions of the problem often make it impossible to give a virtual
displacement, which will be consistent with the constraints. Thus, if we have a
quadrilateral ABCD, consisting of heavy rods AB, BC, CD, DA suspended from the
point A and the points B and D connected by a rigid rod, it is impossible to give a virtual
displacement which will make thrust of the rod do work. To get over this problem, we
replace the system under consideration by another in which there are forces T and T
acting at B and D in the direction DB and BD respectively. If T in the second system is
equal to the thrust of the rod in the first system, the equation is evidently not affected. We
can now give a virtual displacement to the second system, which will involve T in the
equation of virtual work. T will thus be determined.
Similarly the equations of equilibrium of the surrounding bodies are not altered if
we replace the string joining A and B by a force T acting at A along AB and a force T
acting at B along BA.
We shall show that it is not necessary to consider the separate virtual
displacements of the ends of the string or the rod, but merely its increment in length.
25
Taking the case of a string joining A and B where AB = l, and considering a
displacement of A to A and B to B where AB = l + l, we have that the work done by
the two forces introduced as above
= T.AM – T.BN,
where M and N are the projections of A and B on AB,
B
= T(AB –MN) = T(AB – AB cos),
A
where is the angle between AB and AB,
= T(AB – AB)
to the first order of approximation.
A
M T
T B N
Hence the work done is equal to – T l.
Similarly, the work done by the thrust in an extension of a rod from length l to l +
l is +T l, where T is the thrust in the rod.
Roberval’s Balance:
This balance, which is a common form of letter- weigher, consists of four rods
AB, BE, ED and DA freely jointed at the corners A, B, E and D so as to form a
parallelogram, whilst the middle points, C and F of AB and ED are attached to fixed
points C and F which are in vertical straight line. The rods AB and DE can freely turn
about C and F.
To the rods AD and BE are attached equal scale-pans. In one of these is placed
the substance W, which is to be weighted and in the other the counterbalancing weight P.
A .
D
.
C
F
. B
.
E
We shall apply the Principal of Virtual work to prove that it is immaterial on what
part of the scale-pans the weights P and W are placed.
Since CBEF and CADF are parallelograms it follows that, whatever be the angle
through which the balance is turned, the rods BE and AD are always parallel to CF and
therefore are always vertical.
If the rod AB be turned through a small angle the point B rises as much as the
point A falls. The rod BE therefore rises as much as AD falls, and the right hand scalepan rises as much as the left-hand one falls. In such a displacement the virtual work of the
weights of the rod BE and its scale-pan is therefore equal and opposite to the virtual work
of the weights of AD and its scale-pan. These Virtual work therefore cancel one another
in the equation of virtual work.
Also if the displacement of the right-hand scale-pan be p upwards, that of the lefthand one is p downwards. The equation of virtual work therefore gives
26
P.p + W(- p) = 0, i.e. P = W.
Hence, if the machine balance in any position whatever, the weights P and W are
equal and this condition is independent of the position of the weights in the scale-pans.
The weights therefore may have any position on the scale-pans. It follows that the scalepans need not have the same shape, nor be similarly attached to the machine, provided
only that their weights are the same.
Ex.1 Five weightless rods of equal length are jointed together so as to form a rhombus
ABCD with one diagonal BD. If a weight W be attached to C and the system be
suspended from A, show that there is a thrust in BD equal to W/3.
Sol.
Let the five rods AB, BC, CD, DA,
A
and BD form the rhombus ABCD and the

diagonal BD. Since the system is suspended
from A and there is a weight W attached to
C, AC will be vertical and consequently BD
horizontal.
B T
T
D
Suppose that the rod AB or AD makes
an angle with the horizontal through A. let us
give a small symmetrical displacement such
C
that changes to + . Since A is fixed, we
measure the depth of C, the point of application
W
of W, below A. Now AC is 2a sin where a is the
length of a rod and hence the work done by W during the small displacement is W (2a
sin), the point C moves down and hence the plus sign. The length BD is 2a cos and the
work done by the thrust T in the rod AD is T (2a cos).
By the principle of virtual work,
T (2a cos) + W (2a sin) = 0,
or
2a (W cos - T sin) = 0.
Since 0, it follows that
W cos - T sin = 0,
i.e.
T = W cot.
In the equilibrium position, ABC is an equilateral triangle and is 600. Therefore
T = W/3.
Ex.2 A quadrilateral ABCD, formed of four uniform rods freely jointed to each other at
their ends, the rods AB, AD being equal and also the rods BC, CD, is freely suspended
from the joint A. A string joins A to C and is such that ABC is right angle. Apply the
principle of virtual work to show that the tension of the string is
(W+W )sin2 + W,
where W is the weight of an upper rod and W of a lower rod and 2 is equal to the angle
BAD.
27
Sol.
If the angle BCA be equal to , then
in the position of equilibrium is the
A
complement of . Let AB = AD = 2a and
CB = CD = 2b.

Then if we assume from the very
beginning that = /2 - , we shall have
b = a tan and it would be impossible to make
an increment in without altering b or a. This
B
W
W
D
would introduce into the equation of virtual
work the unknown stress in the sides of the
quadrilateral with which we are not concerned.
We, therefore, use an independent symbol for

the angle ACB and imagine a displacement
which alters the angles of the figure and the
W
C
W
length AC but not the lengths of the sides of
the quadrilateral.
Now AC = 2a cos + 2b cos, the depth of the centre of gravity of AB or AD is a
cos and the depth of the centre of gravity of BC or DC is 2a cos + b cos
The equation of virtual work is then
- T(2a cos + 2b cos + 2W(a cos) + 2W(2a cos + 2b cos
or
T(a sin + b sin - Wa sin - W(2a sin + b sin
But and are connected by the relation
a sin = b sin
so that
a cos = b cos
hence
T(tan + tan = W tan + W(2 tan + tan
Now put cot instead of tan and we find that
T = (W+W )sin2 + W.
Ex.3 A regular hexagon ABCDEF consists of six equal rods, which are each of weight W
and are freely jointed together. The hexagon rests in a vertical plane and AB is in contact
with a horizontal table. If C and F be connected by a light string. Prove that its tension is
W3.
E
D
Sol.
W
Let ABCDEF be the regular hexagon
formed of the six rods AB, BC, CD, DE, EF
W
W
and FA, each of weight W. The rod AB is in
F
C
contact with a horizontal table LABM, and C
T
T
and F are connected by a string. Suppose that
the angle FAL or CBM is .
W
W
Give a small displacement such that 

changes to + This displacement changes
L
A
B
M
W
28
the positions of the middle points of the rods BC, CD, DE, EF and FA (AB remaining in
contact with the table), but the lengths of the rods are not altered. Since AB is fixed, we
find the heights of the middle points of these five rods above AB. The height of the
middle point of AF or BC is a sin, where 2a is the side of the hexagon and hence work
done by the weight of AF or BC during the change of to + is - W(a sin). Minus
sign is prefixed, since the point of application moves upwards.
Again the height of the middle point of CD or EF is 3a sin and the work done by
weight CD or EF is - W(3a sin). Similarly we see that the work done by the weight of
DE is - W(4a sin). Since the middle point of AB is not shifted, no work is done by the
weight of AB.
Lastly the length of the string FC is 2a + 4a cosand hence, the work done by the
tension T in the string is
- T(2a + 4a cos).
The equation of virtual work is then
- 2W(a sin) - 2W(3a sin) - W(4a sin ) - T(2a + 4a cos) = 0, 
or
4a(T sin - 3W cos) = 0.
Since
0,
T = 3 cotW
In the position of equilibrium, FAB = 1200 and accordingly = 600. It follows
that T = W3.
Ex.4 A frame ABC consists of three light rods, of which AB, AC are each of length a,
BC of length 3a/2 freely jointed together. It rests with BC horizontal, A below BC and
the rods AB, AC over two smooth pegs E and F, in the same horizontal line, distant 2b
apart. A weight W is suspended from A, find the thrust in the rod BC.
Sol.
Let AB, BC, AC be three rods forming
a framework, such that AB = AC = a and
B T
M
T
C
BC = 3a/2. Let BAM be equal to , where
AM is the perpendicular from A on BC.
K
Let us imagine a displacement which
alters to . Since BC = 2a sin, increase
E
F
in its length is (2a sin) and the work done by
the thrust in BC

= T.(2a sin).

Since the pegs are fixed, we consider the
A
depth of A below EKF. Now AK = bcot , hence
the work done by W is
W
= W.(b cot).
The reactions at the pegs do not work, accordingly the equation of virtual work is
T.(2a sin) + W.(b cot) = 0,
or, since 0,
T.2a cos - W.b cosec2 = 0.
Therefore
29
2
T = Wb cosec  sec
2a
In the position of equilibrium,
sin = ¾, cos = 7/4
hence
T = Wb 16 4
2a 9 7
T = 32Wb 
97 a
Ex.5 A smooth parabolic wire is fixed with its axis vertical and vertex downwards, and in
it is placed a uniform rod of length 2l with its ends resting on this wire. Show that, for
equilibrium, the rod is either horizontal, or makes with the horizontal an angle given by
cos2 = 2a/l,
4a being the latus rectum of the parabola.
Sol.
Let AB be the rod of length l. Choose
Y
OX and OY as coordinate axes, thus the
equation of the parabola may be written as
B
2
x = 4ay.
If the coordinates of the point A be
G

2

(2at, at ) and if BAC be equal to , then
A
 
C
coordinates of B will be (2at + 2l cos, at2 + 2l sin).
Since B lies on the parabola,
A O M
B
X
2
2
(2at + 2l cos) = 4a(at + 2l sin),
W
giving
t = tan – (l cos)/(2a).
If y be the height of the centre of gravity G of the rod,
y = ½ [at2 + (at2 + 2l sin)] = at2 + l sin
y = a[tan – (l cos)/(2a)]2 + l sin
= (l2 cos2)/(4a) + a tan2.
Now displace the rod so that increases to , the ends of the rod remaining
in contact with the wire. By the principle of virtual work,
- W y = 0, where W is the weight of the rod,
or,
W(l2 cos2)/(4a) + a tan2] = 0,
or
W[- (l2 2cossin)/(4a) + 2a tan2sec2] = 0.
Since W 0 and 0, we must have
sin (cos4 – 4a2/l2) = 0.
Therefore either
sin = 0, i.e., = 0,
or,
cos4 – 4a2/l2 = 0,
i.e.,
cos2 = 2a/l.
30
It is clear that = 0 corresponds to the horizontal position of the rod.
Ex.6 Four equal jointed rods, each of length a, are hung from an angular point, which is
connected by an elastic string with the opposite point. If the rods hang in the form of a
square, and if the modulus of elasticity of the string be equal to the weight of a rod, shew
that the unstretched length of the string is (a2)/3.
Sol.
Let AB, BC, CD and DA be four equal
A
rods of length 2a and let the framework hang
from A, the corners A and C being connected
by an elastic string. Let T be the tension of the
elastic string. If we suppose that the depth of
B W
W
D
the centres of gravity of the two upper rods
below A is x, then the depth of the centres of
gravity of the two lower rods would be 3x and
W
W
the length of the string 4x.
C
Now give a displacement to the figure such that AC is stretched. By the principal
of virtual work,
2W(x) + 2W(3x) – T(4x)= 0,
or,
4(2W – T)x = 0.
Since x 0, T = 2W.
But, by Hooke‟s law,
T = W.[(a2 – l)/l],
where l is the unstretched length of the string, hence
(a2 – l)/l = 2.
Therefore
l = (a2)/3.
Ex.7 One end of a beam rests against a smooth vertical wall and the other on a smooth
curve in a vertical plane perpendicular to the wall; if the beam rests in all positions, show
that the curve is an ellipse whose major axis lies along the horizontal line described by
the center of gravity of the beam.
Sol.
Let OY be the wall and CBD be the
curve on which the end B of the rod AB rests.
Y
During a small displacement of the rod at A
and B do no work, hence by the principle of
A
virtual work, if h be the height of the centre of
G
gravity G of the rod above a fixed horizontal

D
line OX, Wh = 0, which gives that h = constant.
M
B

Hence the centre of gravity of the rod
lies always on a horizontal line for all positions of the
rod. Now take OX and OY as the coordinate
axes. If AB is inclined at an angle to the
O
C
X
horizontal and (x, y) are the coordinates of B,

31
we have
x = 2a cos, where AB = 2a.
(1)
The height of G above OX = y + a sin.
Hence
h = y + a sin.
(2)
Eliminating between (1) and (2), we get
x2 + (y – h)2
= l,
4a2
a2
which represents an ellipse with its centre at (0, h). Its major axis is y = h, the horizontal
line described by the centre of gravity.
Ex.8 An endless chain of weight w rests in the form of a circular bond round a smooth
vertical cone, which has its vertex upwards. Find the tension in the chain due to its
weight, assuming the vertical angle of the cone to be 2.
Sol.
Let AB be the endless chain resting
O
under its own weight on the cone. If its length
be 2x, the radius AC of the circle which it
C
form will be x.
A
B
Suppose that a small displacement is
given to the chain such that x becomes x + x.
Then the work done by the tension is
A
B
 T(2x).
C
We may take the vertex O to be the
fixed position from which the level of the
centre of gravity is to be measured. The centre of gravity of the chain is at depth x cot
from O in the position of equilibrium. During the displacement the centre of gravity goes
lower and accordingly the work done by the weight of the chain is
w(x cot).
Since the reactions at the various points of contact do no work, we have, by the
principle of virtual work,
- T.(2x) + w(x cot) = 0,
i.e.,
x(- 2T + w cot) = 0.
Therefore,
since x 0,
T = (w cot)/(2).
Ex.9 Two small smooth rings of equal weight slide on a fixed elliptical wire, whose
major axis is vertical and they are connected by a string, which passes over a small
smooth peg at the upper focus; show that the weights will be in equilibrium wherever
they are placed.
32
Sol.
Let APAQ be the elliptical wire with
A
its upper focus at S, the major axis AA being
vertical. If P and Q be the positions of the
S
smooth rings, each of mass m, then principle
of virtual work gives
M
P
mg (y) + mg (y) = 0,
i.e.,
y + y = 0,
(1)
Q
N
where y and y represent the lengths SM and
SN respectively.
Taking S as the pole and SA as the
A
initial line, the equation of the elliptical wire
can be written
l
.
r=
1 + e cos
(2)
Let SP be r, then
SQ = a – r ( = r, say),
(3)
a being the length of the string.
From (2)
y = SM = SP cos (1800  ) = (r – l)/e, where we have used (2)
and
y = SN = SQ cos ( ), where is the vectorial angle of Q,
= (r – l)/e = (a – r –l)/e, on using (3).
Hence the left hand side of (1) is
r–l
a–r–l
(
) + (
),
e
e
which is identically zero. Hence the weight will be in equilibrium wherever they are
placed.
Ex.10 A frame consists of the bare forming the sides of a rhombus ABCD with the
diagonal AC. If four equal forces P act inwards at the middle points of the sides, and at
right angles to the respective sides, prove that the tension in AC is
P cos2
sin
where denotes the angle BAC.
Sol.
Let ABCD be the rhombus, connected
Y
by the rod AC forming its diagonal. Let BAC
A
be the angle . For the sake of simplicity we
P

P
can suppose that we give displacement such
that becomes , the point of intersection
H
E
O of the diagonals and the directions of the lines
D
X
OB and OA remaining fixed. This means that
O
B
there is alteration in the length of AC and also in
G
F
the position of the middle points E, F, G, and H of
P
P
the sides.
C
33
Now the length of the diagonal AC is 4a cos, where 2a is the side of the
rhombus. The work done by the tension in the rod AC is
 T(4a cos),
where T is the tension.
To find the work done by the equal forces P acting at the middle points E, F, G
and H, take an axis OX along OB and a perpendicular axis OY along OA as shown in the
figure. By symmetry the force P at the four points will do equal work, so that it will
suffice to find the work done by the force at E. The coordinates of E are
x = a sin , y = a cos
and the components of P at E along the positive directions of the axes are
X = - P cos , Y = - P sin
The work done by P in a small displacement is therefore
X x + Y y = - P cos(a sin) - P sin(a cos)
= - aP (cos2 – sin2)
= - aP cos2 
Hence the equation of virtual work gives
- T (4a cos) – 4aP cos2= 0,
i.e.,
4a [T sin – P cos2= 0,
so that
T = P cos2



sin

Ex.11 A uniform square lamina rests equilibrium in a vertical plane under gravity with
two of its sides in contact with smooth pegs in the same horizontal line at a distance „c‟
apart. Show that the angle  made by a side of the square with the horizontal in a nonsymmetrical position of equilibrium is given by
c(sin + cos) = a,
2a being the length of a side of the square.
Sol.
Let the reactions at the pegs P and Q
meet in the point E, then the vertical through
C
G must pass through E.
Let the body be given a small vertical
G
displacement. If z be the height of C.G. above
D
S
the horizontal line PQ, the equation of virtual
E R B
work is
Q
M

Wdz = 0 or dz = 0
P
45
But z = GM = GN – NM = GN – LP

= AG sin( + 45) – AP sin

A N L
= 2 a sin( + 45) – c cos sin
dz = +2 a cos( + 45) – c (cos2 - sin2)
So 0 = +2 a (cos cos45 – sin sin45) – c (cos2 - sin2)

Hence a = c (cos + sin) which gives the position of equilibrium.
34
Ex.12 Two equal light rods AOB, COD freely jointed at O, their middle points, are rest in
a vertical plane with their ends B, C on smooth horizontal table. A string to the ends of
which equal weight are attached, passes over A and D. Show that, in the position of
equilibrium, in the angle between the rods is tan1(4/3).
Sol.
Since the equal rods are joined at their
middle point and the weights are equal. It is
D
A
clear that the string must be placed symmetrical
over the rods. Let 2 be the supposed angle
between the rods.
a
Let the systems be given a small vertical
W
O
W
displacement so that angle changes. Let 2a be
the length of each rod.
The equation of virtual work is
y
y
- 2W dy = 0, i.e., dy = 0.
Also the length of the string = l.
So
2AC – 2y + AD = l.
B
C
Hence 4a cos – 2y + 2a sin = l,
i.e.
(- 2a sin + a cos) d – dy = 
Therefore 2a sin = a cos
i.e.
tan = ½
So
tan2 2 tan2  2½  4
1 – tan 
1–¼ 3
Hence angle between the rods = tan-1(4/3).
Ex.13 A solid hemisphere is supported by a string fixed to a point on its rim and to a
point on a smooth vertical wall with which the curved surface is in contact. If are the
inclinations of the string and the plane base of the hemisphere to the vertical show that.
tan = 3/8 + tan
Sol.
Let O be the point of suspension and
C the centre of the hemisphere. The normal
O
reaction at F will be perpendicular to the
l
wall and also pass through C. Hence FC is
D
horizontal.

If G is the C.G. of the hemisphere,
K
A
CG = 3/8.

Let OA = l.
Let y be the depth of G below the
F
E H 
C
fixed point O.
y = OK + AE + HG
B
= l cos + a cos + (3/8) a sin (1)
G
The only force doing work will be the
weight of the hemisphere.
The equation of virtual work is
35
wdy = 0; so dy = 0.
But from (1),
dy = - l sin d – a sin d + (3/8) a cos d
Hence l sin d = [(3a/8) cos - a sin] d
Also and are connected by the relation
l sin + a sin = a
[since CF = CE + AK].
Hence l cos d + a cos d= 0.
Therefore
l sin d = a[(3/8) cos - sin] [- (l cos)/(a cos)]d
So
sin cos = - (3/8) cos cos + cos sin
Dividing by cos cos,
we get tan = - 3/8 + tan
or
tan = 3/8 + tan
Ex.14 Six equal heavy beams are freely jointed at their ends to form a hexagon are placed
in a vertical plane with one beam resting on a horizontal plane; the middle points of the
two upper slant beams, which are inclined at an angle  to the horizon, are connected by a
light cord. Show that its tension is 6W cot, where W is the weight of each beam.
Sol.
Let a be the length of each beam and
G1, G2, G3, G4, G5, G6 be their centres of gravity.
G1
The points G2 and G6 are connected by
E
D
a cord of tension T.
T
W
DCH = EFK = 
G2
G6
Let the system be given a slight downward
W
T


W  C
vertical displacement. If the heights of G3, G4,
F 
above AB be each equal to x, that of G2, G6 will be
K
H
3x and that of G1 will be 4x.
G3
G5
The work done by the weights
W
G4
W
= - 2Wdx – 2Wd(3x) – Wd(4x) = - 12Wdx
[- ve, since weights act downward
A
B
and x increases upward].
Work done by tension = - Tdp, where p = G2G6
= - T(- 2a sin d) = 2a + 2a cos
and so dp = - 2a sin d = T.2a sin d.
The equation of virtual work is
- 12Wdx + T.2a sin d= 0
Again, since x = a sin , so dx = a cos d
- 12W a cos d+ T.2a sin d= 0.
Hence T = 6W cot.
Ex.15 Four rods are jointed together to form to form a parallelogram, the opposite joints
are joined by strings forming the diagonals, and the whole system is placed on a smooth
horizontal table. Show that their tensions are proportional to their lengths.
Sol.
Let the string AC = x and string BD = y.
36
Let the system be supposed given a slight displacement so that only the lengths of
the strings change.
The equation of virtual work is
D
C
T x – T y = 0.
(1)
If AB = a and AD = b, then
T
T
2
2
2
2
x + y = 2(a + b ) (by geometry)
Hence 2x x + 2y y = 0
T T
From (1) and (2)
A
B
T/x = T/y.
Hence the tensions are proportional to the lengths of the strings.
Ex.16 A uniform square lamina of side a and weight w is suspended from four points in a
horizontal plane by equal inextensible vertical strings of length l attached to the corners
of a square. Show that the couple that would be necessary to hold the lamina in this
position in which it has been turned through an angle  from the former position is
wa2 sin
2l2 – 2a2 sin2½)
Sol.
ABCD is the square and AL, BM, CN
and DP are the strings. If O is the centre of the
P
square, let O be the displaced position of O
when a couple of moment G is applied to it.
L
N
Draw OK CN
So
KOC = , CN = CN = l
M
 D
K
and
OK = OC = a/2
O 
C
If OO = y, the equation of virtual work is

O
wdy – Gd = 0
(1)
A
C
2
2
2
also
CN = CK + KN
B
So
l2 = 2a2 sin2½ + (l – y)2
(2)
Hence 0 = 2a2.2 sin½ cos½ ½d + 2(l – y)(- dy)
So
(l – y)dy = ½ a2 sin d
Comparing with (1);
w
2G
=
2
l–y
a sin
2
wa sin
G=
2l2 – 2a2 sin2½)
Ex.17 Six equal rods AB, BC, CD, DE, EF, and FA are each of weight W and are freely
jointed at their extremities so as to form a hexagon; the rod AB is fixed in a horizontal
position and the middle point of AB and AD are jointed by a string; prove that its tension
is 3W.
Sol.
Let G1, G2, G3, G4, G5 and G6 be the middle points of the rods. Since, by
symmetry, BC and CD are equally inclined to the vertical the depths of the points C, G3
and D below AB are respectively 2, 3 and 4 times as great as that of G2.
37
Let the system undergo a displacement in the vertical plane of such a character
that D and E are always in the vertical lines through B and A and DE is always
horizontal. If G2 descend a vertical distance x, then G3 will descend 3x, G4 will descend
4x, whilst G5 and G6 will descend 3x and x respectively.
The sum of the virtual works done by
the weights
G1
= W.x + W.3x + W.4x + W.3x + W.x
A
B
= 12 W.x.
G6 T
G2
If T be the tension of the string, the
W
W
virtual work done by it will be
F
C
T (- 4x).
G5
G3
For the displacement of G4 is in a
W
G4 T
W
direction opposite to that in which T acts
E
D
and hence the virtual work done by it is negative.
W
The principle of virtual work then gives
12W.x + T(- 4x) = 0, i.e. T = 3W.
Ex.18 Four equal uniform rods are jointed to form a rhombus ABCD, which is placed in
a vertical plane with AC vertical and A resting on a horizontal plane. The rhombus is
kept in the position which BAC = by a light string joining B and D. Shew that its
tension is 2W tan, where W is the weight of rod.
Sol.
Let x be the height above A of the middle points of AB and AD, so that 3x is
clearly the height of the middle points of BC and CD.
Let BO = y = OD, where O is the centre
C
of the rhombus. Choose as our displacement one
in which becomes and hence x becomes
G3
G2
x + x and y becomes y + y.
WW
Then, T being the tension of BD, the
equation of virtual work is
D
T
O T B
2T(-y) + W(-x) + W(-x) + W[-(3x)] + W[-(3x)] = 0.
G4
G1
So
T = - 4W(x/y),
W
W
Now, if AB = 2a, we have x = a cos and y = 2a sin.
A
Hence
x
- a sin
=
= - ½ tan
y
2a cos
Therefore
T = 2W tan
Ex.19 A uniform beams rests tangentially upon a smooth curve in a vertical plane and
one end of the beam rests against a smooth vertical wall; if the beam is in equilibrium in
any position, find the equation to the curve.
Sol.
Take the wall as the axis of y and any point O on it as the origin. If y be the height
of the centre of gravity of the beam above Ox, the equation of virtual work becomes
W.y = 0. So y = constant = h.
Hence G is the point (a cos, h), where 2a is the length of the rod and is its
inclination to the horizontal.
38
Hence the equation to AG is
y – h = tan (x – a cos)
= x tan – a sin.
For its envelope, differentiating with
respect to , we have x = a cos3 and
y – h = - a sin3.
Hence x2/3 + (y – h)2/3 = a2/3, so that
the required curve is a portion of a four-cusped
hypocycloid.
B
y
G
P
A
O


M
x
1.5 Catenary
Definition:
A string is said to be perfectly flexible if the action across any normal section of it
is a single force acting along the tangent to the string. Such a string, therefore, offers no
resistance to any action tending to bend it at a point i.e., it possesses no rigidity of shape.
The section of the string is considered to be so small that it may be treated as a
curved line.
A chain, whose links are very small and perfectly smooth, may be treated as a
flexible string.
A string or a chain is taken as uniform if its weight per unit length is the same
throughout.
The curve in which a uniform inextensible heavy string hangs freely under gravity
is called a Catenary.
Equation of Catenary:
Theorem:
A uniform heavy inextensible string hangs freely under the action of gravity. Find
the equation of the curve, which it forms.
Proof:
Let A be the lowest point of the string where the tangent is horizontal. Let P be
any point on it. If arc s be measured from A, let arc AP = s.
Consider the equilibrium of the portion AP of the string. The string is under
tension due to rest of the part.
The three forces acting on it are:
(i)Its weight acting vertically down through its C.G.
(ii)Tension T0 at A along the tangent in the sense PA.
(iii)Tension T at P along the tangent in the sense AP.
Since these three forces are in equilibrium, the line of action of the weight must
pass through the point of intersection of tangents at A and P.
39
Resolving horizontally and vertically,
T cos = T0
T sin = ws
where w is the weight per unit length of the string.
tan = ws/T0
If T0 be taken equal to ws i.e., weight of
T
length c of the string, then
P
S
tan = (ws)/wc

s = c tan 


T0
A
Q
which is the required intrinsic equation of the catenary.
Note:
ws
The point A is called the vertex of the catenary and c is called the parameter.
The horizontal line below A at a distance c from it is called the directrix.
The vertical line through the lowest point A is called the axis of the catenary.
If the catenary is formed by suspending the string from two points R and S at the
same horizontal level the distance RS is called the Span.
Relation between x, y and :
The intrinsic equation is
s = c tan
So
ds/d = c sec2
Now dy/d = (dy/ds).(ds/d)
= sin.c sec2
So
dy/d = c tan sec
Integrate
y = c sec + A
Let
y = c when = 0
i.e., the origin is taken at depth c below A.
So
c = c.1 + A, hence A = 0
Therefore y = c sec.
Again dx/d = (dx/ds).(ds/d) = cos c sec2 = c sec
Integrate
x = c log (sec + tan) + B
Let us take the y-axis passing through the lowest point,
therefore
= 0 when x = 0
i.e.,
0 = 0 + B, hence B = 0.
Therefore x = c log (sec + tan).
Cartesian Equation of the catenary:
The intrinsic equation is
s = c tan = c (dy/dx)
40
ds = c d2y
dx
dx2
2
[1 + ( dy)2] = c d y2
dx
dx
So
1 = d2y/dx2
.
2
c
[1 + (dy/dx) ]
Integrating
x/c = sinh– 1 (dy/dx) + A
where A is an arbitrary constant.
Let the y-axis be taken as a vertical line through A
dy/dx = 0 when x = 0, so A = 0
Therefore x/c = sinh– 1 (dy/dx),
hence dy/dx = sinh(x/c)
Integrate again,
we get
y = c cosh(x/c) + B
Let the origin be taken vertically below A at a depth c from it. Hence at A, y = c
when x = 0.
Therefore
c = c + B i.e., B = 0
Hence y = c cosh(x/c) is the required equation of the catenary.
So
Relation between x and s:
or
i.e.,
y = c cosh(x/c)
So
dy/dx = sinh(x/c)
Therefore tan = sinh(x/c)
s/c = sinh(x/c)
s = c sinh(x/c)
Tension at a point:
(a)We know that T cos = T0 = wc.
Therefore T = wc sec
Hence T = wy
[since y = c sec]
Therefore tension at a point is equal to the weight of a length of the string equal to
its height above the directrix.
In other words, the tension at any point P of a catenary is equal to the weight of
the portion of the string whose length is the vertical distance between P and the directrix.
(b)Since T0 = wc, the tension at the lowest point is equal to the weight of string of length
equal to its height above the directrix.
(c )Again, since T cos = T0 = wc, the horizontal component of the tension at any point
is constant and equal to wc.
(d)Since T sin = ws, the vertical component of the tension at any point is equal to the
weight of the length of string lying between the point and the vertex.
Cor:
Since T = wy.
41
If a heavy string hangs over two smooth pegs not in the same vertical line, since
the tension at the pegs must support the vertical part of the string, the ends of the string
must rest on the directrix.
Geometrical Properties of Catenary:
1.From the equation T cos = T0, we get that the horizontal component of the tension at
every point of the curve is the same and is equal to wc or T0.
2.From the equation T sin = ws, we deduce that the vertical component of the tension at
any point is equal to ws, i.e. equal to the weight of the portion of the string lying between
the vertex and the point.
3.The equation T0 = wc, shows that the tension at the lowest point is equal to the weight
of the string whose length is the same as the distance between the origin and the vertex.
4.On squaring each side of
y = c sec and s = c tan,
and substituting we have
y2 = s2 + c2.
5.If denotes the radius of curvature at any point, then
 = ds/d = c sec2
Approximations to the common catenary:
1.The equation of the catenary has been shown to be
y = c cosh(x/c) = ½ c(ex/c + e– x/c)
= c[1 + x2/(2! c2) + x4/(4! c4) + … ],
on expending the exponentials.
If x/c is small, the series on the right can be limited to the first few terms and the
Cartesian equation approximates to
y = c + x2/(2c),
where we have retained the first two terms.
This shows that so long as x is small and c large, the curve coincides very nearly
with a parabola of latus rectum
2c or 2T0/w.
2.We now obtain an approximation to the shape of catenary when x is large, i.e. at points
far removed from the lowest point. When x is large, e– x/c becomes very small, hence
y = ½ c(ex/c + e– x/c) behave as y = ½ c ex/c.
Thus for very large values of x, catenary behaves as an exponential curve.
Sag of a tightly stretched wire:
We shall now consider a wire stretched nearly horizontal – as for instance a
telegraph wire. Thus let B, C be two points in a horizontal line between which a wire is
stretched. Let l be the length, W the total weight, T0 the horizontal tension, k the sag AN
and h the span BC.
42
The tension T0 may be found in terms
of W from the first principles. For instance,
B
N
C
taking moments about C for the portion AC of
the wire, we have
T0.k = ½ W.¼l
approximately, or
T0
T0 = Wl/(8k).
A
We can calculate the increase in the
W/2
length of the wire on account of the sag by
considering the equation s = c sinh(x/c).
Since cos2 = c, c will be large if the curve is flat near its vertex. It follows that
x/c will be small in the case of a tightly stretched string. Retaining only the two terms in
the expansion of sinh(x/c) we get from (1), s = x + x3/(6c2).
Hence
s – x = AC – NC = x3/(6c2) approximately
2 3
= w x2
6 T0
If we put x = ½ h, we obtain that the total increase, due to sagging, in a span of
length h, is
2 3
2s – h = w h 2
24T0
Ex.1Show that the length of an endless chain which will hang over a circular pulley of
radius a so as to be in contact with two-thirds of the circumference of the pulley is
3
.
4
a
+
.
log(2 + 3) 3
Sol.
Let ABLCA be the endless chain
Y
hanging over the circular pulley MBLC of
L
radius a. Since the chain is in contact with
two-thirds of the pulley, this portion
CLB = (2/3).(2a) = 4a/3.
O
The remaining portion CAB will
hang in the form of a catenary with its
N
vertex at the lowest point A. Its length is
C
B
M
therefore twice the length of AB.
The tangent at B is perpendicular to
OB and since
A
60 0
COB = 2/3, NOB = /3,
and hence the tangent at B is inclined at /3
O
X
to the horizontal.
And NB = a cos(/6) = (a3)/2.
Applying the formula x = c log(tan + sec) for the point B, we have
a3
.
c=
2 log(2 + 3)
[since = /3].
Hence the length
AB = c tan
43
3a
.
2 log(2 + 3)
The length CAB is accordingly
3a
.
=
log(2 + 3)
It follows therefore that the total length of the chain is
3
.
4
a
+
.
log(2 + 3) 3
=
Ex.2 A weight W is suspended from a fixed point by a uniform string of length l and
weight w per unit of length. It is drawn aside by a horizontal force p. Shew that in the
position of equilibrium, the distance of W from the vertical though the fixed point is
P sinh–1 W + lw
W
- sinh–1
w
P
P
Sol.
Let BC be the string of length l, suspended from the fixed point B, with weight W
hanging at C and drawn aside by a horizontal force P acting at C. The tension at C
balances the resultant of P and W. Hence the direction of the tangent at C is given by
tanc = W/P.
We shall next find the direction of the tangent at B. If X and Y be the horizontal
and vertical components of the tension at B, we have, for the equilibrium of BC,
X = P,
Y = W + lw,
where lw is the weight of the string BC. Thus
Y
the direction of the tangent at B is given by
tanb = (W + lw)/P.
B
Let the horizontal distance of C from A, X
the lowest point of the catenary formed by BC,
C
be x. Then
P
s = c tanc = c sinh(x/c), where AC = s,
A
whence
x = c sinh–1(tanc) = c sinh–1(W/P).
W
Also the horizontal component of the
tension at any point is P. Hence
P = wc,
therefore
c = P/w
Accordingly
x = (P/w) sinh–1(W/P).
Similarly we can show that the horizontal distance of B from A is given by
x = (P/w) sinh–1[(W + lw)/P].
Hence the required horizontal distance
= x – x
= P sinh–1 W + lw - sinh–1 W
w
P
P
44
Ex.3 The end links of a uniform chain slide along a fixed rough horizontal rod. Prove that
the ratio of the maximum span to the length of the chain is
2
log 1 + (1 + )

where is the coefficient of friction.
Sol.
Let AB be the maximum span. In this
position, the end links A and B are in a state
of limiting equilibrium. If the reaction at A be
R
R
R, then (where tan = ) is the angle that
 

the resultant of R and the frictional force R

makes with the direction of R. For the
R
R
B
A
equilibrium of A, the direction of this resultant
o
should be opposite to the direction of the
tension at A, hence the tangent at A makes an angle 900 – with the horizontal.
The length of the chain
= 2s = 2c tan = 2c cot= 2c/,
and the maximum span
AB = 2x = 2c log(tan + sec) = 2c log (cot + cosec)
2
= 2c log 1 + (1 + )

Hence the required ratio
= 2x
2s
2
=  log 1 + (1 + )

Ex.4 A kite is flown with 600 ft. of string from the hand to the kite, and a spring balance
held in a hand shows a pull equal to the weight of 100 ft. of the string, inclined at 30 0 to
the horizon. Find the vertical height of the kite above the hand.
Sol.
Let PB be the string with the kite at B.
The hand at P experiences a pull of 100w,
where w is the weight per unit length of the
Y
string. PB forms part of the catenary APB,
B
tangent at P giving the direction of the pull.
Since the tension T at any point is equal to
wy, where y is its ordinate, we have at P,
T = wy = 100w,
therefore
y = 100.
Also y = c sec,
A
P
0
hence
c = y cos = 100 cos30
30 0
= 503 ft.
L
O
M
X
45
Further
AP = s = c tan = 503.tan300 = 50 ft.
We have to find the vertical height of B above P. Let this be k. Then the ordinate
of B will be (100 + k) ft. Applying the formula
y2 = c2 + s2,
for the point B, we get that
(100 + k)2 = (503)2 + (650)2, since APB = AP + PB
= 430000 = 431002.
Therefore
k = 100(43 – 1) = 555.7 ft. nearly.
Ex.5 A uniform chain is hung up from two points at the same level and distance 2a apart.
If z is the sag at the middle, shew that z = c [cosh(a/c) – 1].
If z be small compared with a, shew that 2cz = a2 nearly.
A telegraph wire is supported y two poles distant 40 yards apart. If the sag be one
foot and the weight of the wire half an ounce per foot, show that the horizontal pull on
each pole is ½ cwt. nearly.
Sol.
Let CAB be the uniform chain, its equation, referred to OX and OY as axes, is
y = c cosh(x/c),
where OA = c.
Y
The sag is, therefore
C
B
= OD – OA = c cosh(a/c) – c
D
= c [cosh(a/c) – 1],
for the „y‟ of D is the same as the „y‟ of B
A
whose x-coordinate is given to be a.
Now
2
4
O
X
cosh a = 1 + 1 a 2 + 1 a 4 + … ,
c
2! c
4! c
Hence substituting in (1), we get
1 a2 1 a4
z=
+
+…,
2! c
4! c3
When z is small compared with a, c must be large and hence neglecting the
second subsequent terms in the above expansion, we get
2cz = a2 nearly.
In the numerical part,
z = 1 ft. and a = 60 ft., hence
c = a2/(2z) = 6060/2 = 1800 ft.
The required horizontal pull at A
= wc = (1/32).1800 lbs. = ½ cwt. nearly.
Ex.6 A telegraph wire, stretched between two points at a distance „a‟ feet apart, sags n ft.
in the middle; prove that the tension at the ends is approximately w[a2/(8n) + 7n/6].
Sol.
If y be the ordinate at the extreme
y = n + c, so y – c = n
46
Also
y = c cosh(x/c) = c cosh(a/2c), since x = a/2
1 a2 1 a4
= c[1 +
+
+…]
2! 4c2 4! 16c4
Therefore
a2
a4
approx.
n=y–c=
+
8c
24c3
B
a2
a4
cn =
+
8
24c2
But to the first approximation
C = a2/(8n)
Y
a/2
A
n
y
c
O
X
Hence
or
2
4
2
cn = a + a 64n4
8
24a
= a2/8 + n2/6
c = a2/(8n) + n/6
Now T = wy = w(c + n)
a2
7n
=w
+
8n

approx.
Ex.7 A chain ABC is fixed at A and passes over a smooth peg at B and the portion BC is
vertical. The length of each of the portions AB, and BC is 15 ft. and depth of B below the
horizontal through A is 5 ft. Find the horizontal pull at A, having given that wt. of whole
string is 45 lbs.
Sol.
Since BC = 15 ft. and the string passing over a smooth peg retains the same
tension, this tension balances the weight of the string BC. So, if B is the point (x, y) then
wy = w.15 or y = 15 ft.
Let arc DB = s1 and arc DA = s2.
Therefore, at A(x, y), y = 15 + 5 = 20 ft.
Now from the relation,
y2 = c2 + s2
At B; (15)2 = c2 + s12
(1)
A
Y
At A; (20)2 = c2 + s2 2
(2)
Subtracting them, we get
s2
s1
(20)2 – (15)2 = s2 2 – s12
175 = (s2 – s1)(s2 + s1)
D
15 ft.
But
s2 + s1 = 15, therefore
s2 – s1 = 175/15 = 35/3 ft.
Hence s1 = 5/3 ft.
O
C
X
Now from (1)
225 = c2 + 25/9
So
c = 205/3 ft.
Hence w = wt. per unit foot of length = 45/30 = 3/2 lbs.wt.
Therefore, tension at the lowest point = wc
= (3/2)205/3 lbs.wt. = 105 lbs.wt.
47
Ex.8 A uniform chain of length 2l and weight W is suspended from two points, A and B,
in the same horizontal line. A load P is now suspended from the middle point D of the
chain and the depth of this point below AB is found to be h. Show that each terminal
tension is
Pl
W(h2 + l2)
½
+
h
2hl
Sol.
Let the weight P be suspended from the middle point D of the string. Let C be the
lowest point of the catenary of which DA is part. (DA and DB are not parts of the same
catenary)
Let arc CD = s and ordinate of D be y. Let T, T be the tension at D towards the
two parts. Then the resultant of T, T at angle (180 – 2) = 2T cos(90 - ) = 2T sin.
For equilibrium at the point D,
2T sin = P
2wy sin = P
or
2Wys
B
A
=P
2ly
or
s = lP/W
h
l
Also at D,
y2 = c2 + s2
at A,
(y + h)2 = c2 + (s + l)2
T
T
180 – 2
Subtract,

2hy + h2 = 2ls + l2
2
2
D
So
y = l – h + 2l lP
2h
2h W
2
2
2
or
C
P lbs
C
y= l –h + l P
2h
hW
Therefore, tension at A
= w(h + y)
W
l2 – h2 + l2P
=
h+
2l
2h
hW
Pl
W(h2 + l2)
=½
+
h
2hl
Ex.9 A uniform chain, of length 2l, has its ends attached to two points in the same
horizontal line at a distance 2a apart. If l is only a little greater than a, shew that the
tension of the chain is approximately equal to the weight of a length
a3 . 1/ 2
6(l – a)
of the chain, and that “sag”, or depression of the lowest point of the chain below its ends
is ½[6a(l – a)]nearly.
Sol.
Since l is very little greater than a, the tension of chain must be very great and
hence c must be large.
Now from
48
s = c tan = ½ c[ex/c – e–x/c], we get
l = ½ c[ea/c – e–a/c], we get
or
2a
2a3
l = ½ c[
+ 3
+…]
c
6c
or
a3
a5
l=a+
+
+
2
6c
120c4
…
A first approximation is then l – a = a3/(6c2), and hence
a3 . 1/ 2
c=
6(l – a)
so that the tension at the lowest point is equal to a weight of this length of the chain.
The ordinate of the end of the chain is,
y = ½ c[ea/c + e–a/c], we get
or
2a2
2a4
y = ½ c[2 + 2 +
+…]
2c
24c4
or
a2
a4
y=c+
+
+…
2c
24c3
Hence the sag of the lowest point
= y – c = a2/(2c) approx.
1/ 2
= ½ a2 6(l3 – a)
a
= ½[6a(l – a)]
It follows easily that if d be the sag of the lowest point, then the tension there is
approximately equal to a length a2/(2d) of the chain.
Ex.10 A uniform chain of length l hang between two points A and B which are at a
horizontal distance „a‟ from one another, with B at a vertical distance b above A. Prove
that the parameter of the catenary is given by 2c sinh(a/2c) = (l2 – b2).
Sol.
Let C be the vertex of the catenary of which AB is a part. Let A be the point (x, y)
and arc AC = s. So
y = c cosh(x/c) = ½ c[ex/c + e–x/ c]
(1)
At B; y + b = c cosh[(x + a)/c]
Y
= ½ c[e(x + a)/c + e–(x + a)/ c]
(2)
s = c sinh(x/c) = ½ c[ex/c – e–x/ c]
(3)
a
B
s + l = c sinh[(x + a)/c]
b
= ½ c[e(x + a)/c – e–(x + a)/ c]
(4)
C
l
From (1) and (2) subtracting:
A
b = ½ c[e(x + a)/c + e–(x + a)/ c] – ½ c[ex/c + e–x/ c]
From (3) and (4) subtracting:
l = ½ c[e(x + a)/c – e–(x + a)/ c] – ½ c[ex/c – e–x/ c]
O
X
Therefore,
l + b = c[e(x + a)/c – ex/ c] = c ex/c[ea /c – 1]
l – b = – c[e–(x + a)/ c – e–x/ c] = – c e–x/ c[e–a / c – 1]
Multiplying them, we get
l2 – b2 = – c2[– ea / c – e– a / c + 2] = c2[ea / c + e– a / c – 2]
= c2[ea / 2c  e– a / 2c]2 = c2 4 sinh2(a/2c), i.e. (l2 – b2) = 2c sinh(a/2c)
49
Ex.11 A uniform string, of length 18a ft. and weight 3W lbs., lies on a horizontal smooth
table along with a rod of weight 2w 1bs., to the ends of which its ends are attached. When
the middle point of the string is raised to a height 7a, the pressure on the table just
vanishes. Show that the length of the rod is 16a log 2 ft.
Sol.
Let C be the vertex of the catenary of which AD is a part. When the pressure of
the rod on the table just vanishes, the tension at A and B just balance the weight of the
rod.
Therefore, resolving the forces vertically;
2T sin = 2W,
wy(s/y) = W
where A is the point (x, y) and arc AC = s
D
But
w = 3W = W
18a
6a
9a
9a
So
3Ws = W
7a
18a
Therefore
s = 6a
T
T
A  B
Now at A;
y2 = c2 + (6a)2
and at D;
(y + 7a)2 = c2 + (15a)2
C
Subtract; 14ay + 49 a2 = 189a2
or
y = 10a, so
c = 8a
Again since s = c sinh(x/c)
At A;
6a = 8a sinh(x/c)
Therefore
x/c = sinh–1(3/4) = log[3/4 + (1 + 9/16)]
So
x = c log2
Again at D; 15a = 8a sinh[(x + b)/c]
So
(x + b)/c = sinh–1(15/8) = log[15/8 + (1 + 225/64)]
= log4 = 2 log2
Therefore
x + b = 2c log2, so b = c log2
Hence length of the rod = 2c log2 = 16a log2
Ex.12 A heavy uniform string 90 inches long hangs over two smooth pegs at different
heights. The parts, which hang vertically, are of lengths 30 and 33 inches. Prove that the
vertex of catenary divides the whole string in the ratio 4:5, and find the distance between
the pegs.
Sol.
Let s1 and s2 be the actual distances of the two pegs from the vertex C of the
catenary, so that s1 + s2 = 27. If c be the parameter of the catenary then by y2 = c2 + s2, we
get s12 + c2 = 302 and s22 + c2 = 332; for the ends of the strings must lie on the directrix of
the catenary by the property (a) of Art 4.6, since the tension of the string is unaltered by
its passing round a smooth peg.
Hence, easily, s1 = 10, s2 = 17 and c = 202, so that
s1 + 30
4
=
s2 + 33
5
50
Also from y = c cosh(x/c), we have
2
2
x = c loge y + (y – c )
c
Hence, when y = 30 or 33, x = 102 loge2 or 102 loge(25/8).
Therefore,
x1 + x2 = 202 loge2.5 = 28.2840.9163 = 25.92,
so that the horizontal and vertical distance between the pegs are 25.92 and 3 inches.
Ex.13 A box kite is flying at a height h with a length l of wire paid out, and with the
vertex of the catenary on the ground; shew that at the kite the inclination of the wire to
the ground is 2 tan–1(h/l), and that its tensions there and at the ground are
w(l2 + h2)
2h
and
w(l2 – h2)
2h
where w is the weight of the wire per unit of length.
Sol.
Let P is the kite and C is on the ground, so that NP = h + c and hence, from the
triangle NYP,
Y
2
2
2
2
2
(h + c) = NY + YP = c + l
So
(l2 – h2)
A
B
c=
2h
and
NP = c + h
T
2
2
(l + h )
ws
P
=
2h
Also cos = c/(c + h)
T0
C Y
(l2 – h2)
c
= 2

l + h2
so that
tan(/2) = h/l
O T
N
X
Also the required tensions at P and C are w.PN and w.c.
Ex.14 A uniform chain, of length 2l and weight W, is suspended from two points, A and
B, in the same horizontal line. A load P is now suspended from middle point D of the
string; if AB=2a, find the depth below AB of the position of D.
Sol.
Let C be the lowest point of the catenary of which DA is a part; let its parameter
be c, let the arc CD = s and let the ordinate of D be y. Then
P = vertical component of the tensions at D
= 2T sin
= 2Wys = Ws
2ly
l
Also
y2 = c2 + s2
and
(y + h)2 = c2 + (s + l)2
(1)
if h be the required depth.
51
Theses equations give s = lP/W, and
2
y = l (W + 2P) – h
2hW
2
Also, if x be the abscissa of D, then by
y = c cosh(x/c) and s = c sinh(x/c), we have
y + s = c ex/c and y + h + s+ l = c e(x + a)/c
So
ea/c = 1 + h + l
y+s
On substituting from (1) and (2), we have an equation to give h.
(2)
Ex.15 A chain of length 2l, is hung over two small smooth pulleys which are in the same
horizontal line at a distance 2a apart; to find the positions of equilibrium and to determine
whether they are stable.
Sol
Since the tension of the chain is
A
a
a
A
unaltered by passing over the pulley and since
on one side it is equal to the weight of the
free part AN and on the other it is equal to
C
the weight of the chain that would stretch
vertically down to the directrix, it follows
c
that N and also N lie on the directrix of the
catenary.
x
Hence
N
O
N
l = arc CA + line AN
= ½ c[ea/c – e–a/c] + ½ c[ea/c + e–a/c]
= c ea/c
(1)
R
where c is the parameter of the catenary.
Equation (1) cannot be solved
algebraically, but a graphic solution may be
Y
P
obtained as follows when a and l are given
numerically.
Q
X
Put a/c = X; then e = (l/a)X
(2)
Draw the curve Y = eX and
the straight line Y = (l/a)X.
The points Q and R in which they cut give,
on measurement of their abscissa, approximate
O
X
solutions for X and hence for c. It is clear
that there will be two real, coincident or
imaginary solutions according as
A
a
a
A
<
(l/a) >
= tan POX,
where OP is the tangent from O to the curve.
Now P is given by
C
Y/X = tan POX = dY/dX = eX =Y,
and is therefore the point (1, e), so that
c
tan POX = e.
52
There are therefore two, one or no possible catenaries according as
l <= ae.
>
One catenary will be somewhat as drawn in the first figure and the other as in the
annexed one. The parameter c of the first case is clearly greater than in the second.
Stability or instability:
The height of the centroid of the catenary above its directrix = (cx + ys)/(2s).
So its depth below AA
= y – cx + ys = ys – xc
2s
2s
Hence the depth below AA of the C.G. of the whole chain
2s.(ys – xc) 2y.y
+
2s
2
=
2s + 2y
2
y
+
ys
–
xc
=
2(y + s)
Now in the above case x = a; y = ½ c[ea/c + e–a/c], s = ½ c[ea/c – e–a/c] and
l = y + s = c ea/c.
Therefore depth below AA of the centre of gravity of the whole chain
a/c
–a
= ye a/c
2e
(l/c).(c/2)[l/c
+ c/l] – a
=
2l/c
2
2
l
+
c
–
2ac
=
4l
2
(l
–
a2) + (c – a)2
=
4l
Hence, the greater c, the greater is the depth of the centre of gravity below AA‟.
Hence the first form of the possible curves the stable one, and the second is the unstable.
Ex.16 A uniform heavy string of given length l is attached to two points P and Q, the
latter point being horizontal and vertical distances, h and k, from P; to find the parameter
c of catenary in which it rests.
Sol.
Let P be the point (x, y) referred to the directrix Ox and the vertical line through
the lowest point of the Catenary. Then,
y = ½ c[ex/c + e–x/c]
(1)
(x + h)/ c
–(x + h)/ c
y + k = ½ c[e
+e
]
(2)
and
l = SQ – SP = ½ c[e(x + h)/ c – e–(x + h)/ c] – ½ c[ex/c – e–x/c]
(3)
Therefore
l + k = ce(x + h)/c – cex/c = cex/c(eh/c – 1)
(4)
and
l – k = – ce–(x + h)/c + ce–x/c = ce–x/c(1 – e–h/c)
(5)
Hence
l2 – k2 = c2 (eh/c – 2 + e–h/c)
giving
53
(l2 – k2)= c (eh / 2c – e–h / 2c)
(6)
This equation can not be solve algebraically. A graphic solution may be obtained
by putting h/2c = X, so that (6) gives
2
2
sinhX = (l – k ) X
h
(7)
and hence X is given as the point in which the straight lines
2
2
Y = (l – k ) X
h
meet the curve Y = sinhX.
On drawing the curve we obtain two equal and opposite values for X, and hence
two equal and opposite values for c, provided that
(l2 – k2) > 1
h
i.e. provided that l is greater than length PQ.
The value of c being now know to any degree of approximation, equation (4)
gives x and then equation (1) gives y. The solution is therefore complete.
It is clear that only the positive value of c need be taken. For a negative value of c
would make y negative.
Supposing that a value of X1 as an approximate solution of (7) has been obtained
by graphical methods, a further approximation may be obtained by analysis. For, taking
the upper sign in (6) and putting
(l2 – k2)
= 
h
we have to solve sinh X = X, where X1 is an approximate solution.
Putting X = X1 + , where  is small, we have
sinh(X1 + ) = (X1 + ),
i.e.
sinhX1 + coshX1 + … = (X1 + ), by Taylor‟s theorem.
Hence, neglecting square of , we have
2
2
1
= X1 – sinhX1 = (l – k ) X1 – h sinhX
coshX1 – 
h coshX1 – (l2 – k2)
so that X1 +  is a second approximation.
1.6 Unit Summary:
1. A system of forces acting in one plane at different points of a rigid body can always be
reduced to a single force through any given point and a couple.
2. A system of forces acting in one plane at different points of a rigid body can be
reduced to a single force, or a couple.
3. If G = 0, the given forces reduce to the single force R. Hence in every case the forces
can be reduced either to a single force, or a couple.
4. If three forces, acting in one plane upon a rigid body, keep it in equilibrium, they must
either meet in a point or be parallel.
5. If a straight line CD drawn from the vertex C of a triangle ABC to the opposite side
AB divides it into two segments in the ratio of m: n, then
(m + n) cot = m cot – n cot and (m + n) cot = n cotA – m cotB,
54
where and are the angles which CD makes with CA and CB and the angle which
CD makes with the base AB.
6. The resultant R and the couple G must separately vanish.
7. A system of forces in a plane will be in equilibrium if the algebraic sum of the
moments of all the forces with respect to each of three non-collinear points is zero.
8. A system of forces in a plane will be in equilibrium if the algebraic sum of the
moments about each of any two different points is zero and the algebraic sum of the
resolved parts of the forces in any given direction not perpendicular to the line joining the
given points is zero.
9. Resultant forces and couple corresponding to any base point O of a system of coplanar
forces.
10. Equation to the resultant of a system of forces in one plane.
11. In any given problem, we imagine the body to be displaced a little and find out the
work done during the displacement. The condition of equilibrium is obtained by equating
to zero the sum of the work done. Since the body is not actually displaced, the work done
is called virtual work. The virtual work that is calculated is the amount of work that
would have been done if the displacements had actually been made.
12. The necessary and sufficient condition that a rigid body acted upon by a number of
coplanar forces be in equilibrium is that the algebraic sum of the virtual works done by
the forces in any small displacement consistent with the geometrical conditions of the
system is zero.
13. If the distance between two particles of a system is invariable, the work done by the
mutual action and reaction between the two particles is zero.
14. The reaction R of any smooth surface with which the body is in contact does no work.
15. If the force between two bodies of a system is a mutual pressure at a point of contact,
then no work is done in any virtual displacement of the system by the action and reaction
if the same points of the bodies remain in contact during the displacement.
16. When a body rolls without sliding on any fixed surface the work done in a small
displacement by the reaction of the surface on the rolling body is zero.
17. If any body is constrained to run round a point or on an axis fixed in space, the virtual
work of the reaction at the point or on the axis is zero.
18. The curve in which a uniform inextensible heavy string hangs freely under gravity is
called a Catenary.
19. We know that T cos = T0 = wc.
Therefore T = wc sec
Hence T = wy
[since y = c sec]
Therefore tension at a point is equal to the weight of a length of the string equal to
its height above the directrix.
20. Since T0 = wc, the tension at the lowest point is equal to the weight of string of length
equal to its height above the directrix.
21. Again, since T cos = T0 = wc, the horizontal component of the tension at any point
is constant and equal to wc.
22. Since T sin = ws, the vertical component of the tension at any point is equal to the
weight of the length of string lying between the point and the vertex.
23. From the equation T cos = T0, we get that the horizontal component of the tension
at every point of the curve is the same and is equal to wc or T0.
55
24. From the equation T sin = ws, we deduce that the vertical component of the tension
at any point is equal to ws, i.e. equal to the weight of the portion of the string lying
between the vertex and the point.
25. The equation T0 = wc, shows that the tension at the lowest point is equal to the weight
of the string whose length is the same as the distance between the origin and the vertex.
26. On squaring each side of
y = c sec and s = c tan,
and substituting we have
y2 = s2 + c2.
27. If denotes the radius of curvature at any point, then
 = ds/d = c sec2
1.7 Assignments:
1. The wire passing round a telegraph pole is horizontal and the two portions attached to
the pole are inclined at an angle of 600 to one another. The pole is supported by a wire
attached to the middle point of the pole and inclined at 600 to the horizon; show that the
tension of this wire is 43 times that of the telegraph wire.
2. AB is a diameter of a circle and BP and BQ are chords at right angles to one another;
show that the moments of forces represented by BP and BQ about A are equal.
3. Three forces act along the sides of a triangle; show that, if the sum of two of the forces
be equal in magnitude but opposite in sense to the third force, then the resultant of the
three forces passes through the center of the inscribed circle of the triangle.
4. At what height from the base of a pillar must the end of a rope of given length be fixed
so that a man standing on the ground and pulling at its other end with a given force may
have the greatest tendency to make the pillar overturn.
5. Forces proportional to AB, BC and 2CA act along the sides of a triangle ABC taken in
order; show that the resultant is represented in magnitude and direction by CA and that its
line of action meets BC at a point X where CX is equal to BC.
6. A smooth rod, of length 2a, has one end resting on a plane of inclination  to the
horizon, and is supported by a horizontal rail, which is parallel to the plane and at a
distance c from it. Show that the inclination of the rod to the inclined plane is given by
the equation
c sin = a sin2 cos().
7. Two equal circular discs of radius r, with smooth edges, are placed on their flat sides in
the corner between two smooth vertical planes inclined at an angle 2, and touch each
other in the line bisecting the angle. Show that the radius of the smallest disc that can be
pressed between them, without causing them to separate, is r (sec – 1).
8. The altitude of a cone is h and the radius of its base is r; a string is fastened to the
vertex and to a point on the circumference of the circular base, and is then put over a
smooth peg; show that, if the cone rest with its horizontal, the length of the string must be
(h2 + 4r2).
9. A cylinder, of radius r, whose axis is fixed horizontally, touches a vertical wall along a
generating line. A flat beam of uniform material, of length 2l and weight W, rests with its
56
extremities in contact with the wall and the cylinder, making an angle of 450 with the
vertical. Show that, in the absence of friction, l/r = (5 – 1)/10, that the pressure on the
wall is ½ W, and that the reaction of the cylinder is ½5W.
10. A solid cone of height h and semi-vertical angle , is placed with its base against a
smooth vertical wall and is supported by a string attached to its vertex and to a point in
the wall; Show that the greatest possible length of the string is h[1 + (16/9) tan2].
11. A regular hexagon is composed of six equal heavy rods freely jointed together, and
two opposite angles are connected by a string, which is horizontal, one rod being in
contact with a horizontal plane; at the middle point of the opposite rod is placed a weight
W1; if W be the weight of each rod, shew that the tension of the string is
3W + W1
3
12. Six equal heavy rods, freely hinged at their ends, form a regular hexagon ABCDEF
which when hung up by the point A is kept form altering its shape by two light rods BF
and CE. Prove that the thrusts of these rods are (53W)/2 and (3W)/2, where W is the
weight of either rod.
13. A prism whose cross section is an equilateral triangle rests with two edges on smooth
planes inclined at angles  to the horizon. If be the angle which the plane containing
these edge makes with the vertical shew that
tan = 23 sin sin + sin ()
3 sin()
14. A smooth rod passes through a smooth ring at the focus of an ellipse whose major
axis is horizontal, and rests with its lower end on the quadrant of the curve, which is
furthest, removed from the forces. Find its position of equilibrium, and shew that its
length must at least be (3a/4) + (a/4)(1 + 8e2), where 2a is the major axis and e is the
eccentricity.
15. A heavy rod AB of length 2l rests upon a fixed smooth peg at C and with its end B
upon a smooth curve. If it rests in all positions, shew that the curve is a conchoids whose
polar equation, with C as origin is
r = l + a/sin
16. A heavy uniform string, of length l, is suspended from a fixed point A, and its other
end B is pulled horizontally by a force equal to the weight of a length a of the string.
Shew that the horizontal and vertical distances between A and B are
a sinh–1(l/a) and (l2 + a2) – a .
17. A string, of length l, hangs between two points (not in the same vertical) and makes
angles with the vertical at the points of support. Shew that, if k is the height of one
point above the other and the vertex of the catenary does not lie between them, then.
k cos[()/2] = l cos[()/2].
18. A uniform chain, of length l is suspended from two points A and B, in the same
horizontal line so that either terminal tension is n times that at the lowest point. Shew that
the span AB must be
l
. loge[n + (n2 – 1)]
2
(n – 1)
If l = 100 feet and n = 3 show, from the tables, that the length is about 62.3 feet.
57
19. A heavy chain, of two length 21 , has one end tied at A and the other is attached to a
small heavy ring which can slide on a rough horizontal rod which passes through A. If
the weight of the ring be n times the weight of the chain, shew that its greatest possible
distance from A is.
2l
. log [ + (1 + 2)]
e

where 1/(2n + 1) and is the coefficient of friction.
1.8 References:
1. Loney, S. L.: An Elementary Treatise on Statics, Cambridge University Press, 1956
2. Varma, R. S.: Text-Book on Statics, Pothishala (Private) Limited Lajpat Road,
Allahabad – 2, 1962.
UNIT –II
Forces in Three Dimensions, Stable and
Unstable Equilibrium
STRUCTURE
2.1 Introduction
2.2 Objectives
2.3 Forces in three dimensions
2.4 Poinsot‟s central axis
2.5 Null lines and planes
2.6 Stable and unstable equilibrium
2.7 Unit Summary
2.8 Assignments
2.9 References
58
2.1 Introduction:
This unit provide us to understanding the basic concepts of Forces in
three dimensions, Poinsot‟s central axis, Null lines and planes, Stable and
unstable equilibrium. This unit gives a brief idea to understand the above
topics.
In this unit we shall study the basic ideas of Forces in three dimensions, Poinsot‟s
central axis, Null lines and planes, Stable and unstable equilibrium. It is hopped the unit
help students in studying.
2.2 Objectives:









At the end of the unit the students would be able to understand the concept of:
Forces in three dimensions
Parallelopiped of Forces
Moment of a force about a point
Couples
General Conditions of Equilibrium
Poinsot‟s Central Axis
Null Lines and Planes
Stable and Unstable Equilibrium
Conditions of stability for a body with one degree of freedom
59
2.3 Forces in three dimensions:
Parallelopiped of Forces:
Forces in three dimensions may be compounded by the law of parallelopiped of
forces, which is an extension of law of parallelogram of forces in two dimensions. It may
be stated as follows:
If three forces acting at a point O are
represented, in magnitude and direction, by
straight lines OA, OB, OC their resultant is
represented in magnitude and direction, by
the diagonal OD of the parallelopiped whose
edges are OA, OB and OC.
For the two forces represented by OA
Z
and OB are equivalent to force represented
by OE, the diagonal of the parallelogram
C
OAEB. Again the resultant of the forces
D
represented by OE and OD, the diagonal of
the parallelogram OEDC, represents OC.
O
A X
Hence the resultant of forces represented by
OA, OB, OD represents OC.
B
E
60
If the parallelopiped is rectangular, so
Y
that OA, OB, OC are taken along the
rectangular axes of OX, OY, OZ respectively
and the forces. OA, OB, OC be respectively
X, Y and Z and their resultant OD be R, then R2 = X2 + Y2 + Z2
and acts along OD whose direction cosines are cos AOD, cos BOD, cos COD.
i.e.
OA
X
OB
Y
l = OD = R
m = OD = R
OA
Z
n = OD = R
i.e.
X/R, Y/R, Z/R
Conversely a force R acting along a line with direction cosine l, m, n has as
components along axes of co-ordinates
X = lR, Y = mR. Z = nR.
Forces in three dimensions:
When forces in different direction in space are acting upon a rigid body, we say that the
body is under the action of forces in three dimensions.
Moment of a force about a point:
The moment of a forces F about any point is equal to F.p, where p is perpendicular
distance of the point from the line of action of force. It is said to be +ive or –ive according as the
force has a tendency to rotate the body about the point in anticlockwise or clockwise direction.
However, in three dimension those forces which try to rotate the body from axis
of x to axis of y, from axis of y to axis of z and from axis of z to axis of x will have + ive
moments about the axis, similarly the moments of couples.
Couples:
We already know that two equal and unlike parallel forces form a couple whose
moment is equal to the product of the force and the arm of the couple. (By arm we mean
the perpendicular distance between the lines of action of the forces of the couple.)
A couple is represented by a line perpendicular to the plane of the couple and
whose length is proportional to the magnitude of the moment. If G be the moment of the
couple and L, M, N be its components on the axis of co-ordinates then G2 = L2 + M2 + N2
and d.c.‟s of its axis are
L/G, M/G, N/G
Conversely a couple G about a line whose d.c.‟s are l, m, n is equivalent to three
couples about the axis whose moments are lG, mG, and nG.
Theorem: Any given system of forces acting at any given point of a rigid body can be
reduced to a single force acting through an arbitrarily chosen point and a couple whose
axis passes through that point.
In other words, a system of forces acting on a rigid body can be reduced in
general to a force acting at an arbitrarily chosen point of the body and a couple.
Or
The resultant of any given system of forces acting at points of a rigid body.
Proof:
Let any arbitrarily chosen point O be taken as origin and three mutually
perpendicular lines through it as axes of co-ordinates.
61
Again suppose that at any point
P1 (x1, y1, z1) acts one of the given forces
z
Z1
whose components parallel to axes are
X1, Y1, Z1.
From P1, draw P1M1 perpendicular
P1
Q1 X1
to x-y-plane and from foot of the
perpendicular M1 draw M1N1 parallel to
Z1
axis of y meeting the axis of x in N1, so that
Y1 z1
Z1
ON1 = x1, M1N1 = y1 and P1M1 = z1. Now
draw a line Q1N1S1 through N1 parallel to
O
x
axis of z.
x1
N1
Introduce force each equal to Z1 along
y
y1
N1Q1, N1S1, OZ and OZ. These being two
Z1
M1
Z1
sets of equal and opposite forces do not alter
the effect of given forces. Thus we have five
z
S1
equal forces parallel to axis of z.
(i) Z1 along P1Z1 and Z1 along N1S1 from a couple whose moment is Z1.M1N1 = Z1. y1
in a plane perpendicular to Ox. Also it has tendency to rotate the body from axis of y to
axis of z and as such its moment is +ive and its axis is Ox.
(ii) Z1 along N1Q1 and Z1 along OZ form a couple whose moment is Z1.ON1 = Z1x1 in a
plane at right angles to Oy. Also it has a tendency to rotate the body from axis of x to axis
of z and as such its moment is – ive, i.e. its moment is – Z1x1 about Oy and its axis is Oy.
(iii) A single force Z1 along OZ
Thus the component Z1 of a forces at P1 is equivalent to
A couple of moment +y1Z1 about Ox
A couple of moment –x1Z1 about Oy
(1)
A single force Z1 along Oz.
Similarly the component X1 of a force is equivalent to
A couple of moment +z1X1 about Oy
A couple of moment –y1X1 about Oz
(2)
A single force X1 along Ox
Again the component Y1 of a force is equivalent to
A couple of moment +x1Y1 about Oz
A couple of moment –z1X1 about Ox
(3)
A single force Y1 along Oy
Combining (1), (2) and (3), we can say that the components X1, Y1, Z1 of a force
at P are together equivalent to
A couple of moment (y1Z1- z1Y1) about Ox
A couple of moment (z1x1 –x1Z1) about Oy
(4)
A couple of moment (x1Y1- y1X1) about Oz
and forces X1, Y1, Z1 along Ox, Oy, and Oz respectively.
The result D is quite symmetrical and there should be no difficulty in
remembering it.
In a similar manner we can replace other forces acting at other points like (x 2, y2,
z2), (x3, y3, z3) etc. whose components are X2, Y2, Z2; X3, Y3, Z3 etc. by couples about the
axes of co-ordinates and forces along the axes.
62
Thus the whole system is equivalent to
A couple of moment (y1Z1- z1Y1) about Ox = L
A couple of moment (z1X1- x1Z1) about Oy = M
A couple of moment (x1Y1- y1X1) about Oz = N
A single force X1 + X2 +……= X1 along Ox = X
(5)
A single force Y1 + Y2 +……= Y1 along Oy = Y
A single force Z1 + Z2 +…… = Z1 along Oz = Z
Three forces X, Y, Z, acting at O, are together equivalent to a single force R
through O where R2 = X2+Y2+Z2 and the direction – cosines of its line of action are X/R,
Y/R, Z/R.
Similarly the three couples of moment L, M, N are together equivalent to a single
couple of moment G, where
G2 = L2+M2+N2
and the direction-cosines of its axes are L/G, M/G, N/G.
Thus the entire system has been reduced to a single force R acting at arbitrarily
chosen point O together with a couple of moment G whose axis passes through O.
General Conditions of Equilibrium:
The system will be in equilibrium when R = 0 and G = 0, which in other words
means that X = 0, Y = 0, Z = 0. Also L = 0, M = 0, N = 0.
The sum of the resolved parts of the system of forces parallel to any three axes of
co-ordinates must vanish separately, and also the sums of their moments about the three
axes must vanish separately.
Elements of System:
The six quantities X, Y, Z, L, M, N. are called the elements of the system. It is
very easy to write down their values. If the forces are acting at (x 1, y1, z1) and its
components along the axes be given, then write
x1 y1 z1
X1 Y1 Z1
Then
X = X1, Y = Y1, Z = Z1
and
y1
z1
L1 =
i.e. (y1Z1 – z1Y1); so L = L1
Y1
Z1
z1
x1
M1 =
i.e. (z1X1 – x1Z1); so M = M1
Z1
X1
x1
y1
N1 =
i.e. (x1Y1 – y1X1); so N = N1
X1
Y1
Another form:
If the force P acts along a line
x – x1
y – y1
z – z1
=
=
l
m
n
then its components X1, Y1, Z1 along the axes will be Pl, Pm, Pn respectively and we
should in this case write
63
x1 y1 z1
Pl Pm Pn
X = X1 = Pl, Y = Y1 = Pm, Z = Z1 = Pn
y1
z1
L1 =
i.e. L = L1
Pm
Pn
z1
x1
M1 =
i.e. M = M1
Pn
Pl
x1
y1
N1 =
i.e. N = N1
Pl
Pm
then
and
and
and
Rule:
Write down in a line the co-ordinates of the point at which the forces is acting. In
the second line write down the components of the force parallel to axes of co-ordinates
and then form determinates of 2nd order to get the values of L1, M1, N1 etc. Similarly treat
other force.
2.4 Poinsot’s Central Axis:
Theorem: Any system of forces acting on a rigid body can be reduced to a single force
together with a couple whose axis is along the direction of the force.
Proof:
We know that a system of forces acting on a rigid body can be reduced to a single
forces R acting at O along OA together with a couple G whose axis OD is a line passing
through O.
A
D
A
G cos
R
G
R

O
G sin
B
O
C
B
C
Fig. (1)
Fig. (2)
A
G cos
R
A


A
R
O
G cos R
B
O
B
64
O
C
OR
Fig. (3)
C
Fig. (4)
Let the axis OD of the couple be inclined to the direction of R at an angle Draw
OB perpendicular to OA so that OA, OD and OB all lie in one plane AOB. Now draw
OC perpendicular to plane AOB.
The couple G about OD as axis is equivalent to a couple G cos about OA as axis
and a couple G sin about OB as axis (Fig. 2). The latter couple G sin about OB as axis
acts in the plane AOC because its axis OB is perpendicular to plane AOC. Hence the
couple G sin can be replaced by two equal and unlike parallel forces of moment G sin.
Let one of the forces be R acting at O in a direction opposite to R along OA and
the other will also be R acting parallel to OA at some point O in OC such that its
moment
R.OO = G sin, so OO = (G sin)/R (Fig. 3)
The two equal forces R at O in opposite directions each other and we are now to
deal with force R at O and a couple of moment G cos about OA as axis. The axis of a
couple can always be transferred to a parallel axis; hence we can choose OA as the axis
of couple G cos. Thus we are left with a force R along OA and a couple of moment G
cos whose OA is along the direction of force R.
Note:
1.Poinsot’s Central axis:
The line OA, the axis of the single couple and also the line of action of single
forces R to which a system of forces is reduced is called Poinsot‟s Central Axis.
The moment of the resultant of couple about the central axis i.e. G cos is less
than the moment of the resultant couple corresponding to any other point O which is not
on the central axis because G cos <G.
Also central axis for a system of forces is unique.
2.Wrench:
A single force R together with a couple K whose axis coincides with the direction
of the force when taken together is called a wrench of the system. The single force R is
called the intensity of the wrench.
3.Pitch:
The ratio K/R i.e. moments of the couple about central axis divided by forces is
called the pitch and is of a linear magnitude. Also it is clear that when the pitch is zero
the wrench reduces to a single force. When the pitch is infinite then R = 0 and hence the
wrenches reduce to a couple K only.
4.Screw:
The straight line, along which the single forces acts, when considered together
with the pitch, is called a screw. Hence screw is a definite straight line associated with a
definite pitch.
Theorem: Find the condition that a given system of forces should compound into a
single force.
Proof:
65
We know that the forces are equivalent to a single force R acting at an arbitrary
origin O and a single couple G. If be the angle between R and the axis of G, then R is
equivalent to a force R cos along the axis OB of the couple, and a force R sin in the
plane of the couple. This force R sin, together with the parallel forces of the couple, are
equivalent to a parallel force R sin that does not pass through O and therefore cannot in
general, compound with R cos into a single force.
But, if R cos = 0, i.e. if cos = 0, then we are left with a single force R sin.
Hence must be 900, i.e. the angle between the straight lines whose direction cosines are
(X/R, Y/R, Z/R) and (L/G, M/G, N/G) must be a right angle.
So
X L + Y M + Z N = cos900 = 0
R G R G R G
Therefore XL + YM + ZN = 0 is the required condition.
Theorem: Whatever origin or base point and axes are chosen for any system of forces
the quantities
X2 + Y2 + Z2 and LX + MY + NZ
are invariable, where X = X1, etc. and L = (y1Z1 – z1Y1) etc.
Proof:
We know that X2 + Y2 + Z2 is the square of the resultant force R corresponding to
the Central Axis and is therefore invariable.
Again, if (l, m, n) are the direction cosines of the resultant force and (l1, m1, n1)
those of the axis of the resultant couple, then
X L + Y M + Z N = ll + mm + nn
1
1
1
R G R G R G
= the cosine of the angle between the resultant force and
the axis of the resultant couple = cos
Therefore
LX + MY + NZ = R.G cos = R.K,
where K is the moment of the couple about the Central Axis.
Hence I = LX + MY + NZ is an invariant.
It follows that if K be zero, that is, if the given system reduces to a single force,
then LX + MY + NZ = 0.
This second invariant will be zero also when the resultant force R is zero. In this
case the first invariant is zero also.
The pitch, p, of the resultant Wrench of the system = K/R
= the invariant I of the system divided by the square of the invariant R.
Theorem: Equation of Central Axis of any given system of forces.
Proof:
Let (f, g, h) be the coordinates referred to the axes Ox, Oy, Oz of any point Q.
The moment about a line through Q parallel to Ox is clearly obtained by putting x 1 – f, y1
– g, z1 – h instead of x1, y1, z1.
Hence the moment
= [(y1 – g)Z1 – (z1 – h)Y1]
= (y1Z1 – z1Y1) – gZ1 + hY1 = L – g Z + h Y
So the moments about lines through Q parallel to the other axes are
66
M – h X + f Z and N – f Y + g X.
Also the components of the resultant force are the same for all points such as Q
and are thus X, Y and Z.
If Q be a point on the central axis, the direction cosines of the axis of the couple
corresponding to it are proportional to those of the resultant force. Hence
L – gZ + hY = M- hX + fZ = N – fY + gX
X
Y
Z
LX
+
MY
+
NZ
K
=
=
X2 + Y2 + Z2
R
Hence the equation of the locus of the point (f, g, h), i.e. the required equation of
the central axis is
L – yZ + zY = M- zX + xZ = N – xY + yX
X
Y
Z
= K/R = the pitch p of the wrench.
Ex.1 A single force is equivalent to component forces X, Y, Z along the axes of
coordinates and to couples L, M, N about three axes. Prove that the magnitude of the
single force is (X2 + Y2 + Z2) and the equations to its lines of action are
yZ – zY =
zX – xZ
= xY – yX = l
L
M
N
Sol.
If the system reduces to a single force, then LX + MY + NZ = 0 and the equation
of central axis i.e. line of action of single force are
L – yZ + zY = M- zX + xZ = N – xY + yX = 0
X
Y
Z
or
L = yZ – zY , M = zX – xZ , N = xY – yX
or
yZ – zY =
zX – xZ
= xY – yX = l
L
M
N
Since the couples do not affect the magnitude of the single force and hence in
magnitude R = (X2 + Y2 + Z2).
Ex.2 Forces P, Q, R act along three non-intersecting edges of a cube; find the central
axis.
Sol.
Let the three edges be OA, CA, CO, whose equations are
x
y
z
=
=
1
0
0
x
y
z–a
=
=
0
1
0
x–a
y–a
z
=
=
0
0
1
z
C
(0, 0, a)
B
Q
Y
Z
67
A
O
O
P
R
x
A (a, 0, 0)
B (0, a, 0)
y
R
X
C (a, a, 0)
So X = P, Y = Q, Z = R.
Also moments of couples are
0, 0, 0; – aQ, 0, 0; aR, – aR, 0.
So
L = a(R – Q), M = – aR, N = 0.
For equation of central axis write
x y z i.e. x y z
X Y Z
P Q R
L – yR + zQ = M – zP + xR = N – xQ + yP
P
Q
R
a(R – Q) – yR + zQ = – aR – zP + xR = – xQ + yP
P
Q
R
Ex.3 OA, OB, OC are the three coterminous edges of a cube and AA, BB, CC and OO
are diagonal. Along BC, CA, AB and OO act forces equal to X, Y, Z and R
respectively. Show that they are equivalent to a single resultant if
(YZ + ZX + XY) 3 + R (X + Y + Z) = 0.
Sol.
Take O as origin and the axes along OA, OB, OC. If a be the edge of a cube, then
coordinates of A, B, C are as marked and those of O are (a, a, a).
z
C
(0, 0, a)
B
Q
Y
Z
A
O
O
R
x
P
B (0, a, 0)
y
A (a, 0, 0)
R
X
C (a, a, 0)
The lines of action of the three forces are parallel to axes and hence their
equations are
x
y–a
z for X,
=
=
1
0
0
x
y
z–a
for Y,
=
=
0
1
0
x–a
y
z
for Z.
=
=
68
0
x
=
1/3
Since d.c.‟s are
a
,
a/3
or
0
y
=
1/3
1
z
for R,
1/3
a
,
a/3
a
a/3
The components of the forces parallel to axes are
X, 0, 0; 0, Y, 0; 0, 0, Z; R/3, R/3, R/3.
So
X = X + R/3, Y = Y + R/3, Z = Z + R/3
It is easy to calculate L1, M1, N1 etc. and consequently
L = – aY, M = – aZ, N = – aX
The system will reduce to a single force if
LX + MY + NZ = 0
– aY (X + R/3) – aZ (Y + R/3) – aX (Z + R/3) = 0
(YZ + ZX + XY) 3 + R (X + Y + Z) = 0.
Ex.4 Equal forces act along two perpendicular diagonals of opposite faces of a cube of
side a. Show that they are equivalent to a single force R acting a line through the centre
of the cube and a couple ½ aR with the same line for axis.
Sol.
Let the axes be along the edges OA, OB and OC. Let ON be one diagonal and LM
be the perpendicular diagonal of the opposite face. Suppose equal forces P act along these
diagonals whose equations in terms of d.c.‟s are
x
y
z
=
=
a/
a/2
0
x
y–a
z–a
=
=
a/2
–a /2
0
z
C
(0, 0, a)
M (a, 0, a)
P
L
(0, a, a)
P (a, a, a)
o
O
x
A (a, 0, 0)
B (0, a, 0)
y
and
P
N (a, a, 0)
The components of forces parallel to axes are
Pa , Pa , 0
2 2
Pa , –Pa , 0
2 2
Therefore
X = X1 = Pa2, Y = 0, Z = 0
(1)
69
Clearly L, M, N for forces P along first line are zero and for the second line we
write
0
a
a
Pa –Pa
0
2
2
So
Pa2 , M = Pa2 , N = – Pa2
L=
2
2
2
(2)
2
2
2
R = (X + Y + Z ) = Pa2.
(3)
If the axis of the couple coincides with the line of action of R, then it is central
axis and the moment of the couple about axis is K where
LX + MY + NZ = KR.
Pa2 .Pa2 + 0 + 0 = K.Pa2
2
Therefore, moment of couple
Pa2
a
aR
K=
= .Pa2 =
2
2
2
[by (3)].
To write down the equation of central axis, we have
x y z i.e. x
y z
X Y Z
Pa2 0 0
L – yZ + zY = M- zX + xZ = N – xY + yX
X
Y
Z
and
Pa2 – 0
Pa2 – (z.Pa2)
– Pa2 – (0 – yPa2)
2
= 2
= 2
Pa2
0
0
Above gives y = a/2 and z = a/2 which is a line through the centre O of the cube
whose co-ordinates are (a/2, a/2, a/2).
Ex.5 Two forces act, one along the line y = 0, z = 0 and the other along the line x = 0, z =
c. As the forces vary, show that the surface generated by the axis of their equivalent
wrench is (x2 + y2)z = cy2.
Sol.
A single force R together with a couple K whose axis is along the line of action of
R is called a wrench of the system. The axis is central axis.
Let the forces be P and Q acting along the lines
x
y
z
=
=
1
0
0
x
y
z–c
=
=
0
1
0
so that their components parallel to the axis are Pl, Pm, Pn etc. i.e. P, 0, 0 and 0, Q, 0
respectively.
Therefore
X = X1 = P, Y = Y1 = Q, Z = Z1 = 0.
(1)
In order to find the values of L, M, N we note that force P acts at (0, 0, 0) and X 1,
Y1, Z1 are P, 0, 0.
So
0
0
0
P
0
0
i.e.
L1 = 0, M1 = 0, N1 = 0.
70
For the force Q it acts at 0, 0, c and X2, Y2, Z2 are 0, Q, 0.
Therefore
0
0
0
0
Q
0
So
L2 = 0 – cQ, M2 = 0, N2 = 0.
Hence L = L1 = –cQ, M = M1 = 0, N = N1 = 0.
(2)
The equations of central axis are
L – yZ + zY = M – zX + xZ = N – xY + yX
X
Y
Z
(3)
In order to write down the values of expressions within brackets, we write
x y z i.e. x y z
X Y Z
P Q 0
The three brackets in order are the values of determinants
y
z
z
x
x
y
Q
0 ,
0
P ,
P
Q
i.e.
– zQ, Pz, xQ – yP.
(4)
Hence equations of central axis by putting the values from (1), (2), (4) in (3) are
– cQ – (– zQ) = – PZ =
– (xQ – yP)
P
Q
0
From last, we get xQ – yP = 0; So P/Q = x/y.
(5)
From 1st two, we get
(z – c)Q
– Pz
P
Q
(6)
(5) and (6) together give the equation to central axis. In order to find the surface
generated by central axis, let us eliminate P and Q from (5) and (6) and we get
(z – c)y =
– xz
x
y
or
y2(z – c) = – x2.z
or
z(x2 + y2) = cy2
Ex.6 Forces X, Y, Z act along the three lines given by the equations y = 0, z = c; z = 0, x
= a; x = 0, y = b. Prove that the pitch of the equivalent wrench is
aYZ + bZX + cXY
X2 + Y2 + Z2
If the wrench reduces to a single force, show that the line of action of the force
lies on the hyperboloid (x – a)(y – b)(z – c) – xyz = 0.
Sol.
The three lines along which forces X, Y, Z act are
x
y
z – c for X,
=
=
1
0
0
x–a
y
z
for Y,
=
=
0
1
0
x
y–b
z
for Z.
=
=
0
0
1
The components parallel to the axes are
X, 0, 0; 0, Y, 0; 0, 0, Z
So
X = X1 = X, Y = Y, Z = Z
(1)
Also L1, M1, N1 for force X are given by
0
0
c
71
i.e
X
0
0
L1 = 0, M1 = cX, N1 = 0.
Similarly
L2 = 0, M2 = 0, N2 = aY,
L3 = bZ, M3 = 0, N3 = 0.
So
L = L1 = bZ, M = cX, N = aY.
K
LX + MY + NZ
Pitch of the wrench =
=
R
X2 + Y2 + Z2
bZX + cXY + aYZ
=
X2 + Y2 + Z2
Write
x
y
z
X
Y
Z
Equations of central axis are
(2)
[by (1) and (2)]
L – yZ + zY = M – zX + xZ = N – xY + yX
X
Y
Z
LX + MY + NZ
=
X2 + Y2 + Z2
In case the system reduces to a single force, then LX + MY + NZ = 0 and hence
each of the three ratios vanish.
Hence putting the values of L, M, N we have
bZ – yZ + zY = 0, cX – zX + xZ = 0, aY – xY + yX = 0
or
0.X + zY + (b – y)Z = 0,
(c – z)X + 0Y + xZ = 0,
yX + (a – x)Y + 0Z = 0.
The locus of central axis i.e. the line of action of single force in this case is
obtained by eliminating X, Y, Z from the above three and is
0
z
b–y
c–z 0
x
=0
y
a–x 0
– z(0 – xy) + (b – y)[(c – z) (a – x)] = 0
or
(x – a)(y – b)(z – c) – xyz = 0.
Ex.7 Forces X, Y, Z act along three straight lines y = b, z = – c; z = c, x = – a and x = a, y
= – b respectively. Show that this will have a single resultant if a/X + b/Y + c/Z = 0 and
that the equations of its line of action are two of the three
y
z
a
=0


Y
Z
X
z
x
b
=0


Z
X
Y
x
y
c
=0


X
Y
Z
Sol.
The three lines are
x
y–b
z + c for X,
=
=
1
0
0
72
i.e
or
or
or
x+a
y
z–c
for Y,
=
=
0
1
0
x–a
y+b
z
for Z.
=
=
0
0
1
The components of forces parallel to axes are
X, 0, 0; 0, Y, 0; 0, 0, Z.
So
X = X1 = X, Y = Y, Z = Z
(1)
Also L1, M1, N1 for force X are given by
0
b
–c
X
0
0
L1 = 0, M1 = – cX, N1 = – bX .
Similarly
L2 = – cY, M2 = 0, N2 = – aY,
L3 = – bZ, M3 = – aZ, N3 = 0.
So
L = L1 = – (cY + bZ), M = – (cX + aZ), N = – (bX + aY). (2)
The system will reduces to a single forces if
LX + MY + NZ = 0
Therefore
– X(cY + bZ) – Y(cX + aZ) – Z(bX + aY) = 0
[by (1) and (2)]
2(aYZ + bZX + cXY) = 0 or a/X + b/Y + c/Z = 0
(3)
Now write
x
y
z
X
Y
Z
Equations of central axis are
L – yZ + zY = M – zX + xZ = N – xY + yX
X
Y
Z
– (cY + bZ) – yZ + zY = 0
(4)
– (cX + aZ) – zX + xZ = 0
(5)
– (bX + aY) – xY + yX = 0
(6)
Dividing (4) by YZ, we get
– c/Z – b/Y – y/Y + z/Z = 0 or a/X – y/Y + z/Z = 0. [by (3)]
y
z
a
=0


Y
Z
X
(7)
Similarly dividing (5) and (6) by ZX and XY, respectively and using relation (3),
we get
z
x
b
=0


Z
X
Y
(8)
and
x
y
c
=0


X
Y
Z
(9)
Thus the equations of line of action of the single force are any two of three
relations (7), (8) and (9).
Ex.8 Three forces each equal to P act on a body, one at the point (a, 0, 0) parallel to Oy,
the second at the point (0, b, 0) parallel to Oz and the third at the point (0, 0, c) parallel to
Ox, the axes being rectangular. Find the resultant wrench in magnitude and position.
Sol.
The lines of action of the equal forces are
x–a
y
z
=
=
73
0
1
0
x
y–b
z
=
=
0
0
1
x
y
z–c
=
=
1
0
0
The components parallel to the axis are
0, P, 0; 0, 0, P; P, 0, 0.
So
X = X1 = P, Y = P, Z = P.
(1)
For L1, M1, N1 we write
a
0
0
0
P
0
So
L1 = 0, M1 = 0, N1 = aP.
Similarly L2 = Pb, M2 = 0, N2 = 0; L3 = 0, M3 = cP, N3 = 0.
Therefore
L = L1 = Pb, M = Pc, N = Pa.
(2)
In order to find the magnitude of wrench we should find the values of both K and
R.
or
or
or
or
or
Therefore
R2 = X2 + Y2 + Z2 = 3P2, so R = P3
(3)
2
Also KR = LX + MY + NZ, so KP3 = P (a + b + c).
Hence
K = P(a + b + c)
3
(4)
(3) and (4) give the resultant wrench in magnitude.
Equations to the central axis are given by
x
y
z
x
y
z
X
Y
Z
or
P
P
P
L – yZ + zY = M – zX + xZ = N – xY + yX = K
X
Y
Z
R
Pb – yP + zP = Pc – zP + xP = Pa – xP + yP = a + b + c
P
P
P
3
b–y+z
c
–
z
+
x
a
–
x
+
y
a
+
b
+
c
=
=
=
1
1
1
3
a
+
2b
+
3c
b
+
2c
+
3a
c
+
2a
+
3b
x+
=y+
=z+
3
3
3
Above represent a line equal inclined to the axes and passing through the point
 a + 2b + 3c ,  b + 2c + 3a ,  ac + 2a + 3b
3
3
3
Ex.9 Equal forces act along the coordinate axes and along the straight line
x –  = y  = z
l
m
n
Find the equation of the central axis of the system.
Sol. The given lines are
x
y
z
=
=
1
0
0
74
x
y
z
=
=
0
1
0
x
y
z
=
=
0
0
1
and
x –  = y  = z
l
m
n
and let the equal force be P, so that their components parallel to the axis are
P, 0, 0; 0, P, 0; 0, 0, P; Pl, Pm, Pn.
So
X = X1 = P (1 + l), Y = P (1 + m), Z = P (1 + n)
(1)
L, M, N for P along first line
0
0
0
P
0
0
So
L1 = 0, M1 = 0, N1 = 0.
Similarly along other two axes their values are zero. For forth line we write



Pl
Pm
Pn
Therefore L = P (n – m), M = P (l – n), N = P (m – l)
(2)
For writing the equation of central axis we write
x
y
z
x
y
z
X
Y
Z
or
P (1 + l)
P ( 1 + m)
P (1 + n)
or
L – yZ + zY = M – zX + xZ = N – xY + yX
X
Y
Z
The factor P will cancel throughout and so we do not write it.
(n – m) – y (1 + n) + z (l + m) = (l – n) – z (1 + l) + x (1 + n)
1+l
1+m
= (m – l) – x (1 + m) + y (1 + l)
1+m
Ex.10 Two equal forces act one along each of the straight lines
x  a cos =
y b sin =
z
a sin
b cos
c
show that their central axis must, for all values of , lie on the surface
y (x/z + z/x) = b (a/c + c/a).
Sol.
P being the force, then at the point (a cos, b sin, 0) we have a force whose
components are proportional to a sin.P, – b cos.P, cP and at the point (– a cos, b sin,
0) a force whose components are proportional to a sin.P, b cos.P, cP.
and
Hence X = X1  2a sin.P,
Y = 0 and Z 2cP,
L = (y1Z1 – z1Y1)  2bc sin.P,
M = 0 and N – 2abP.
The equation of central axis then become
bc sinyc = – ab + ya sin
a sin
c
za sin = x.0.
75
Substituting the value of sin from the second of these equations in the first we
have, as the locus of the central axis,
y (x/z + z/x) = b (a/c + c/a).
Ex.11 OBDC is a rectangle such that OB = b and OC = c, also OA is perpendicular to its
plane. Along, OA, CD and BD act forces X, Y, Z respectively. Show that the component
force R and couple K of the resultant wrench are (X2 + Y2 + Z2) and
X (Zb – Yc)
(X2 + Y2 + Z2)
Show also that with OA, OB, OC as axes of x, y, z the equation to the central axis
are
x =
X
y  ZK =
Y
XYR
z + KY
Z
XZR
Sol.
Given forces X, Y, Z act along OA, CD and BD respectively. Now introduce two
forces each equal to Y along OB and BO and two forces each equal to Z along OC and
CO.
We have three forces X, Y, Z acting at along OA, OB and OC which can be
compound to a single force R = (X2 + Y2 + Z2).
The force Y along BO and Y along CD being equal and opposite form a couple of
moment – Yc about a line OA as axis, which is perpendicular to its plane. Similarly the
force Z along BD and Z along CO form a couple of moment Zb about a line OA as axis.
Thus the moment of the couple is L = (Zb – Yc) about OA as axis
Moment about OB and OC are each zero.
C
So M = 0, N = 0.
Y
Also
LX + MY + NZ = KR.
D
c
2
2
2
Hence (Zb – Yc) X = K (X + Y + Z ).
Z
Therefore
Z
– Yc)
Y O
K = X (Zb
(X2 + Y2 + Z2)
b
Thus K and R the two elements of the
B
resultant wrench are known. The equations
of central axes are
L – yZ + zY = M – zX + xZ = N – xY + yX = K
X
Y
Z
R
From 2nd and 4th ratio, we have
– zX
xZ
K
+
=
Y
Y
R
Multiplying by Y/(XZ).
Hence
–z
x
KY
+
=
Z
X
XZR
Y
X
A
76
or
or
x
z
KY
=
+
X
Z
XZR
Similarly taking 3rd and 4th ratio and multiplying by Z/(XY), we get
x
y
KZ
=

X
Y
XYR
Ex.12 If a force (X, Y, Z) act along a generator of the hyperbolic paraboloid
x2
y2
2z
 2 =
a2
b
c
and be equivalent to an equal force (X, Y, Z) at the origin together with a couple L, M, N,
show that aL bM = 0, bX aY = 0 and cN abZ = 0.
Sol.
The generators of the given hyperboloid are given by
x + y = 2z , x  y = 
a
b
c
a
b
(1)
and
x  y = 2z , x  y = 
a
b
c a
b
(2)
(1) and (2) give two generators of the opposite system. Let us reduce the
generator (1) to symmetrical form. Putting z = 0, we get
x = a/2, y = – b/2, z = 0
and direction-cosines of this can be shown to be proportional to a, b, c.
Thus actual direction cosines are aA, bA, cA, where
1
A=
2
2
(a + b + c22)
Hence in symmetrical form the equations of the generator are
x – a
y + b
2
2
z–0
=
=
aA
bA
cA
Now if be the force whose components are X, Y, Z, then
X = PaA, Y = PbA, Z = PcA.
(1)
For L, M, N, we have
a  b
0
2
2
X
Y
Z
L = – bZ/2, M = – aZ/2, N = aY/2 + bX/2
(2)
aL – bM = – abZ + ab Z = 0
2
2
bX – aY = b.PaA – a.PbA = 0.
cN – abZ = ½ c(a.PbA + b.PaA) – ab.PcA.
= ca.PbA – ab.PcA = 0.
Similarly if we take the other generator (2) and reduce to symmetrical form, then
proceeding as above we can show that aL + bM = 0, bX + aY = 0, cN + abZ = 0.
77
Ex.13 Forces act along generators of the same system of a hyperboloid. Their magnitudes
are such that if they were transferred parallel to themselves to act in point, they would be
in equilibrium. Show that they are in equilibrium when acting along the generators.
Sol.
We have written the values of X1, Y1, Z1, L1, M1, N1.
So
X = R1a sin1.A1, Y = – R1b cos1.A1, Z = R1c.A1.
L = R1bc sin1.A1 = bcX/a
M = R1ca cos1.A1 = caY/b
N = – R1ab.A1 = – abZ/c
X, Y, Z are the component forces acting at O along the axes. If X, Y, Z are in
equilibrium, then their resultant R = (X2 + Y2 + Z2), should vanish i.e. X = 0, Y = 0, Z =
0.
Clearly if X, Y, Z are zero, then L, M, N also vanish. Thus all the six elements X,
Y, Z, L, M, N vanish and hence the forces will be in equilibrium when acting along the
generators.
Ex.14 Forces act along generators of the same system of a hyperboloid and the pitch p of
the equivalent wrench is given. Prove that the central axis is that generator of the
hyperboloid
(bc/a – p) x2 + (ca/b – p) y2 – (ab/c + p) z2 = (bc/a – p)(ca/b – p)(ab/c + p)
which intersects xy-plane at the point
x = (ac – bp) Z1 cos , y = (bc – ap) Z1 sin
c
Z1
c
Z1
Sol.
The values of X, Y, Z, L, M, N are
X = R1a sin1.A1, Y = – R1b cos1.A1, Z = R1c.A1.
L = R1bc sin1.A1 = bcX/a
M = R1ca cos1.A1 = caY/b, N = – R1ab.A1 = – abZ/c
The pitch of the wrench i.e. K/R = p given.
Equation of central axis is
L – yZ + zY = M – zX + xZ = N – xY + yX = K = p
X
Y
Z
R
Therefore
(bc/a – p) X + zY – yZ = 0,
(1)
(ca/b – p) Y + xZ – zX = 0,
(2)
(ab/c + p) Z + yX – xY = 0,
(3)
Eliminating X, Y, Z the locus of the central axis
(bc/a – p)
z
–y
–z
(ca/b – p)
x
=0
y
–x
(ab/c + p)
Expending the above determinants we get the locus as given. In order to find the
point on xy-plane put z = 0 in (1) and (2), we get
y = (bc/a – p) (X/Z)
(bc/a – p)R1a sin1.A1
=
R1c.A1
(bc
–
ap)
Z1 sin
y=
78
c
Z1
Similarly the value of x is as given.
2.5 Null Lines and Planes:
1.Null Lines:
Corresponding to a given system of forces referred to a base or origin O, all those
lines which radiate from O, such that the moment of the given system of forces about
these lines is zero are called null lines at O.
2.Null Plane:
The plane in which all these null lines lie is called null plane of the point O.
3.Null Point:
The point O itself is called null point.
Suppose that, corresponding to any origin or base point O, the resultant force is R
and the resultant couple is G. Take any line through O perpendicular to the axis of G;
then the sum of the moments of the forces of the system about this line is zero; for the
axis of G has no component along it and R meets it.
For this reason the line is called a null line and its locus, which is the line
perpendicular to the axis of G, is called the null plane of O. Also the point O is called the
null point of the plane.
Theorem1: Find the equation of the null plane of a given point (f, g, h) referred to any
axes Ox, Oy, Oz.
Proof:
Let X, Y, Z be the component forces along Ox, Oy, Oz and L, M, N the
component couples about them.
The component couples about lines parallel to the axes through (f, g, h) are
L – gZ + hY, M – hX + fZ, N – fY + gX,
and these are proportional to the direction cosines of the axis of the resultant couple at (f,
g, h), which is the normal to the null plane there.
Hence the equation to the null plane is
(x – f)(L – gZ + hY) + (y – g)(M – hX + fZ) + (z – h)(N – fY + gX) = 0.
i.e.
x (L – gZ + hY) + y (M – hX + fZ) + z (N – fY + gX) = fL + gM + hN. … (1)
The above equation of null plane can be put in determinant form as follows:
x
y
z
f
g
h
= L(x – f) + M(y – g) + N(z – h)
X
Y
Z
Theorem2: Find the null point of a given plane lx + my + nz = 1.
Proof:
To obtain the null point of the plane
lx + my + nz = 1,
(i)
compare this plane with (1) of the above theorem, we get
(L – gZ + hY)
= (M – hX + fZ)
= (N – fY + gX) = fL + gM + hN.
l
m
n
79
Since the point (f, g, h) also must lie on the plane (i), we have, on solving
f
g
h
1
=
=
=
X – nM + mN
Y – lN + nL
Z – mL + lM
lX + mY + nZ
giving the null point of (i).
Theorem3: Find the condition that the straight line
x–f
z–h
= y–g =
l
m
n
may be a null line for the same system of forces.
Proof:
The component couples about lines through (f, g, h) parallel to the axes are
L – gZ + hY, M – hX + fZ, N – fY + gX,
Hence the moment of the couple about the given line
= l (L – gZ + hY) + m (M – hX + fZ) + n (N – fY + gX)
and hence is zero if
X(mh – ng) + Y(nf – lh) + Z(lg – mf) = Ll + Mm + Nn,
i.e. if
X
Y
Z
l
m
n
= Ll + Mm + Nn
f
g
h
This is therefore the condition that the given line may be a null line of the system.
Theorem4: Show that a given system of forces may be replaced by two forces, one of
which acts along a given line OA.
Proof:
With O as origin, or base point, let R and G be the resultant force and couple.
Through OA and R let a plane be drawn and let it cut the plane of the resultant couple
(i.e. the plane COD perpendicular to the axis of G) in OB. Resolve R into two forces,
one, P1 along OA, and the other, P2, along OB.
The force P2 along OB, when
compounded with the two forces in the plane
A
BOC which form the resultant couple, will
G
R
give a force P2 in this plane which is parallel
to OB.
It follows that the given system of
O
C
forces is equivalent to some force P1 acting
along the given straight line OA, together with
a second force P2 which acts somewhere in the
D
B
null plane of O.
Note:
Such forces as P1 and P2 are called conjugate forces and their lines of action are
called conjugate lines.
Whatever point O we take on OA the force P2 will still lie in its null plane, so that,
as O moves along OA, its null plane continually turns round so that it always passes
through the line conjugate to OA. Hence the conjugate line of OA may be determined by
taking any two convenient points on it, and obtaining the equations to their null planes by
Theorem 1. The conjugate line is then the intersection of these two planes.
80
Theorem5: Find the equation of a line conjugate to a given line
x–f
z–h
= y–g =
l
m
n
Proof:
In order to find the line conjugate to given line we have to find the equation of
null planes of any two conveniently chosen points on the given line. These two points
together will give the conjugate line.
Let one point be (f, g, h) and the other point be at infinity whose coordinates are
(l, m, n) where is infinite.
If the system reduces to X, Y, Z, L, M, N, then the null plane of (f, g, h) is
x
y
z
f
g
h
= L(x – f) + M(y – g) + N(z – h)
(1)
X
Y
Z
and the null plane of (l, m, n) is
x
y
z
l
m
n
= L(x – l) + M(y – m) + N(z – n)
X
Y
Z
Dividing by , we get
x
y
z
l
m
n
= L(x/ – l) + M(y/ – m) + N(z/ – n)
X
Y
Z
Now put infinite and the null plane becomes
x
y
z
l
m
n
= – (Ll + Mm + Nn)
(2)
X
Y
Z
The line of intersection of null planes (1) and (2) is a line conjugate to given line.
Ex.1 A system of forces given by (X, Y, Z, L, M, N) is replaced by two forces, one acting
along the axis of x and another force. Show that the magnitudes of the forces are
LX + MY + NZ
L
and
[(MY + NZ)2 + L2 (Y2 + Z2)]1/ 2
L
and also find the equation of the line of action of the other force.
Sol.
The equations of axes of x are
x
y
z.
=
=
1
0
0
Let there be a force P acting along it at (0, 0, 0), so that its components parallel to
axes of co-ordinates are P, 0, 0.
The components of couples at (0, 0, 0) are all zero. Since the system is X, Y, Z; L,
M, N, therefore the components parallel to axes of the other forces are X – P, Y, Z.
If the other forces act at the point (f, g, 0), then for moments of couples we write
f
g
0
X–P Y
Z
81
L = gZ, M = – fZ, N = fY – g(X – P)
or
– MY  L(X – P)
N=
Z
Z
Therefore
LX + MY + NZ = P
L
or
LX +MY + NZ  MY + NZ
X–P=X–
L
L
Y = Y, Z = Z.
Therefore resultant
= [(X – P)2 + Y2 + Z2]
2
2
2
2 1/ 2
= [(MY + NZ) + L (Y + Z )]
L
In order to write down the equations of its line of action we write down the
equations of null planes of point (0, 0, 0) and a point (, 0, 0) where is infinity on the
given line.
Null plane of (0, 0, 0) is LX + MY + NZ = 0
(1)
Null plane of (, 0, 0) is
x
y
z

0
0
= L(x – ) + M(y – 0) + N(z – 0)
X
Y
Z
or
–  (yZ – zY) = L(x – ) + My + Nz.
Dividing by  and put  = .
Therefore
– (yZ – zY) = L(– 1) or yZ – zY = L.
(2).
Equations (1) and (2) together give the equations of the line of action of the other
force.
So
Ex.2 Show that among the null lines of any system of forces four are generators of any
hyperboloid, two belonging to one system of generators and two to the other system.
Sol.
Let the hyperboloid be
x2
y2  z2.
= 1,
2 +
a
b2
c2
and referred to its centre and axes let the system be given by (X, Y, Z; L, M, N). Any
generator is
x – a cos
y – b sin
z.
=
=
a sin
– b cos
c
this is a null line of the system if
X(– bc sin) + Yca cos + Zab = La sin – Mb cos + Nc,
i.e. if
sin [X/a + L/(bc)] – cos [Y/b + M/(ca)] = Z/a – N/(ab),
which clearly gives two values of , in general. Hence two generators belonging to one
system are null lines. Similarly for the other system.
Ex.3 A straight line is given by the equations
Ax + By + Cz = D, x + By + Cz = D
82
Show that its conjugate is given by equating to zero any two of the determinants
L
M
N
Lx + My + Nz
A
B
C
D
=0
A
B
C
D
where L, M, N, are the component couples at the point (x, y, z) and L, M, N those at the
origin.
Sol.
Let x, y, z be the current coordinates. If (x, y, z) be any point, then directionratios of null line are x – x, y – y, z – z. Also L, M, N being the moments of couple at
the point (x, y, z), they represent the direction ratios of the axis of the couple which by
definition is perpendicular to null line.
Hence by applying the condition of perpendicularity, the equations of null plane is
L(x – x) + M (y – y) + N (z – z) = 0
or
Lx + My + Nz = Lx + My + Nz
(1)
But L = L – yZ + zY, M = M – zX + xZ, N = N – xY + yX, where L, M, N are
component couples at origin.
Therefore
Lx + My + Nz = (L – yZ + zY)x = Lx + My + Nz.
(2)
Hence from (1) and (2), the equation of null plane is
Lx + My + Nz = Lx + My + Nz
(3)
Now the conjugate line whose equations are
Ax + By + Cz = D, Ax + By + Cz = D
lie in the null plane (3).
Therefore
Ax + By + Cz = D,
(4)
Ax + By + Cz = D,
(5)
Hence the conjugate line i.e. locus of (x, y, z) is obtained by eliminating x, y, z
from (3), (4) and (5) or the conjugate line is given by equating to zero any two of the
following determinants:
L
M
N
Lx + My + Nz
A
B
C
D
=0
A
B
C
D
2.6 Stable and Unstable Equilibrium:
Consider an object supported at one point say O. There will be equilibrium so
long as its centre of gravity G lies in the vertical line through the point of support, for, its
weight then acts through the point of support and there is no moment tending to turn the
body about that point. But the nature of equilibrium differs greatly with the position of G.
We shall distinguish three cases:
1. Suppose that the centre of gravity lies below the point of support. If the body be
slightly displaced from this position of rest, then in the displaced position there are two
equal and opposite forces acting on the body, one through O and other through G. These
83
form a couple which has a tendency to restore the body to its original position of rest. In
this case the body is said to be in stable equilibrium.
2. Next suppose that the centre of gravity lies above the point of support. If the body be
slightly displaced, a couple acts, which has a tendency to restore the body still further
from the position of rest. Such equilibrium is called unstable equilibrium.
3. If the centre of gravity is at the fixed point of support, the object will be at rest in any
position. This is called neutral equilibrium.
As another illustration consider a sphere of radius a whose centre of gravity lies at
a distance b (b < a) from its geometric centre. Suppose that the sphere rests on a
horizontal plane, the centre of gravity being either below O as at G1 or above O as at G2.
In the position of equilibrium as given by the first figure, we have two forces
acting on the body, (i) its weight acting through G1 or G2 (ii) the reaction of the
horizontal plane on the sphere acting through A. These forces act in the same vertical line
and there is equilibrium.
G2
G2
O
O
G1
G1
A
B
Suppose that the body, with its centre of gravity at G1, is slightly displaced so that
it assume the position shown in the second figure, the point of contact with the plane
begin B. In this position there are two equal and opposite forces acting on the sphere, the
weight of the sphere through G1 vertically downwards and the reaction, equal in
magnitude to the weight, through B vertically upwards. These forces form a couple,
which has a tendency to restore the body to the original position of rest. The sphere with
its centre of gravity vertically below O is in a position of stable equilibrium.
In case the centre of gravity of the sphere is at G2 in the first position and the
sphere is displaced slightly to assume the form as shown in the second position, there is a
couple acting on the sphere which has a tendency to turn it further away from its position
of equilibrium. Under this condition the position of the sphere is that of unstable
equilibrium.
If, however, the centre of gravity of the sphere be at O, the body will be in
equilibrium when displaced. The sphere with its centre of gravity at its geometrical centre
is in a state of neutral equilibrium.
Definitions:
If, when a body is slightly displaced from an equilibrium position, the force acting
on the body tend to make it return towards its position of equilibrium, then the
84
equilibrium is said to be STABLE. If the force acting on the body tends to move the body
away from the position of equilibrium, the equilibrium is said to be UNSTABLE. If the
forces acting on the body in the displaced position are in equilibrium, the equilibrium is
said to be NEUTRAL.
Thus the equilibrium of a pendulum is stable when it is displaced from its vertical
position of equilibrium, for its returns towards the vertical position again. Any top-heavy
thing or a stick placed vertically on a finger is an instance of unstable equilibrium. A
sphere made of wood or rubber and floating on water is an example of neutral
equilibrium. Again a cone on its base is in a state of stable equilibrium, on its vertex in an
unstable equilibrium, and when resting along a generator is in neutral equilibrium.
Conditions of stability for a body with one degree of freedom:
A body is said to have one degree of freedom when it is so constrained that only
one geometrical quantity is needed in order to fix its position.
Now consider a body with one degree of freedom. The work done on the body in
bringing it from any standard position may be written in the form
W = f(),
where is the quantity defining the position of the body. Now if F be the resultant force
on the body and s the displacement of the body in the direction of F in the position ,
then
dW = F ds,
that is,
f () d = F ds.
Therefore
F = f ().[d/ds]
(1)
In a position of equilibrium, F = 0, hence
f ().[d/ds] = 0.
Since there are always a number of geometrical quantities, each of which could be
used to fix the position of a body with one degree of freedom, we choose to be that one
which continually increases or continually decreases as the body is moved in the same
direction. With the assumption
d/ds 0,
hence for equilibrium
f () = 0.
(2)
Suppose a root of this equation is 0. Then we have to test whether the
equilibrium is stable or unstable for = 0. Now if the force and the displacement have
opposite signs, the equilibrium will be stable.
Thus for stability dF/ds must be negative. But from (1)
dF = f ( )( d )2 + f ( ) d2
0
0
ds 0
ds
ds2
Hence for stability
f (0)(d )2
ds
must be negative, since f () = 0, i.e., f () = 0 must be negative.
It follows therefore that, if W = f(), the conditions for stable equilibrium are
85
f () = 0
and
f () = 0 is negative,
and these are the conditions for the existence of a maximum of W for = 0.
Similarly it may be prove that if W is a minimum, the equilibrium is unstable. If
W is neither a maximum nor a minimum the equilibrium is either neutral or unstable. The
equilibrium is neutral if dF is absolutely zero. But if dF has the same sign as ds for
displacements in one direction and the opposite sign for displacements in the other
direction, then, after any small displacement, the body is certain to move on the unstable
side after a short time and hence the equilibrium is really unstable.
When the forces are of the conservative type, we have seen that the potential
energy V of the body is given by
V=C–W
where C is a constant.
It follows that in the case of a conservative system, the equilibrium is stable or
unstable according as the potential energy is a true minimum or maximum. For example,
whenever gravitational energy is the only form of potential energy involved, the height of
the centre of gravity must be a minimum.
When the force of gravity is the only external force acting on a body, we arrive at
the following rule for deciding whether a given equilibrium position is stable or unstable:
Suppose that a body is tilted from its position of equilibrium, the measure of the
tilt being given by . Express the height of the centre of gravity of the body in terms of
the variable . If this height is given by
z = f(),
positions of equilibrium are given by
dz/d = f ().
We than find the sign of d2z/d2 for the roots of
f () = 0.
If the sign is positive for a root, the height of the centre of gravity is minimum and
the equilibrium is stable for that , in case the sign is negative the centre of gravity is at a
maximum height and the equilibrium is unstable.
In some cases, it is convenient to find, instead of the height of the centre of
gravity above a fixed plane, the depth y of the centre of gravity below a fixed plane. In
such cases, if the depth x is minimum, the position of equilibrium is unstable and if the
depth y is maximum the equilibrium is stable.
Theorem: A heavy body rests on a fixed body. Find the nature of equilibrium.
Proof:
We shall suppose that (i) the upper body is free to roll, without sliding, on the
lower body, (ii) the portions of the two bodies in contact being spheres of radii r and R
respectively and (iii) the straight line joining their centres is vertical initially. Let O and
O be the centres of the spherical surfaces. The adjoining figure is a section of the bodies
through G, the centre of gravity of the upper one. Let the upper body BAC be displaced
into the position BAC, the new point of contact being A. Let G be the new position of
the centre of gravity of the upper body which now lies on OA, the new position of OA.
86
B
A
B
O
C
G
O
A
A
C

O
Further let
AOA = , AOA = .
Since we suppose the upper body displaced by rolling on the lower, we have
arc AA = arc AA;
therefore
R = r.
(1)
If z be the height of G above O, then
z = OO cos – OG cos (+ )
= (R + r) cos – (r – h) cos[(r + R)/r], by (1),
where AG = h.
Now for equilibrium,
dz/d = 0,
hence
– (R + r) sin + (r – h)[(r + R)/r] sin[(r + R)/r] = 0.
This is satisfied by = 0.
Now we have to find the sign of d2z/d2 for = 0.
But
d2z =  (r + R) cos + (r – h)( r + R )2 cos ( r + R  ),
d2
r
r
which, for = 0 reduce to
2
 (r + R) + (r – h)( r + R ) ,
r
2
i.e., to
( r + R ) [ rR - h].
r
r+R
The equilibrium is stable or unstable
According as z is minimum or maximum,
i.e.
”
” d2z/d2 is positive or negative,
i.e.
”
” rR/(r + R)
> or < h,
i.e.
”
” (1/h) > or < 1/r + 1/R.
Particular cases:
1.If the lower surface be concave instead of convex near A, then R is negative and the
equilibrium is stable or unstable according as
(1/h) > or < 1/r + 1/R.
2.If the lower surface near A be plane, R is infinite and the equilibrium is stable or
unstable according as
h < or > r.
87
3.If the upper body have a plane face in contact with the lower body, r is infinite and the
equilibrium is stable or unstable according as
h < or > R.
In the critical case, when
1/h = 1/r + 1/R, the determination of stability is difficult.
Ex.1 A uniform rod AB of length 2a is hinged at A, a string attached to the middle point
of G of the rod passes over a smooth pulley at C at a height a, vertically above A and
supports a weight P hanging freely, find the position of equilibrium and determine their
nature.
Sol.
Let AB be the rod, its weight W acting
C
through the middle point G. The string of length
l attached to G passes over the pulley at C,
AC being equal to AG and carries a weight P at
A
the other end. Hence the tension in the string is P.

If the angle DAB be denoted by , then
P
G
ACG = ½, since AC = AG. Now the depth of
D
G below C = a + a cos and the depth of P below
W
B
C = l – CG = l – 2a cos ½.
Hence if y be the depth, below the pulley,
of the centre of gravity of the system consisting of
the weights P and W,
(P + W)y = P(l – 2a cos½ ) + Wa(1 + cos)
(1)
and accordingly,
(P + W)[dy/d] = P(a sin½ ) + W(- a sin)
and
(P + W)[d2y/d] = P[(a/2) cos½] + W(- a cos)
= P[(a/2) cos½] + Wa(1- 2a cos2½).
Positions of equilibrium are given by dy/d = 0, i.e., by
P sin½ = W sin = 2W sin½cos½
which is satisfied by
sin½0,i.e., = 0, or cos½= P/(2W).
For = 0,
(P + W)[d2y/d] = a[½P – W].
Hence the position of equilibrium given by = 0 is stable when P < 2W and
unstable when P > 2W.
The second solution cos½= P/(2W) is possible only when P < 2W and then
(P + W)[d2y/d] = aW[P2/(4W2) – P2/(2W2) + 1]
= aW[1 - P2/(4W2)],
which is positive, since P < 2W. The depth y is then minimum and the position of
equilibrium given by
cos½= P/(2W)
is therefore unstable.
88
Ex.2 A homogeneous body, consisting of a cylinder and a hemisphere joined at their
bases, is placed with the hemispherical end on a horizontal table; is the equilibrium stable
or unstable.
Sol.
Let G1 and G2 be the centres of gravity of the hemisphere and cylinder, and let A
be the point of the body which is initially in contact with the table and let O be the centre
of the base of the hemisphere.
G
G
G1
2
O
A
If h be the height of the cylinder and r be the radius of the base, we have
OG1 = 3r/8 and OG2 = h/2
Also the weights of the hemisphere and cylinder are proportional to (2/3)r3 and
2
.r h. The reaction of the plane, in the displaced position of the body, always passes
through the centre O.
The equilibrium is stable or unstable according as G, the centre of gravity of the
compound body, is below or above O, i.e., according as
OG1 wt. of hemisphere is <
> OG2 wt. Of cylinder,
i.e. according as
3r/8  (2/3)r3 is <
> h/2  .r2h,
i.e. according as
r2/2 is <> h2,
i.e. According as
r is <> 2h,
i.e.
>
< h 1.42 …
Ex.3 A body consisting of a cone and a hemisphere, on the same base, rests on a rough
horizontal table, the hemisphere being in contact with the table; show that the greatest
height of the cone, so that the equilibrium may be table, is 3 times the radius of the
hemisphere.
Sol.
If O is the centre of base, then the C.G. of cone ABD and hemisphere BCD must
be below O, for stable equilibrium.
The weight of the cone, (1/3)r2h, acts at G1, where OG1 = h/4.
Weight of BCD, (2/3)r3, acts at G2 where OG2 = 3r/8. For the C.G. of the body
to be at O,
(1/3)r2h h/4 = (2/3)r3 3r/8
Hence h < r3 (for stable equilibrium)
A
h
89
G1
B
x
O
D
G2
C
Ex.4 A solid consists of a cylinder and a hemisphere of equal radius, fixed base-to-base,
find the ratio of the height to the radius of the cylinder, so that the equilibrium may be
neutral, when the spherical surface rests on a horizontal plane.
Sol.
Let ABC be a hemisphere and ABDE
E
D
a cylinder. If r is the radius of the hemisphere
and h the height of the cylinder, their weights
will be in the ratio of their volumes i.e. in the
G2
3
2
ratio (2/3)r : r h.
For neutral equilibrium, the C.G. of the
body must be at O.
A
O
B
3
2
(2/3)r OG1 = r h OG2
or
(2/3)r3 2r/3 = r2h h/2
or
r = h
G1
C
Ex.5 A hemisphere rests in equilibrium on a sphere of equal radius; show that the
equilibrium is unstable when the curved and stable when the flat, surface of the
hemisphere rests on the sphere.
Sol. In the figure, when the curved surface of the hemisphere is placed on the sphere,
h = 5r/8 and R = r.
So 1/h = 8/(5r) and 1/r + 1/R = 2/r.
Hence 1/h < 1/r + 1/R, [since 8/(5r) < 2/r] therefore the equilibrium is unstable and if
the plane surface of the hemisphere is placed on the sphere
h = 3r/8 and r = 
So 1/h = 8/(3r) and 1/r + 1/R = 1/R = 0.
Hence 1/h > 1/r + 1/R,
Therefore the equilibrium is stable.
C
O1
L
G2
G1
O2
A1
A2
90
O
Ex.6 A heavy right cone rests with its base on a fixed rough sphere of given radius, find
the greatest height of the cone if it is in stable equilibrium.
Sol. Let the greatest height of the cone be
A
h. If B is the centre of the base of the cone and
G
C.G. of the cone be at G, then BG = h/4.
For equilibrium to be stable, by
B
particular case 3 of Art. 2.3, h/4 < R or h < 4R.
Ex.7 A heavy uniform cube balances on the highest point of a sphere, whose radius is r.
If the sphere be rough enough to prevent sliding and if the side of the cube be r/2, show
that the cube can rock through a right angle without falling.
Sol.
Let ABC be the given sphere of radius r.
The cube is placed in equilibrium at A. The side
of the cube is r/2.
G
In the position of the equilibrium, GA will
A
be vertical and
GA = r/4.
K
So
GA > r.
O
Hence the equilibrium will be stable.
C
Now if the cube begins to swing and swings
upto an arc AK, then
B
arc AK = r (if AGK = ).
Now if = /4, then AK = r/4 and the value
of AG = r/4.
In this case GK is vertical.
Hence the cone can swing making an angle /4 on both sides, or through one right
angle, without falling.
Ex.8 A uniform beam, of thickness 2b, rests symmetrically on a perfectly rough
horizontal cylinder of radius a; show that the equilibrium of the beam will be stable or
unstable according as b is less or greater then a.
Sol.
By particular case 3, the equilibrium will be stable or unstable if b < a or b > a.
Ex.9 A lamina in the form of an isosceles triangle, whose vertical angle is , is placed on
a sphere, of radius r, so that its plane is vertical and one of its equal sides is in contact
with the sphere; show that, if the triangle be slightly displaced in its own plane, the
equilibrium is stable if sin be less than 3r/a, where a is one of the equal sides of the
triangle.
Sol.
Let ABC be the isosceles triangle in which AB = a and BAC = .
So
AX = a cos(/2) [CX = BX].
C
91
If G is the C.G. of triangle ABC and D
the point of contact of the sphere with AB,
then GD will be vertical and AG = (2/3)AX,
or
AG = (2/3)a cos(/2),
and
GD = AG sin(/2)
= (2/3)a cos(/2) sin(/2),
= (a/3) sin
=h
[By particular case 3 of Art. 2.3]
Hence for the equilibrium to be stable, h < r.
or
(a/3) sin < r
or
sin < (3r/a)
X
G
A
D
B
T

P


W
Ex.10 A weight W is supported on a smooth inclined plane by a given weight P,
connected with W by means of a string passing round a fixed pulley whose position is
given. Find the position of equilibrium of W on the plane and show that it is stable.
Sol.
If the inclination of the plane is and is the angle between the string and the
plane and tension of the string is P, then resolving the forces parallel to the plane,
W sin = P cos
or
cos = (W sin)/P
or
= cos– 1[(W sin)/P]
Hence when we know the position of the pulley and the value of , the position of
W can be found.
Now, if W is displaced slightly downwards,  will decrease; hence P cos will
increase.
So W will move upwards and if W is slightly moved upwards, then increasing,
P cos will decrease and hence W will move downwards. Therefore the equilibrium will
be stable.
Ex.11 A rough uniform circular disc, of radius r and weight p, is movable about a point
distance c from its centre. A string, rough enough to prevent any slipping, hangs over the
circumference and carries unequal weights W and w at its ends. Find the position of
equilibrium and determine whether they are stable or unstable.
A
Sol.
Let ABD be the circular disc with centre
O. C is the point round which the disc can rotate.
C
O
If CO (= c) is inclined at angle to vertical,
taking moments about C,
p
W(r – c sin) = w(r + c sin) + pc sin


r
or
Wr – wr = sin (Wc + wc + pc)
B
D
So
(W
–
w)r
.
w
W
sin =
(W + w + p)c
So  = sin– 1[ (W – w)r
.]
92
(W + w + p)c
The moment increases with increase in and decreases with decrease in . Hence
the equilibrium is stable.
Ex.12 A solid sphere rests inside a fixed rough hemispherical bowl of twice its radius.
Show that, however large a weight is attached to the highest point of the sphere, the
equilibrium is stable.
Sol.
Let w be the weight of the sphere of radius r and let W be the weight placed at the
highest point of the sphere. If h is the length of C.G. of the body from the lowest point of
the sphere, then
h = w.r + W.2r
w+W
Therefore the equilibrium will be stable if
1/h > 1/r – 1/(2r)
or
2r > h
or
2r > (w + 2W)r
w+W
or
2w + 2W > w + 2W
which is always true.
Therefore the equilibrium is stable.
Ex.13 A thin hemispherical bowl, of radius b and weight W, rests in equilibrium on the
highest point of a fixed sphere, of radius a, which is rough enough to prevent any sliding.
Inside the bowl is placed a small smooth sphere of weight W. Show that the equilibrium
is not stable unless
w < W.[(a – b)/2b]
Sol.
If the bowl is displaced, the small sphere moves in such a way that its weight still
acts through the centre of the bowl.
Hence to know the stability of equilibrium the weight of the small sphere may be
taken to act at the centre of the bowl.
Height of the C.G. of the body,
h = w.b + W(b/2)
w+W
The equilibrium will be stable if
1/h > 1/a + 1/b
or
w+W
. > 1/a + 1/b
wb + W(b/2)
or
(w + W)a > (w + W/2)(a + b)
or
Wa/2 > wb + Wb/2
or
w < W.[(a – b)/2b]
Ex.14 A square lamina rests in a vertical plane on two smooth pegs, which are in the
same horizontal line. Show that there is only one position of equilibrium unless the
distance between the pegs is greater than one quarter of the diagonal of the square, but
93
that, if this condition is satisfied, there may be three positions of equilibrium and the
symmetrical position will be stable, but the other two positions of equilibrium will be
unstable.
Sol.
Let ABCD be the square and P and Q
the pegs. Let the diagonal AC = 2d and let it
C
be inclined at an angle to the horizontal AX.
The height GN(= z) of the centre of gravity G
above PQ is given by
z = AG sin – AP sin(- 450)
D
G
B
= d sin – c cos(- 450) sin(- 450) ,
N
if PQ = c,
Q
P
i.e.
z = d sin + (c/2) cos2
(1)

So dz/d = d cosc sin2
(2)
A
X
2
2
and
d z/d = - d sin- 2c cos2 (3)
Now since the pegs are smooth, the equation of virtual work reduces to W.z = 0.
Hence, by (2), the position of equilibrium are given by
cos(d – 2c sin) = 0
(4)
The solutions of this equation are = 900 and sin = d/(2c).
This latter equation has real roots only when 2c > d, i.e. when PQ > ¼ AC.
Take the case when 2c > d.
There are then three positions; the first when AC is vertical and the other two
when AC is inclined at either side of the vertical at an angle sin–1[d/(2c)] to the
horizontal.
When = 900, then by (3), d2z/d2 = - d + 2c = positive.
Therefore z is a minimum and the equilibrium is stable.
When
sin = d/(2c), then
d2z /d2 = - d sin –2c + 4c sin2
= (d2 – 4c2)/(2c) = negative.
In this case z is a maximum and the equilibrium is unstable.
Next take the case when 2c < d.
In this case there is only one position of equilibrium given by = 900, and then
d2z/d2 = - d + 2c = negative
z is now a maximum and the equilibrium is unstable.
Ex.15 A rod SH, of length 2a and whose centre of gravity G is at a distance d from its
centre, has a string, of length 2c sec, tied to its two ends and the string is then slung
over a small smooth peg P; find the position of equilibrium and show that the position
which is not vertical is unstable.
Sol.
Since SP + PH = 2c sec, the peg P must be somewhere on an ellipse of foci S
and H and semi-major axis c sec.
Also its semi-minor axis = (c2 sec2 – CH2) = c tan.
Hence the equation to the ellipse is
94
x2 sin2 + y2 = c2 tan2, or, referred to polar
coordinates through G,
P
2
2
2
2
2
2
sin  (r cos + d) + r sin  = c tan (1)
If we find the value of for which r
T
r T
A
is a maximum or minimum and take the

corresponding point P of the ellipse for the
H
G
C
position of the peg and make PG vertical, we A
S
shall have the slant position of equilibrium.
(1) gives
cos2r2 cos2 - 2 cos.dr sin2= r2 - c2 tan2+ d2 sin2
So cos d sin tan + [r2 – (c2 – d2) tan2] .
r cos
The least value of r is clearly (c2 – d2) tanand then
So
cos d tan
(c2 – d2)
Since in this case r is a minimum the centre of gravity is at its minimum depth
below the peg and therefore at the maximum height above the horizontal and the
equilibrium is unstable. The other two positions of equilibrium are when P is at A or A
and the rod is then clearly vertical.
If GP is a minimum it is clear that GP must be a normal at P; so that P may also
be found from the fact that its normal passes through a known point G on the major axis.

2.7 Unit Summary:
1. Forces in three dimensions may be compounded by the law of parallelopiped of forces,
which is an extension of law of parallelogram of forces in two dimensions.
2. When forces in different direction in space are acting upon a rigid body, we say that the body is
under the action of forces in three dimensions.
3. The moment of a forces F about any point is equal to F.p, where p is perpendicular distance of
the point from the line of action of force. It is said to be +ive or –ive according as the force has a
tendency to rotate the body about the point in anticlockwise or clockwise direction.
4. However, in three dimension those forces which try to rotate the body from axis of x to
axis of y, from axis of y to axis of z and from axis of z to axis of x will have + ive
moments about the axis, similarly the moments of couples.
5. We already know that two equal and unlike parallel forces form a couple whose
moment is equal to the product of the force and the arm of the couple.
6. Any given system of forces acting at any given point of a rigid body can be reduced to
a single force acting through an arbitrarily chosen point and a couple whose axis passes
through that point.
7. The system will be in equilibrium when R = 0 and G = 0, which in other words means
that X = 0, Y = 0, Z = 0. Also L = 0, M = 0, N = 0.
95
8. The sum of the resolved parts of the system of forces parallel to any three axes of coordinates must vanish separately, and also the sums of their moments about the three axes
must vanish separately.
9. Any system of forces acting on a rigid body can be reduced to a single force together
with a couple whose axis is along the direction of the force.
10. A single force R together with a couple K whose axis coincides with the direction of
the force when taken together is called a wrench of the system. The single force R is
called the intensity of the wrench.
11. The ratio K/R i.e. moments of the couple about central axis divided by forces is called
the pitch and is of a linear magnitude. Also it is clear that when the pitch is zero the
wrench reduces to a single force. When the pitch is infinite then R = 0 and hence the
wrenches reduce to a couple K only.
12. The straight line, along which the single forces acts, when considered together with
the pitch, is called a screw. Hence screw is a definite straight line associated with a
definite pitch.
13. Corresponding to a given system of forces referred to a base or origin O, all those
lines which radiate from O, such that the moment of the given system of forces about
these lines is zero are called null lines at O.
14. The plane in which all these null lines lie is called null plane of the point O.
15. The point O itself is called null point.
16. If, when a body is slightly displaced from an equilibrium position, the force acting on
the body tend to make it return towards its position of equilibrium, then the equilibrium is
said to be STABLE. If the force acting on the body tends to move the body away from
the position of equilibrium, the equilibrium is said to be UNSTABLE. If the forces acting
on the body in the displaced position are in equilibrium, the equilibrium is said to be
NEUTRAL.
17. If the lower surface be concave instead of convex near A, then R is negative and the
equilibrium is stable or unstable according as
(1/h) > or < 1/r + 1/R.
18. If the lower surface near A be plane, R is infinite and the equilibrium is stable or
unstable according as
h < or > r.
19. If the upper body have a plane face in contact with the lower body, r is infinite and
the equilibrium is stable or unstable according as
h < or > R.
In the critical case, when
1/h = 1/r + 1/R, the determination of stability is difficult.
2.8 Assignments:
1. Two forces P and Q act along the straight lines whose equations are y = x tan, z = c
and y = – x tan, z = – c respectively. Show that their central axis lies on a straight line
z =
P2 – Q2
y = x(P – Q) tan
2
P+Q
c
P + 2PQ cos2 + Q2
For all values of P and Q prove that this line is a generator of the surface
96
(x2 + y2) z sin2 = 2cxy.
2. Two equal forces act along generators of the same system of the hyperboloid
(x2 + y2)/a2 – z2/b2 = 1 and cut the plane z = 0 at the ends of perpendicular diameters of
the circle x2 + y2 = a2. Show that the pitch of the equivalent wrench is (a2b)/(a2 + 2b2).
3. A force parallel to the axis of z acts at the point (a, 0, 0) and an equal force
perpendicular to the axis of z acts at the point (– a, 0, 0). Show that the central axis of the
system lies on the surface z2 (x2 + y2) = (x2 + y2 – ax)2.
4. A force F acts along the axis of Z and a force mF along a straight line intersecting the
axis of x at a distance c from the origin and parallel to the plane of yz. Show that as this
straight line turns round the axis of x, the central axis of the force generates the surface
[m2z2 + (m2 – l)y2] (c – x)2 = x2z2.
5. Three forces act along the straight lines x = 0, y – z = a; y = 0, z – x = a; z = 0, x – y =
a. Show that they cannot reduce to a couple. Prove also that if the system reduces to a
single force, its line of action must lie on the surface
x2 + y2 + z2 – 2yz – 2zx – 2xy = a2.
6. Forces act along generators of the same system of a hyperboloid. Shew that two
generators of the same system are null lines of the system of forces.
7. Shew that null planes of a series of points which lie in a straight line AB passes
through a second straight line CD; and that if the series of lines AB be generators of a
hyperboloid, the lines CD will also be generators of a hyperboloid.
8. A system of forces is reduced to two forces one of which acts along as assigned line.
Shew (i) that the four lines of action of two such pairs of forces are generators of the
same system of a hyperboloid of one sheet; (ii) that lines meeting to such forces and the
central axis generate a hyperbolic paraboloid, one set of whose generators is
perpendicular to the central axis.
9. A solid homogeneous hemisphere of radius r has a solid right cone of the same
substance constructed on its base; the hemisphere rests on the convex side of a fixed
sphere of radius R, the axis of the cone being vertical. Show that the greatest height of the
cone consistent with stability for a small rolling displacement is
r[{(3R + r)(R – r)} – 2r]
R+r
10. Three equal particles repelling each other forces proportional to the nth power of the
distance connected together by three equal elastic strings. Find the position of
equilibrium and shew that it is stable if n < p/(p – a), where a is the unstretched and p the
stretched length of any string.
11. A solid ellipsoid, whose axes are of lengths 2a, 2b, 2c, rests with the “c-axis” vertical
on a rough horizontal plane. The centre of gravity is on the vertical axis at a distance h
from the bottom vertex. Show that the equilibrium is stable if h is less than both a2/c and
b2/c.
12. A solid frustum of a paraboloid of revolution, of height h and latus rectum is 4a, rests
with its vertex on the vertex of a paraboloid of revolution, whose latus rectum is 4b; show
that the equilibrium is stable if h < 3ab/(a + b).
13. Show that a sphere partially immersed in a basin of water cannot rest in stable
equilibrium on the summit of any convex part of the base.
97
2.9 References:
1. Loney, S. L.: An Elementary Treatise on Statics, Cambridge University Press, 1956
2. Varma, R. S.: Text-Book on Statics, Pothishala (Private) Limited Lajpat Road,
Allahabad – 2, 1962.
UNIT –III Velocity and Acceleration, Simple
Harmonic Motion and Elastic String
STRUCTURE
1.1 Introduction
1.2 Objectives
1.3 Velocity and Acceleration Along Radial and Transverse Direction
1.4 Along Tangential and Normal Direction
1.5 Simple Harmonic Motion
1.6 Elastic String
1.7 Unit Summary
1.8 Assignments
1.9 References
98
1.1 Introduction:
This unit introduces the basics of Velocity and Acceleration Along Radial and
Transverse Direction, Along Tangential and Normal Direction, Simple Harmonic Motion,
Elastic String. It also help us to understand the basic concepts of above topics.
In this unit we shall study the basic ideas of Velocity and Acceleration Along
Radial and Transverse Direction, Along Tangential and Normal Direction, Simple
Harmonic Motion, Elastic String. It is hopped the unit help students in studying.
1.2 Objectives:
At the end of the unit the students would be able to understand the concept of:









Velocity and Acceleration
Radial and Transverse Velocity and Acceleration
Angular Velocity and Acceleration
Tangential and Normal Direction
Simple Harmonic Motion
Geometrical Representation of the S. H. M.
Hooke‟s Law
Horizontal Elastic String
Vertical Elastic String
99
1.3 Velocity and Acceleration Along Radial and Transverse Direction
Velocity and Acceleration:
Let a particle move along a straight line starting from a given point O on the line
and let it come to the position P in time t where OP = x. Further suppose in a subsequent
interval t where t is small, the particle moves through a distance PQ(= x) and comes
to the position Q. Thus (x/t) is the average velocity of the particle during the interval
t. As t and consequently x becomes smaller and smaller, the point Q approaches the
point P and the quotient (x/t) measure the rate of displacement of the particle. This
quotient gives the velocity of the particle in the limit when t0. Thus if v be the
velocity of the particle at time t, we have
displacement in time t
v = lim ---------------------------t0
t
.
.
.
.
x
dx
O
P
Q
X
= lim ----- = ------t0 t
dt
.
It is usual to denote differential coefficients with regard to time by dots; thus x
..
means (dx/dt) and x means (d2x/dt2).
.
Thus v = (dx/dt) = x
Velocity of a particle has got magnitude as well as direction. The magnitude is
called the speed. If both speed and direction remain the same throughout a certain
interval the velocity is uniform throughout that interval. If either of these changes, the
velocity becomes variable.
Acceleration of a moving particle is defined as the rate of change of velocity. If v
be the velocity of the particle at time t when it is at the point P and v + v be its velocity
at time t + t when it is at Q, then v is the change of velocity in interval t. thus if f be
the acceleration of the particle at time t, we have
change of velocity in time t
f = lim ---------------------------------t0
t
v
dv
= lim ----- = ------t0 t
dt
But
Also
..
v = (dx/dt)
so f = (d2x/dt2) = x
f = lim(v/t) = lim (v/x) . lim (x/t)
100
= (dv/dx) . (dx/dt) = v (dv/dx)
Thus any one of the three expressions (dv/dt) or (d2x/dt2) or v (dv/dx) may be
taken as the acceleration of the moving particle.
Acceleration also has magnitude as well as direction. Negative acceleration is also
known as retardation. Retardation implies decreases in the magnitude of velocity.
.
It should be very carefully noted that (dx/dt) or x is the velocity of the particle in
..
the sense in which x increases, similarly, v (dv/dx) or x is the acceleration of the particle
in the sense x increasing.
Radial and Transverse Velocity and Acceleration:
A particle is moving in a plane curve, to find components of velocity and
acceleration at time t along and perpendicular to the radius vector drawn from a fixed
point in the plane.
Consider the fixed point O as the
pole and line OX as the initial line. Let P
v + v
be the position of the particle at time t, its
u + u
coordinates be (r,) and Q be the position
v
Q
at time t + t, its coordinates (r + r, + ),
u
so that the chord PQ is the displacement in
r + r
time t. Draw QM perpendicular from Q to
P
M
r
OP so that PM and QM are the components

of the displacement PQ along and perpendicular
O

to OP. Let u, v be the components of velocity
A
along and perpendicular to OP.
Then
Displacement along OP in time t
u = lim ----------------------------------------t 0
t
PM
OM – OP
= lim ------ = lim ----------t 0 t
t 0
t
OQ cos  – OP
= lim -----------------t 0
t
(r + r) 1 – r
= lim ---------------, Small quantities of above the first order
t 0
t
being neglected.
r
dr
.
= lim
--- = ---- = r
t 0 t
dt
displacement perp. to OP in time t
101
v = lim --------------------------------------------t 0
t
QM
OQ sin 
= lim
------ = lim --------t 0 t
t 0 t
(r + r) sin 
(r + r) sin 
= lim ------------- = lim
-------------- . ---t 0
t
t 0
t
t

sin 
= lim (r + r) ---- because lim
-------- = 1
t 0
t
0 

r
= lim -----, neglecting the other term
t 0 t
d
.
= r ----- = r
dt
Thus
the
components
of velocity along and perpendicular to the radius vector are
.
.
r and r, in the sense in which r and increase. These are called the radial and transverse
or cross-radial components of velocity.
Now let the components of velocity along and perpendicular to OQ be u + u, v +
v; (u, v) being those along and perpendicular to OP.
Thus the change of velocity along OP in time t
= (u + u) cos – ( v + v) sin – u
= (u + u) . 1 – ( v + v) . – u, neglecting higher power of 
= u – v, neglecting the other term.
Similarly the change of velocity perpendicular to OP in time t
= (u + u) sin + (v + v) cos – v
= (u + u) + (v + v) . 1 – v, as before
= u + v, neglecting the other term.
Therefore, radial acceleration
Change of velocity along OP in time t
= lim
----------------------------------------------t 0
t
= lim
u – v
----------
102
t 0
t
du
d
= ------ v ------dt
dt
d
= --dt
d2r
= ---  r
dt2
dr
--dt
d d
 r ---- . ---dt dt
d
.
..
---- = r – r 2
dt
Transverse acceleration
Change of velocity perp. to OP in time t
= lim ------------------------------------------------t 0
t
u + v
= lim -----------t 0
t
d
dv
= u ----- + --dt
dt
dr d d
= ----- ---- + --dt dt dt
d
r ---dt
dr d dr d
d2
= --- . --- + ---- . --- + r ---dt dt dt
dt
dt2
d2 
dr d
= r --- + 2 --- . --dt2
dt dt
..
..
= r + 2r

also
1 d
d
= --- ---- r2 ---r dt
dt
103
1 d
.
= --- ---- (r2)
r dt
.. .
Thus the components of acceleration are r – r along OP in the sense r increasing
1 d
.
and --- --- (r2) perp. to OP in the sense increasing.
r dt
.
Cor.
. If .the particle describe a circle of radius a, then r = a = constant, so that u = r = 0, v =
r  = a;
.
.
.
.
..
Radial acc. = r – r .. = 0 –. a 2..= - a 2, i.e., a 2 towards the center and
.
Transverse acc. = r + 2r = a, i.e., tangentially.
Ex.1 If the radial and transverse velocities of a particle are always proportional to each
other, show that the path is an equiangular spiral.
Sol.
dr
d
Here
--- = kr --- where k is some constant
dt
dt
or
or
dr
---- = kd
r
Integrating we get, log r = k + C, where C is some constant.
r = aek, where a is also a constant.
This is an equiangular spiral.
Ex.2 The velocities of a particle along and perpendicular to the radius from a fixed origin
are r and ; find the path and show that acceleration, along and perpendicular to the
radius vector, are 2r – (22)/r and (+ /r)
Sol.
dr
d
Here ---- = r and r ----- = 
dt
dt
rd

dr
d
Dividing we get, ----- = ----- or --- ---- = ----dr
r
r2 
Integrating we get
1
- ---- . ---- = log + C, where C is a constant.
r
This gives the path.
.
..
Radial acceleration = r – r 
104
22
.
..
.
.
= r – r . ------, since r = r so r = r = 2r
r2
 
.
=  r – ----, and = ----r
r
Transverse acceleration
. . ..
= 2r + r
 ..
= 2r . ---- + r
. r
By differentiating r..= 
.
. .
we get
r + r = 


..
i.e.
r = ---- - r . ------ = ------ . ( - r )
r
r
r

Hence transverse acceleration = 2 + ----- ( - r)
r
= [2 + (/r) - ] = [+ (/ r)]
2
Ex.3 A straight line of constant length moves with its ends on two fixed rectangular axes
OX, OY and P is the foot of the perpendicular
from O on the .straight line. Show that
.
velocity of P perpendicular to OP is OP. and along OP is 2CP. where C is the middle
point of the line and is the angle COX.
Sol. Since AOB is a right angled triangle, C is the middle point of the hypotenuse, OC =
CA = CB = a if AB = 2a
COA = = CAO
POX = 90 – 
Y
and
PO = OC cos (90 – 2) = a sin2 
B
so polar coordinate of P are (a sin2, 90 – )
Also CP = a cos2
so
Velocity of P along OP
P
d
.
= ---- (a sin2) = 2a cos2 


 C
dt

O
X
.
= 2.CP






A
d
Velocity of P perp. to OP = a sin2 ----- (90 – )
dt.
=
OP.,
.
i.e.
velocity along PA = OP.
105
Ex.4 A boat which is rowed with constant velocity U starts from a point A on the bank of
a river which flows with a constant velocity V; and it points always towards a point B on
the other bank exactly opposite to A; find the equation of the path of the boat. If V = U,
show that the path is a parabola whose focus is B.
Sol.
Let P be the position of the boat at time t. It has two velocities, U towards B and
V downstream. Take the opposite bank as initial line and. B as pole and P the point (r, ),
.
its components of velocity are r along BP produced and r perpendicular to BP.
Hence resolving along and perp. to BP, we get
.
r = V cos – U
.
and
r = - V sin 


B
X

rd
V sin 
u
r
Dividing we get ----- = ------------v
P
dr
V cos – U
or
(dr/r) = [- cot+ (U/V)cosec]dA
Integrating we get log r = log cosec + (U/V) log tan/2 + C1 where C1 is
constant.
This gives the path.
If V = U, it becomes log r = log cosec + log tan/2 + C1
or
log r = log [1/(2cos2/2)] + C1
or
2r cos2/ 2 = A where A is also a constant
i.e.,
A/r = 2 cos2/2 = 1 + cos which is a parabola, with B as focus.
Ex.5 A and B are two fixed points on the circumference of a circle and the distances from
A and B of any other point P on the circumference are r and s respectively. If u and v are
the components of P‟s velocity, as it moves round the circumference, along AP and BP
respectively prove that being the angle APB.
. .
u sin2 = r – s cos,
u
v
. .
2
v sin  = s – r cos 



r
s
Sol.

Resolving along AP,
s
r
.
r = u – v cos(180 - ) = u + v cosB
A
and resolving along BP
.
s = v – u cos(180 - ) = v + u cos
. .
so
r – s cos = u sin2
. .
and
s – r cos = v sin2
Angular Velocity and Acceleration:
The angular velocity of a point P about another point O is the rate of change of
the angle, which OP makes with some fixed direction.
A particle is moving in a plane curve.
Take a line OX fixed in the plane of the curve
as initial line and O as pole. Let P be the
Q
106
position of the particle at time t, being given
by the angle XOP = .
P
Let Q be the position at time t + t, so

that the angle described in time t is .

Thus the average angular velocity of P
O
X
about O is (/t).
As t becomes smaller and smaller, Q approaches P and (/t) becomes the rate
of change of . This is angular velocity.
.
Thus angular velocity of P about O = lim (/t) = (d/dt) = 
t0
..
Similarly angular acceleration is d(d/dt) = (d2/dt2) = v
.
dt
r
.
Let v be the velocity of the particle
r
at P. It is along the tangent to the path at P.
P
Components of velocity
of P along and perp.

.
.
to OP are r and r.
r
Resolving
perp.
to
OP,
we
get
.
r = v sin
O 
X
where is the angle the tangent at P makes
p
with OP.
.
Thus = (vsin/r)
i.e. angular velocity
of P about O = [(velocity of P resolved perp. to OP)/OP]
.
Also
=
(vsin)/r
and sin = (p/r) where p is perp. from O to the tangent at P. So
.
2
= (vp)/r .
Ex.1 A particle describe an equiangular spiral r = ae in such a manner that its
acceleration has no radial component. Prove that its angular velocity is constant and that
the magnitude of the velocity and acceleration is each proportional to r.
Sol.
.
..
Here r – r 2 = 0
.
 .
.
From r = ae
,
r
=
ae

=
r

..
.  ..
.. . .
r
=
r
+
r
=
r
+ r
.. .. . 2
so
r = r –..r = 0 as given above.
Hence
= 0
.
so
= constant = K say
.
.
.
Then r = Kr so that v2 = r2 + r = 2K2r2
so v varies as r
Since radial acc. is zero,
is transverse
. ..only. acc.
.
.
= 2r + r = 2r = 2K2r
so acc. Varies as r.
Ex.2 A rod moves with its ends on rectangular axes OX, OY. If x, y be a point P on the
rod and if the angular velocity of the rod is constant, show that components of
107
acceleration of P along the axes are – x and – y and the resultant acceleration is
OP.2 towards O.
Sol.
If C be the middle point then OC = CA = CB = a,
so that AOC = = OAC.
.
It is given that = const. = 
Let CP = fixed = c for any particular
point P chosen. Then
B
x = a cos + c cos = (a + c) cos
y = (a – c) sin .




C
.
x = - (a + c) sin = - (a + c) sin
.
y = (a – c) cos
..
x = - (a + c) 2cos = - x 2

P
..
y = - (a – c) 2sin = - y 
A
Resultant acc. makes angle with x-axis as
O
.. ..
tan-1(y/x) = tan-1(y/x) = 






where is the angle OP makes with x-axis.
..
..
Hence acc. is along PO as both x and y are negative, and is equal to
.. 2 + ..y2) = 2(x2 + y2) = 2.OP.
(x
1.4 Along Tangential and Normal Direction
A particle is moving in a plane curve; to find components of its acceleration along
the tangent and the normal to the curve at any instant.
Let A be a fixed point on the curve and P be the position of the particle at time t
where AP = s. Let v be the velocity of the particle at P, it being entirely along the tangent
.
at P; so that v = s.
Let Q be the position of the particle at time t + t and v + v be the velocity there
i.e. along the tangent at Q.
Let the tangents at P and Q make angles
and + with a fixed line OX so that 
is the angle between the tangents. 
Thus the change of velocity along the
Y
tangent at P in times t
v + v
= (v + v)cos – v
= (v + v).1 – v,
N
Q
v
neglecting terms of second order
= v

and the change of velocity along the normal at
P in time t
A
P
= (v + v)sin – 0
+ 
= (v + v), neglecting second order terms O
X
= v, neglecting the other terms.
Thus tangential acceleration
change of velocity along the tangent in time t
108
= lim -----------------------------------------------------t0
t
v
dv
d2s
= lim ----- = ----- = ----t0 t
dt
dt2
or
= (dv/ds).(ds/dt) = v. (dv/ds)
Normal acceleration
change of velocity along the normal in time t
= lim ------------------------------------------------------t0
t
v
v
s
= lim ------- = lim -------- . -----t0 t
s
t
v
s
= lim ------- . lim ---s0 s
t0 t
= (v/).v = (v2/)
where is the radius of curvature at P.
Thus for a particle moving in a plane curve, component of acceleration along the
tangent is (dv/dt) or (d2s/dt2) or v(dv/ds), in the sense in which s increase; and the
component of acceleration along the normal is (v2/) in the inward sense.
..
Cor. For a particle moving in a circle of radius a, s = a, so that
tangential
acc. = s
..
.
.
.
=a, in the sense increasing and normal acc. = (v2/) = (s2/a) = (a/a) = a, towards
the center.
.
Ex.1 Prove that the acceleration of a point in a curve with uniform speed is 2. A point
describes a cycloid s = 4a sin with uniform speed v. Find its acceleration at any point s.
Sol. If v = const.
Components
of acc. are (dv/dt) = 0 and (v2/).
.
Now = (d/dt)
= (d/ds).(ds/dt) = (v/)
.
So
v =  .
.
Hence acc. = (/) = 2.
For s = 4a sin, = (ds/d) = 4a cos
So acc. = (v2/) = (v2/4acos) = [v/{4a(1 – sin2)}]
= [v2/{4a(1 – s2/16a2)}] = [v2/(16a2 – s2)]
Ex.2 A point P describes, with a constant angular velocity of OP, an equiangular spiral of
which O is the pole. Find its acceleration and that its direction makes the same angle with
the tangent at P as the radius vector OP makes with the tangent.
Sol.
109
cot
Equation

. to equiangular spiral is r = ae
Here = constant = c, say
.
.
So
r = a cot ecot = cr cot
.. .
r = cr. cot = c2cot2r
..r – r = c2 cot2 r – c2r = (cot2 – 1)c2r
So
..
..
Also 2r + r = 2c2r cot + 0 = 2c2r cot, so = 0
These are the components of acceleration in polar coordinates hence the resultant
acc. = (cot2 – 1)2 c4r2 + 4c4r2 cot2] = c2r cosec2. If be the angle the direction of the
resultant acc. makes with OP then tan = (2c2r cot)/[(cot2 – 1)c2r] = (2cot)/(cot2 –
1) = tan2, so =2 but the tangent makes an with OP, so direction of the acceleration
makes the same angle with the tangent.
Ex.3 A particle is moving in a parabola with uniform angular velocity about the focus;
prove that its normal acceleration at any point is proportional to the radius of curvature of
its path at that point.
Sol.
Equation
pole is p2 = ar.
. to parabola with the. focus as
Here = const. = c, say but = (vp/r2), so c = (vp/r2); so v = (cr2/p)
Also = r(dr/dp)
Differentiating p2 = ar with respect to p, we get
2p = a(dr/dp), so (dr/dp) = (2p/a)
So = r.(2p/a) = (2pr/a) = [(2r.ar)/(ap)] = (2r2/p)
Now normal acc. = (v2/) = [(c2r4/p2)/(2pr/a)] = [(ac2r3)/(2p3)] = [(ac2r3)/(2ar.p)]
= [(c2r2)/(2p)] = (c2/4). i.e. varies as.
Ex.4 A particle describes a parabola with uniform speed; show that its angular velocity
about the focus S, at any point P, varies inversely as SP3/2.
Sol.
Here v = const. = c, say.
Equation
with S as pole is p2 = ar.
. to parabola
Now = (vp/r2) = (cp/r2) = [c(ar)/r2] = (ca/r3/2) (1/r3/2)
Ex.5 Prove that the angular acceleration of the direction of motion of a point moving in a
plane is (v/).(dv/ds) – (v2/2).(d/ds).
Sol.
If the tangent at any point of the path makes angle with the fixed line OX, then
 gives the direction
of motion.
..
To find. 
Now = (d/ds).(ds/dt) = v.(d/ds) = (v/)
.. d v.
v.
d. v.
= ds d

=v
dt 
dt ds 
ds 
2 2
= (v/).(dv/ds) – (v / ).(d/ds).
110
Ex.6 Prove that if the tangential and normal accelerations of a particle describing a plane
curve be constant throughout the motion the angle through which the direction of
motion turns in time t is given by = A log(1 + Bt).
Sol.
..
.
Here s = const. = a say and (s2/) = (v2/) = const. = b, say
.
.
.
.
So s = at + c and [s2/(ds/d) = b or s.[(ds/dt)/(ds/d)] = b or s.(d/dt) = b
So (at + c).(d/dt) = b.
Hence (d/dt) = [b/(at + c)]
Therefore = A log (1 + Bt).
Ex.7 A particle moves in a catenary (s = c tan), the direction of its acceleration at any
point make equal angles with the tangent and the normal to the path at the point. If the
speed at the vertex where = 0 be u, show that the velocity and acceleration at any other
point are given by ue and (2/c).uecos2
Sol.
Since the acc. bisects the angle between the tangent and the normal, the tangential
and normal components of acc. are equal.
Hence v.(dv/ds) = (v2/) or (dv/ds) = (v/) or (dv/ds) = v.(d/ds)
So (dv/v) = d. Integrating log v = + const.
Therefore v = Ae but when = 0, v = u
Hence A = u therefore v = ue
Tang. acc. = v.(dv/ds) = v.(dv/d).(d/ds) = (ue.ue)/(c sec2), since s = c tan
= (u2e2/c).cos2 so (ds/d) = c sec2 = 
Normal acc. = (v2/) = (u2e2/(c sec2) = (u2e2.cos2)/c
Hence resultant acc. = [{(u2e2 cos2)/c}2 + {(u2e2 cos2)/c}2]
= (2/c).u2e2 cos2
1.5 Simple Harmonic Motion:
A particle is to say to execute Simple Harmonic Motion if it moves in a straight
line such that its acceleration is always directed towards a fixed point in the line and is
proportional to the distance of the particle from the fixed point.
Let O be the fixed point on a line BOA
and P be the position of the particle at time t
where OP = x, so that the acceleration of the
..
particle in the sense OP is x.
. . .
Now the given acceleration is towards
B
O x P
A
O and is proportional to x. Let it be x, where
is constant.
..
Since x is in the direction of OP produced and x is towards O, the equation of
motion is
..
x = - x.
..
Taking v(dv/dx) instead of x; we can write the above equation as
v(dv/dx) = - x.
(1)
111
Integrating with respect to x, we get
(v2/2) = - (x2/2) + (C/2) where C is a constant
or
v2 = - x2 + C.
If A be the extrme position of the particle i.e., it is at rest at A i.e., when x = a, v =
0 where OA = a, we get
0 = - a2 + C, so C = a2
Hence v2 = (a2 – x2)
i.e.
v = (a2 – x2)
(2)
If the particle moves from A towards O, v is negative
.
Hence
x = v = - (a2 – x2)
or
(dx/dt) = - (a2 – x2)
or
dt = - [dx/(a2 – x2)]
Integrating we get .t = cos-1(x/a) + C1, where C1 is a constant. Initially at A, t =
0, x = a i.e., the particle started from A, then
0 = cos-11 + C1, so C1 = 0
Hence .t = cos-1(x/a)
or
x = a cos.t
(3)
If the particle moves from O towards A, v is positive so that
.
x = (a2 – x2)
or
.dt = [dx/(a2 – x2)]
Integrating, we get, .t = sin-1(x/a) + C2, where C2 is a constant.
If the particle starts from O, t = 0, x = 0, 0 = sin-10 + C2, so C2 = 0
Hence x = a sin.t
(4)
Thus the solution of (1) is x = a cos.t or x = a sin.t according as the starting
point is A or O.
From (2),
v = 0 when x = a.
Thus if B is a point on the other side of O such that OB = OA = a, the particle
comes to rest also at B. When x = 0, v = .a, i.e., at O, the velocity is a.
Consider the solution x = a cos.t
The motion starts from A under an attraction towards, O. When the particle
reaches O, x = 0. So cos.t = 0. Hence t = (/2) i.e., t = [/(2)] is the time required
in moving from A to O.
As the particle reaches O, the attraction ceases but the particle has a velocity .a
towards the negative side of O hence the particle passes O and moves towards the
negative side. As soon as the particle comes to the left side of O, attraction changes
direction and becomes towards O; hence the velocity will go on decreasing as the particle
moves towards the left, till at B, the velocity becomes zero so that the particle stops. But
the particle is being attracted towards O hence starts moving towards O and reaches O
with a velocity .a, due to which it passes O and moves towards A and again stops at A
where its velocity becomes zero. The motion is then repeated. Thus the motion is from A
to B and back to A and so on. The motion is oscillatory. Time from O to B is equal to
that from A to O hence the period i.e., the time from A to B and back to A is 4.[/(2)
= (2/). The distance a (= OA) i.e., the distance of the center from one of the position
of rest is called the amplitude.
112
Thus the period which is equal to (2/) is independent of the amplitude i.e.,
whatever be the amplitude the period is the same. Thus the simple harmonic motion is
oscillatory and periodic, the period being independent of amplitude.
The frequency is the number of complete oscillation in the one second, so that if
n be the frequency and T the periodic time, then
n = (1/T) = /(2)
..
The equation (1), namely x = - x, can be solved as a differential equation. The
most general solution of this equation is
x = A cos.t + B sin.t
(5)
A, B are constants to be determined from initial conditions. In the first case when the
.
motion starts from A, the initial conditions are t = 0, x = a, x = 0.
Now
t = 0, x = a given a = A.
.
Differentiating (5), x = - Asin.t + Bcos.t
(6)
.
The condition t = 0, x = 0 gives 0 = 0 + B, so B = 0.
Hence the solution is x = a cost
In the second case when the motion starts from O, the first condition is t = 0, x =
0. So 0 = A, A = 0
Hence x = B sint.
To determine B, we must know the velocity of projection from O.
Let us take the case of particle, projected from A with velocity V along OA
.
produced, so that the initial conditions are t = 0, x = a, x = V.
Hence from (5) and (6), we get
a =A
V = B. So B = (V/)
Hence the solution is x = a cos.t + (V/) sin.t
Also the general solution of (1) can be written as
x = a cos (t + )
This is the periodic with period (2/).
The quantity is called the epoch, the angle (t + ) is called the argument.
The particle is at its maximum distance at time t0 where (t0 + ) = 0 i.e., t0 = - (/).
Hence the time that has elapsed since the particle was at its maximum distance is equal to
t – t0 = t + (/) = (t + )/. This is the phase at time t.
Geometrical Representation of the S. H. M.:
Let a particle P move on a circle with
constant angular velocity and let M be the
foot of the perpendicular from P on any diameter
OA. If a be the radius of the circle, the only
acceleration of P is 2a towards O. If AOP =  
and OM = x, the component of this acceleration
along OA
= 2a.cos = 2a.(x/a) = 2x towards O.
Hence the equation of motion of the point M is
..
x = - 2x.
P


O M
A
113
Thus is S. H. M.
Thus if a particle describes a circle with constant angular velocity the foot of the
perpendicular from it on any diameter executes a simple harmonic motion.
Ex.1 A particle is moving with S. H. M. and while making an excursion from one
position of rest to the other, its distances from the middle point of its path at three
consecutive seconds are observed to be x1, x2, x3, prove that the time of a complete
revolution is 2/cos-1[(x1 + x3)/2x2].
Sol.
From
x = a cost
x1 = a cost, x2 = a cos (t + 1), x3 = a cos(t + 2)
So
x1 + x3 = 2a cos(t + 1) cos
Hence
(x1 + x3)/(2x2) = cos
Therefore
= cos-1[(x1 + x3)/(2x2)]. T = (2/)
Ex.2 A particle starts from rest under an acceleration K2x directed towards a fixed point
and after time t another particle starts from the some position under the same
acceleration. Show that the particles will collide at time (/K) + (t/2) after the start of the
first particle provide t < (2/K).
Sol.
..
x = - K2x,
So period = (2/K)
The condition t < (2/K) indicates that the second starts before the first has made
one complete oscillation. Let them meet after time t of the starts of the second then
a cos K(t + t) = a cos Kt
So
K(t + t) = 2 – Kt, hence t = (/K) – (t/2).
Therefore
t + t = (/K) + (t/2)
Ex.3 A horizontal shelf is moved up and down with S. H. M. of period ½ sec. What is the
amplitude admissible in order that a weight placed on the shelf may not be jerked off?
Sol.
(2/) = ½. So = 162
Weight will be jerked off when the max. acc. of S. H. M. is greater than g and if it
is not to be jerked off, max. acc. of S. H. M. must be g i.e.,
a = g. So a = (g/162)
Ex.4 In a S. H. M. of amplitude a and period T, prove that
T
v2dt = (2a/T)
0
Sol.
.
x = a cost, v = x = - asint,
T = (2/)
T
T
T
v dt = a sin tdt = a sin2(t / T)dt
2
2
2
2
114
0
0
0
2
= a sin2z (T/2)dz,
Put (2t / T) = z
0
= [a2T/(2)]. = (a2T/2).(42/ T 2 ) = (22a2/T)
2
Ex.5 Show that, if a smooth straight turned be cut between any two points on the earth‟s
surface; it should be traversed by a particle starting from rest under the action of gravity
in about 42 ½ minutes.
Sol.
Let O be the center of the earth, AB the
tunnel and OC perpendicular from O to AB.
Let P be the position of the particle at
time t where CP = x. As the force of attraction
B
C P
A
x
inside the earth varies as its distance from the
center, the force on P = .OP along PO.
O
Its component along PC
= PO cos OPC
= PO. (x/PO) = - x
..
Hence the equation of motion of P is x = - x.
It is S. H. M. of periodic time T = (2 /)
So
Time from A to B = ½ T = (/).
Now attraction at A is g, so that a = g, where a = earth‟s radius.
So
= (g/a)
Hence Required time = (/) = (a/g)
= 3.1416  [(4000 x 1760 x 3)/32.2] secs.
= 42 ½ minutes nearly.
Ex.6 If in a simple harmonic motion u, v, w be the velocities at distances a, b, c from a
fixed point on the straight line, which is not the center of force, show that the period T is
given by the equation
u2
v2
w2
(/ T2).(b – c).(c – a).(a – b) = a
b
c
1
1
1
Sol.
Let the intensity of the force be and amplitude be so that velocity at distance x
from center of force is given by
v2 = (2 – x2)
Let d be the distance of the fixed point from the center of force.
Hence
u2 = [2 – (a + d)2]
v2 = [2 – (b + d)2] and T =(2/), so = (/ T2)
w2 = [2 – (c + d)2]
Thus
(u2/) + a2 + 2ad + d2 – 2 = 0
115
and
or
(v2/) + b2 + 2bd +d2 – 2 = 0
(w2/) + c2 + 2cd + d2 – 2 = 0
Eliminating d and (d2 – 2), we get
(u2/) + a2
a
1
2
2
(v /) + b
b
1 =0
2
2
(w /) + c
c
1
u2
(1/) v2
w2
a
b
c
1
1
1
a2
+ b2
c2
a
b
c
1
1 = 0, which gives the answer.
1
Ex.7 A body is attached to one end of an inextensible string and the other end moves in a
vertical straight line with n complete oscillations per second. Show that the string will not
remain tight during the motion unless
n2 < g /(2a)
where a is the amplitude of the motion.
Sol. The maximum acceleration of the upper end executing S. H.M. is a and its period is
T = (2/). Also T = (1/n).
Hence = (42/ T2) = 42n2
So that the maximum acc. = 42n2a.
The maximum acc. of the particle is g which is possible when the string is not
tight hence the string will not remain tight if the acc. of the upper end is greater than g
i.e.,
if 42n2a > g
i.e., the string will not remain tight unless n2 < g/(42a)
Ex.8 A particle P of unit mass free to slide on a straight wire is attracted towards a point
O of the wire with a force .OP. If the wire be mad to revolve about O with constant
angular velocity in a horizontal plane, show that the motion of the particle on the wire
is simple harmonic with a period 2/(- 2) provided 2 < ; and prove that when 2 =
½ , the path of P in space is a circle.
Sol. When the wire has turned through an angle ,
P .
let P be the position of the particle where OP = r so
r .
that attraction on P.is r towards O.
Equation
of
r
.
..
2
motion of P is r – r = - r. But = constant = .

..
..
2
2
So r – r = - r. or r = - (-  )r.
O
This is the S. H. M. of periodic time 2/(- 2) if > 2.
..
If = 22 then the equation becomes r = - 2r
Solution
of which is r = a cos(t + )
.
But = So = t + C. But C = 0 since = 0 = t, = t.
Hence r = a cos( + ) which is a circle.
Hooke’s Law:
If an elastic string is fixed at one end and pulled at the other, it is found to
increase in length. The extension is directly proportional to the product of the tension and
116
the natural length and inversely as the area of the cross-section. Thus if x be the
extension, l the natural length, A the area of the cross-section and T the tension, then
x = [(T l)/(A)] or T = (Ax)/ l]
where is a constant depending on the material of the string.
If we take A = unit area, then T = (x/l); i.e. the tension of the elastic string is
proportional to the extension. This Hooke‟s Law and is called the modulus of elasticity.
1.6 Elastic String
Horizontal Elastic String:
One end of an elastic string is fixed to
x
a point on a smooth horizontal table and
.
.
.
. . .
to the other end a particle is attached; the
B
A
O
ATP B
particle is pulled out to a distance and then
let go; to discuss the motion.
Let O be the fixed point and a be the natural length of the string (= OA). Let the
particle be pulled out to a point B where AB = b and then released.
Let P be the position of the particle at any subsequent time t, and let AP = x, so
that x is the extension at time t.
If T be the tension of the string towards A, then by Hooke‟s law
T = (x/a), where is the modulus of elasticity.
If m be the mass of the particle, the equation of motion is
..
mx = - T
..
i.e.
mx = - (x/a)
..
or
x = - [/(am)].x
This shows that the motion is simple harmonic about A.
The period of oscillation = 2(am/
Writing the above equation in the form v(dv/dx) = - [/(am)].x and integrating
with respect to x, we get (v2/2) = - [/(am)].(x2/2) + (C/2) where C is a constant
i.e.
v2 = - [/(am)]x2 + C
At the extreme point B where x = b, v = 0
So
0 = [/(am)].b2 + C, therefore C = [/(am)].b2
Hence
v2 = [/(am)](b2 – x2).
When the particle reaches A, for which x = 0, v = [/(am)].b.
As soon as the particle reaches A, the string gets back its natural length,
consequently T = 0 and therefore simple harmonic motion ceases but the particle has a
velocity b.[/(am)] at A and hence with this velocity the particle moves backwards till it
comes to a point A on the other side of O where OA = OA = a.
This velocity remains constant through out, as the string remains slack and hence
the time taken by the particle to move from A to A, a distance = 2a, is (2a/b).(am/).
As soon as the particle goes beyond A, the string becomes extended so that the
tension comes into play and the motion is again simple harmonic till the particle reaches
a point B where OB = OB, where the motion stops. The particle then retraces its path
under S. H. M. till it comes to A and then under uniform motion till the point A and then
again S. H. M. till the point B where the motion again stops. This motion is repeated.
117
Thus the motion from B to A, A to B, B to A and A to B is simple harmonic
and the total period for which is 2[(am)/].
The time from A to A and A to A is (4a/b).[(am)/].
Hence the total time for one complete oscillation is
2[(am)/] + (4a/b)[(am)/].
Ex.1 A particle of mass m executes S. H. M. in the line joining the points A and B on the
smooth table and is connected with these points by elastic strings whose tension in
equilibrium are each T; show that the time of an oscillation is 2(mll)/{T(l + l)}]
where l, l are the extension of the strings beyond their natural lengths.
Sol.
Let a, b be the natural lengths of two
strings and O the equilibrium position of the
particle.
In equilibrium position
T = 1(l/a) = 2(l/b)
(1)
When the particle is slightly displaced
from O, let P be its position at time t where
OP = x. Then if T1, T2 be the tension in the
strings PA and PB, we have
i.e.,
A
T O T
B
a+l
b + l
A
T1
T2
OxP
B
T1 = 1[(l + x)/a], T2 = 2[(l – x)/b].
..
Equation of motion is mx = T2 – T1 = - x [(1/a) + (2/b)] from (1)
= - Tx[(1/l) + (1/l)], also from (1)
..
x = - [T(l + l)/(mll)].x
This is a S. H. M. of periodic time = 2[(mll)/{T(l + l)}].
Ex.2 A particle m is attached to a light wire, which is stretched tightly between two fixed
points with a tension T. If a, b are the distances of the particle from the two ends, prove
that the period of a small transverse oscillation of m is
2[(mab)/{T(a + b)}]
Sol.
When the particle is slightly displaced
at right angles to AB and it is at a position P at
time t where OP = x, the tension T remains
practically the same. Sum of components of
two T‟s along PO
= T.[x/(a2 + x2)] + T.[x/(b2 + x2)]
B
= T.[(x/a) + (x/b)] neglecting other terms.
Hence the equation of motion is
..
mx = - Tx[(1/a) + (1/b)] = - T.[(a + b)/(ab)].x
..x = - [T(a + b)/(mab)].x
i.e.
P
T
T
x
x
x
b
O
a
A
118
This is a S. H. M. of periodic time = 2[(mab)/{T(a + b)}].
If the length of the wire be given, i.e. (a + b) is given, then time period is
maximum when ab is maximum, i.e. when a = b. Hence for a wire of given length the
period is longest when the particle is attached to the middle point.
Vertical Elastic String:
A particle is suspended from a fixed
O
O
O
point by an elastic string; to discuss the motion.
O is the fixed point, OA = a the natural
a
length of the string. A particle of mass m is
attached to the free end and hangs in equilibrium
at a depth b below A and is at B so that AB = b.
A
A
A
If T0 be the tension in that position then
b T0
T0 = mg
mg
i.e.
(b/a) = mg
(1)
B
B
The particle is then taken to a depth c below
T x
B to the position C so that BC = c and then released.
mg P
Let P be the position of the particle at any
subsequent time t when BP = x; so that if T be the
C
tension in that position
T = [(b + x)/a] = mg + (x/a), from (1).
..
The equation of motion is mx = mg – T = mg – mg – (x/a) = - (x/a)
..
i.e.
x = - [/(am)]x.
(2)
This is a S. H. M. having B, the position of equilibrium, as the centre of
oscillation.
The periodic time = 2[(am)/] = 2 (b/g) from (1).
The amplitude of this S. H. M. is c and if BC<AB, i.e. c<b, the motion is
completely simple harmonic about the point B.
If, however, c<b, the particle in its upward motion goes above A, but at A the
string gets its natural length and so the tension becomes zero hence the S. H. M. ceases,
but the particle has a velocity at A with which it rises against gravity.
Now integrating (2), we get
v2 = C – [/(am)]x2 = C – (g/b)x2; where C is a constant.
But when x = c, v = 0, then 0 = C – (g/b)c2. So C = [(gc2)/b]
Hence v2 = (g/b)(c2 – x2).
Therefore at A, where x = - b, v2 = (g/b)(c2 – b2).
This is the velocity at A.
Hence the height through which the particle rises above A is
[v2/(2g)] = [(c2 – b2)/(2b)]
This will be the case provided the height risen above O does not exceed a, the
natural length of the string, otherwise the motion again S. H. M.
Hence the condition is [(c2 – b2)/(2b)] < 2a, i.e. c2 < (b2 + 4ab).
119
Ex.1 A heavy particle is attached at one point of a uniform light elastic string. The ends
of the string are attached to two points in a vertical line. Show that the period of vertical
oscillation in which the string remains taut is 2[(mh)/(2)], where is the coefficient
of elasticity of the string and h the harmonic mean of the unstretched lengths of the two
parts of the string.
Sol.
O, B the points where the ends are
attached and A the point where the particle
is attached and let OA = a, AB = b and let
l1, l2 be the natural lengths of these two
portions.
Therefore the equilibrium
[(a – l1)/l1] = mg + [(b – l2)/l2]
(1)
Let the particle be slightly displaced.
When it is at P where AP = x, the lengths of
the two portions are (a + x), (b – x) and T1
and T2 be their tensions
O
T1
A
x
P
T2
B
T1 = [(a + x – l1)/l1], T2 = [(b – x – l2)/l2]
Therefore the equation of motion is
..
mx = mg + T2 – T1 = mg + [(b – x – l2)/l2] – [(a + x – l1)/l1]
= - x [(1/l1) + (1/l2)] = - x [(l1 + l2)/l1l2] = - (2/h).x
So period = 2[(mh)/(2)].
Ex.2 Two bodies M and M are attached to the lower end of an elastic string whose upper
end is fixed and are hung at rest, M falls off; show that the distance of M from the upper
end of the string at time t is
a + b + c cos(g/b).t
where a is unstretched length of the string, b and c the distances by which it would be
extended when supporting M and M respectively.
Sol.
Mg = (b/a), Mg = (c/a)
..
Mx = Mg – T = Mg – (x/a) = Mg – (Mg/b).x
..x = - (g / b).(x – b)
So
Therefore
x – b = A cos(g/b).t + B sin(g/b).t
Initially
x = b + c, t = 0, x. = 0, So B = 0, A = c
Hence x = b + c cos(g/b).t
Required depth = a + x = etc.
O
O
A
b+c
A
M
M M
Ex.3 A tight elastic string of natural length l has one extremity fixed at a point A and the
other attached to a stone the weight of which in equilibrium, would extend the string to a
120
length l1; show that if the stone be dropped from rest at A, it will come to instantaneous
rest at a depth (l12 – l2) below the equilibrium position and this depth is attained in time
(2l/g) + [(l1 – l)/g].[ - cos-1(l1 – l)/(l1 + l)}]. Prove also that if the greatest depth
below A be lcot2(/2), the modulus of elasticity is ½ mg tan2 and the above time is
(2l/g)[1 + ( - ) cot].
Sol.
Let P be the position of the particle at time t, where AP = x.
The equation of motion is
..
mx = mg – T = mg – [(x – l)/l].
..
In equilibrium position, x = 0, x = l1
So
0 = mg – [(l1 – l)/l].
..
Hence mx = mg – [(mg)/(l1 – l)].(x – l) = - [(mg)/(l1 – l)].(x – l1)
..
i.e.
x = - [g/(l1 – l)].(x – l1).
Integrating this equation we get
.
x2 = - [g/(l1 – l)].(x – l1)2 + C
where C is constant.
Now the simple harmonic motion beings
when the particle has fallen freely through a
distance l so that when x = l, x. 2 = 2gl.
So
2gl = - [g/(l1 – l)].(l – l1)2 + C
or
C = 2gl – g(l – l1) = g(l + l1)
.2
Hence x = g(l + l1) – [g/(l1 – l)].(x – l1)2
.
The particle will come to rest when x = 0, so
2
2
2
(x – l1) = (l1 – l )
i.e.
x – l1 = (l12 – l2), which is the answer.
Time for free fall is given by l = ½gt2
i.e.
t = (2l/g)
From (1)
x. 2 = [g/(l1 – l)].[l12 – l2 – (x – l1)2]
Time from B to the lowest point is given by
[g/(l1 – l)].t
A
l
B
T
P
(1)
l1 + (l12 – l2)
dx
2
[l1 – l2 – (x – l1)2]
= l

or
l1 + (l12 – l2)
= [sin-1{(x –l1)/(l1 – l )] l
= sin-11 + sin-1[(l1 – l)/(l1 + l)]
= (/2) + (/2) – cos-1[(l1 – l)/(l1 + l)] =  cos-1[(l1 – l)/ (l1 + l)]
Therefore total time of fall
= (2l/g) + [(l1 – l)/g][ - cos-1(l1 – l)/(l1 + l)}]
Further if the greatest depth below the centre be l cot2(/2) we have
l cot2(/2) = l1 + (l12 – l2)
l12 – l2 = l2 cot4(/2) + l12 – 2ll1cot2(/2)
So l1 = (l/2).[1 + cot4(/2)]/cot2(/2)
2
2
121
And
Hence = [(mgl)/(l1 – l)] = [2mgcot2(/2)]/[1 – cot2(/2)]2 = ½ mg tan2
Also time of fall = (2l/g).(1 + {(l1 – l)/2l}.[- cos-1(l1 – l)/(l1 + l)}])
Now [(l1 – l)/2l] = [cot2(/2) – 1]/[2cot(/2)] = cot
[(l1 – l)/(l1 + l)] = [{cot2(/2) – 1}/{cot2(/2) + 1}]2 = cos
Hence time = (2l/g)[1 + (- )cot]
Ex.4 A particle is attached to the mid-point of a light elastic string of natural length a.
The ends of the string are attached to fixed points A and B, A being at a height 2a
vertically above B, and in equilibrium the particle rests at a depth (5a/4) below A. The
particle is projected vertically downwards from this position with velocity (ga). Prove
that the lower string slackens after a time (/g)(a/12) and that the particle comes to rest
after a further time (a/2g) where is the acute angle defined by the equation tan =
(3/2).
Sol.
In eqn./m position C0,
A
A
A
T0 = mg + 
[(5a/4) – (a/2)]/(a/2)
T0
= mg + [(3a/4) – (a/2)]/(a/2)
T1
giving
= mg.
C0
C0
T
When the particle has gone
x
through x below C0,
T0
C
y
..
mx = mg +  – T1
1
= mg + [(3a/4) – x – (a/2)]/(a/2)
– [(5a/4) +x – (a/2)]/(a/2)
= - 4 (mg/a).x using the value of.
B
B
B
..
so
x = - (4g/a).x
(1)
Solution is x = A cos2(g/a).t + B sin 2(g/a).t
.
Initially
t = 0, x = 0, x = (ga)
So
x = (a/2) sin2(g/a).t,
Lower slackens when  = 0, so x = (a/4)
Therefore
(a/4) = (a/2) sin2(g/a).t, hence t = (/12)(a/g).
.
Integrating (1) x2 = (g/a).(a2 – 4x2) since
.
when
x = 0, x = (ga).
.
Therefore when x = (a/4), x2 = (3/4).ga.
At the instant the string slackens, its depth = (5a/4) + (a/4) = (3/2).a.
After this instant, let the particle go down through y, then
..
my = mg – T = mg – [(3/2).a + y – (a/2)].(a/2)
= mg – 2mg.[(a + y)/a] = - mg – 2(mg/a).y
..
y = - 2(g/a).[y + (a/2)]
Solution is
y + (a/2) = A cos (2g/a).t + B sin(2g/a).t
.
.
Initially
t = 0, y = 0, y = [(3/4).ga]
So
A = (a/2), B = [(3/8).a]
.
Hence
y = 0 when tan(2g/a).t = (3/2)
i.e.
tan = (3/2) where = (2g/a).t i.e. t = [a/(2g)]
122
Ex.5 A heavy particle is attached to one end of a fine elastic string, the other end of
which is fixed. The unstretched length of the string is a and its modulus of elasticity is n
times the weight of the particle. The particle is pulled vertically downwards till the length
of the string is a and is then let go from rest. Show that the time it returns so this position
is 2[a/(ng)]1/ 2(- + + tan – tan)where and are positive acute angles given by
sec = (na/a) – n – 1, sec2 = sec2 – 4n.
Sol.
When the depth of the particle below Q is x, we have
..
mx = mg – T = mg – nmg.[(x – a)/a]
..
i.e.
x = - (ng/a).[x – {(n + 1)/n}.a]
(1)
So
x – [(n + 1)/n].a = A cos (ng/a).t + B sin(ng/a).t
When t = 0, x = a and x = 0, giving
x = [(n + 1)/n].a + [a – {(n + 1)/n}.a]cos (ng/a).t
When x = a, we have
a = [(n + 1)/n].a + [a – {(n + 1)/n}.a]cos(ng/a).t
i.e.
- 1 = [(na/a) - (n + 1)]cos(ng/a).t
(2)
= sec cos(ng/a).t
So
cos(ng/a).t = - cos, therefore (ng/a).t = - 
t = (a/ng).( - )
.
Integrating (1) with the condition x = 0 when x = a, we have
x2 = (ng/a).[{a – a(n + 1)/n}2 – {x – a(n + 1)/n}2]
.
So when
x = a, x = u say
u2 = (ng/a)[{a – a(n + 1)/n}2 – (a2/n2)] = (ag/n)tan2
The particle with the velocity u goes against gravity till a height a above O. Then
velocity v is
v2 = u2 – 4ag = (ag/n) tan2 – 4ag = (ag/n) tan2
Time to this point is
t = [(u – v)/g] = (a/ng).(tan – tan)
(3)
The motion thereafter is given by
..
mx = - mg – T = - mg – nmg [(x – a)/a]
..x = - (ng/a).[x – a +(a/n)]
So
Hence x – a + (a/n) = A cos(ng/a).t + Bsin(ng/a).t
.
with conditions t = 0, x = a, x = v = (ag/n).tan
x – a + (a/n) = (a/n) cos(ng/a).t + (a/n)tan sin(ng/a).t
Hence x. = 0 gives
tan(ng/a).t = tan, so t = (a/ng).
(4)
Hence from (2), (3) and (4), we get the answer being twice the sum of three.
Ex.6 A body of mass 5lb. is hung on a light spring and is found to stretched it 6 ins. The
mass is then pulled down a further 2 ins. and released. Find the period of the oscillations
and the kinetic energy of the mass as it passes through its equilibrium position.
Sol.
Since a stretch of 0.5 ft. is due to a force of 5 lb. weight therefore a stretch of x ft.
would correspond to a tension of 10x lb. weight.
123
In the equilibrium position the weight is balanced by the tension and when the
body is at a distance x ft. below the equilibrium position there is therefore an extra
tension of 10xg poundals acting upwards on the body; and it is this extra unbalanced
tension which causes acceleration, so that
..
5x = - 10xg,
..
or
x = - 64x
Hence the period = 2/64 = ¼ = 0.785 sec.
Again the velocity is greatest as the body passes through the equilibrium position
and is then amplitude, where 2 = 64 and amplitude = 2 ins. = ½ ft.
Therefore the velocity v = 4/3 f.s., and the kinetic energy
½ mv2 = ½ 5 16/9 absolute units = 40/9 1/32 = 5/36 foot-pound.
Ex.7 A cage of mass M lb. is being pulled up with uniform velocity u by a long steel
cable when the upper end of the cable is suddenly fixed. Having given that a weight of m
lb. would extend the cable 1 ft., shew that the amplitude of the oscillation of the cage is
u(M/mg).
Sol.
If the cage were hanging from the cable at rest, the cable, though extended by the
weight of the cage, would have a definite length, which we may call the equilibrium
length, and such a position of the cage may be called an equilibrium position. When the
upward motion is uniform, since there is no acceleration, the length of the cable will
remain the equilibrium length and the position of the cage relative to the upper end of the
cage will still be the equilibrium position. We assume that any vertical displacement of
the cage which extends or compresses the cable results in an extra tension or thrust
proportional to the extension or compression. Hence, after the fixing of the upper end of
the cable, the motion of the cage becomes a simple harmonic motion about its
equilibrium position, starting from this position with velocity u.
Since a weight of m lb. would extend the cable 1 ft., therefore a displacement of
the cage through x ft. from the equilibrium position would be opposed by an unbalanced
force of mx pounds weight, so that
..
Mx = - mgx,
..
or
x = - 2x,
where
2 = mg/M
But if a is the amplitude of the oscillation the velocity at the centre of the
harmonic motion is a.
Therefore a = u; but = (mg/M),
so that
a = u(M/mg).
Ex.8 A heavy particle is supported in equilibrium by two equal elastic strings with their
other ends attached to two points in a horizontal plane and each inclined at an angle of
600 to the vertical. The modulus of elasticity is such that when the particle is suspended
from any portion of the string its extension is equal to its natural length. The particle is
displaced vertically a small distance and then released. Prove that the period of its small
oscillation is 2(2l/5g), where l is the stretched length of either string in equilibrium.
Sol. Let m be the mass of the particle and
the modulus of elasticity. Then by supposing
½3l
124
the particle to be suspended from any portion
B
A
of the string, since the extended length is double
½l
l
the natural length we find that = mg.
O
If l0 be the natural length of either string,
we have, in the equilibrium position,
x
y
0
mg = 2[(l – l0)/l0].cos60 = [(l – l0)/l0];
but = mg, therefore l0 = ½ l.
P
Let x denote the vertical displacement and y the length of either string at time t.
..
To find the period of small oscillation we want to obtain an equation of the form
x = - x,
where is a constant. It will therefore be sufficient for our purpose to write down the
equation of motion at time t and neglect all powers of x higher than the first.
..
We have
mx = mg – 2[(y – l0)/l0] cosOPA,
where P is the particle at time t, O is its equilibrium position and PA = PB = y are the
strings.
Now y2 = (x + ½ l)2 + ¾ l2 = l2 + lx + x2;
Therefore
y = l[1 + (x/l)]1/ 2 = l + ½ x,
correct to the first power of x, and
cos OPA = [ ½ l + x]/y = (½ l + x)/(l + ½ x)
= ½ [1 + (2x/l)][1 – x/(2l)]
= ½ [1 + 3x/(2l)],
to the first power of x.
..
Hence mx = mg – [2(l + ½ x – ½ l)/( ½ l)].[ ½ {1 + 3x/(2l)}],
..
therefore
x = g – g (1 + x/l).[1 + 3x/(2l)],
..
or
x = - 5/2(gx/l);
which represents a simple harmonic motion of period 2(2l/5g).
Ex.9 A warship is firing at a target 3000 yards away dead on the beam, and is rolling
(simple harmonic motion) through an angle of 30 on either side of the vertical in a
complete period of 16 secs. A gun is fired during roll 2 secs. after the ship passes the
vertical. The gun was correctly aimed at the moment of firing, but the shell does not
leaves the barrel till 0.03 sec. later. Shew that the shell will miss the centre of the target
by about 4 feet.
Sol.
Let denote the angle turned through by the ship in t seconds after passing the
vertical. Then the change in is simple harmonic, so that it is connected with t by an
equation
..
= - n2
(1)
where 2/n = the complete period = 16 secs.
The complete solution of (1) is
= A sin nt + B cos nt,
but vanishes for t = 0, therefore B = 0,and
= A sin nt.
Also A is the amplitude of the oscillation, i.e. an angle of 30 or /60 radians.
The angular
velocity is therefore given by
.
125
= nA cos nt;
and 2 secs. after passing the vertical the value of this is
(/8).(/60).cos (/4);
or taking 2 as equal to 10, the angular velocity of the ship at the instant of firing the gun
is (1/482) radians per sec.
It follows that during the 0.03 sec. before the shell leaves the barrel the gun
receives an additional angular elevation = 0.03/482 radians.
Now if V be the velocity and  the angular elevation of projection,
(V2 sin2)/g = 9000 feet;
and, if denote the additional range when the elevation is + ,
V2 sin 2( + )/g = 9000 + ;
or
V2(sin2 + 2 cos2)/g = 9000 + ,
so that
= 18000 cot2.
Also the shell will pass over the centre of the target at a height tan ( + )
approximately; and if we neglect the square of the angular elevation, this is equal to
9000, and substituting the value found for this gives 4 feet as the approximate result.
1.7 Unit Summary:
1. Velocity of a particle has got magnitude as well as direction.
2. The magnitude is called the speed.
3. If both speed and direction remain the same throughout a certain interval the velocity is
uniform throughout that interval. If either of these changes, the velocity becomes
variable.
4. Acceleration also has magnitude as well as direction.
5. Negative acceleration is also known as retardation.
6. Retardation implies decreases in the magnitude of velocity.
7. A particle is to say to execute Simple Harmonic Motion if it moves in a straight line
such that its acceleration is always directed towards a fixed point in the line and is
proportional to the distance of the particle from the fixed point.
8. Thus if a particle describes a circle with constant angular velocity the foot of the
perpendicular from it on any diameter executes a simple harmonic motion.
9. Simple harmonic motion is oscillatory and periodic, the period being independent of
amplitude.
10. The frequency is the number of complete oscillation in the one second, so that if n be
the frequency and T the periodic time, then n = (1/T) = /(2)
1.8 Assignments:
1. If the radial and transverse velocities of a point are always proportional to each other
and this holds for acceleration also, prove that its velocity will vary as some power of the
radial vector.
126
2. The velocities of a particle along and perpendicular to a radius vector from a fixed
origin and r2 and 2. Show that the equation to the path is
(/) = ( / 2r2) + C,
and the components of acceleration are
2r3 – 2 (4/r) and r 2 + (3/r)
3. Prove that the path of a point, which possesses two constant velocities, one along a
fixed direction and the other perpendicular to the radius vector drawn from a fixed point,
is a conic section.
4. A particle moves along a circle r = 2acos in such a way that its acceleration towards
the origin is always zero. Prove that
.
(d2/dt2) = - 2 cot.2
5. A particle is performing a S. H. M. of period T about a centre O and it passes through a
point P (OP = b) with velocity v in the direction OP; prove that the time which elapses
before its return to P is (T/).tan-1[(vT)/(2b)].
6. A point in a straight line with S. H. M. has velocities v1 and v2 when its distances from
the centre are x1 and x2. Show that the period of motion is 2[(x12 – x22)/(v22 – v12)].
7. A particle P moves in a straight line OCP being attracted by a force m.PC, always
towards C whilst C moves along OC with a constant acceleration f. If initially C was at
rest at the origin O and P was at a distance c from O moving with velocity V, prove that
the distance of P from O at any time t is
[(f /) + c]cos.t + (V/).sin t – (f /) + ½ f t2.
8. A point executes S. H. M. such that in two of its positions the velocities are u, v and
the corresponding accelerations are ; show that the distance between the positions is
(v2 –u2)/(+ 
and the amplitude of the motion is [(v2 – u2)(2v2 – 2u2)]1/ 2 /( 2 – 2).
9. A body moving in a straight line OAB with S. H. M. has zero velocity when at points
A and B whose distance from O are a and b respectively and has a velocity v when halfway between them. Show that the complete period is (b – a)/v.
10. The particle of masses m1 and m2 are tied to the ends of an elastic string of natural
length a and modulus . They are placed on a smooth table so that the string is just taut
and m2 is projected with any velocity directly away from m1. Prove that the string will
become slack after the lapse of time [(am1m2)/{(m1 + m2)}].
11. A light elastic string of modulus is stretched to double its length and is tied to two
fixed points distant 2a apart. A particle of mass, m, tied to its middle point, is displaced in
the line of the string through a distance equal to half its distance from the fixed points and
released. Prove that the time of a complete oscillation is [(am)/] and the maximum
velocity is [(a)/m] where is the modulus of elasticity.
12. A particle of mass m resting on a smooth horizontal plane is attached to two fixed
points A and B on the plane by elastic string of unstretched lengths a and b respectively
(a > b), the points A and B being (a + b) apart. The particle is held at B and then released.
Prove that the particle will oscillate to and fro through a distance b[(a + b)/a] in a
periodic time (a + b)(m/), where is the modulus of each string.
13. One end of an elastic string, whose modulus of elasticity is and whose unstretched
length is a, is fixed to a point on a smooth horizontal table and the other end is tied to a
particle of mass m which is lying on the table. The particle is pulled to a distance where
127
the extension of the string is b and then let go; shew that the time of a complete
oscillation is 2[+ (2a/b)(am/).
14. An endless cord consists of two portions of lengths 2l and 2l respectively, knotted
together, their masses per unit of length being m and m. It is placed in stable equilibrium
over a small smooth peg and then slightly displayed. Shew that the time of a complete
oscillation is 2[(ml + ml)/{(m – m)g}].
15. A smooth light pulley is suspended from a fixed point by a string of natural length l
and modulus of elasticity ng. If masses m1 and m2 hang at the ends of a light inextensible
string passing round the pulley. Show that the pulley executes S. H. M. about a centre
whose depth below the point of suspension is
l.[1 + (2M/n)]
where M is the harmonic mean between m1 and m2.
16. One end of an elastic string is fixed and to the other end is fastened a particle heavy
enough to stretch the string to double its natural length a. The string is drawn vertically
down till it is four times its natural length and then let go. Show that the particle returns
to this point in time
(a/g).[ + (4/3)].
1.9 References
1. Loney, S.L.:
An Elementary Treatise on the Dynamics of a Particle and of Rigid
Bodies, S.Chand & Company (Pvt.) Ltd, Ram Nagar, New Delhi,
1952.
2. Lamb, Horace: Dynamics: Cambridge University Press, 1961
3. Ramsey, A.S.: Dynamics Part I & II, : Cambridge University Press,1954
UNIT –IV Smooth and Rough plane Curves, Resting
Medium and Particle of Varying Mass
STRUCTURE
2.1 Introduction
2.2 Objectives
2.3 Motion on Smooth and Rough plane Curves
2.4 Motion in a Resting Medium
128
2.5 Motion of Particle of Varying Mass
2.6 Unit Summary
2.7 Assignments
2.8 References
2.1 Introduction:
This unit introduces the basics of Motion on Smooth and Rough plane Curves,
Motion in a Resting Medium, Motion of Particle of Varying Mass. It also help us to
understand the basic concepts of above topics.
In this unit we shall study the basic ideas of Motion on Smooth and Rough plane
Curves, Motion in a Resting Medium, Motion of Particle of Varying Mass. It is hopped
the unit help students in studying.
2.2 Objectives:
At the end of the unit the students would be able to understand the concept of:
 Motion on Smooth Plane Curve
 Motion on Rough Plane Curve
129
 Motion in Resisting Medium
 Particle falls under Gravity
 Particle Projected Upwards
 Particle Projected under Gravity
 Particle Moving under Gravity
 Bead Moves on a Smooth Wire
2.3 Motion on Smooth and Rough plane Curves
Motion on Smooth Plane Curve:
Theorem:
A heavy particle is made to move on a smooth curve in a vertical plane; Discuss
the motion.
Proof:
The external forces are the weight mg
Y
of the particle downward and the normal
reaction R of the curve. If P be the particle at
time t, the equations of motion (tangential and
R
normal) are
mv(dv/ds) = - mg sin
A
P
m(v2/) = R – mg cos
if s be measured from a fixed point A down the
mg

curve below P, the radius of curvature of the
O
X
130
curve at P. Since sin = (dy/ds) we have from the first equation
mv (dv/ds) = - mg (dy/ds)
Integrating ½ mv2 = - mgy + C, where C is a constant. If the particle be projected
from a point where y = y1 with a velocity u then
½ mu2 = - mgy1 + C
Hence ½ mv2 – ½ mu2 = - mg(y – y1) = -mgh
where h is the height of P above the point of projection. This is in fact the equation of
energy and follows at one from the principle of energy namely the change in kinetic
energy is equal to the work done.
Thus for upwards projection
v2 = u2 – 2gh
and for downward projection v2 = u2 + 2g where h is the vertical distance between the
two points. These equations hold good whatever be the paths followed. The second
equation gives R, the reaction on the particle. The particle will leave the curve when R=0.
Ex.1 A particle slides down the smooth curve y = a sinh(x/a), the axis of x being
horizontal and the axis of y downwards, starting from rest at the point where the tangent
is inclined at a to the horizon; shew that it will leave the curve when it has fallen through
a vertical distance a sec.
Sol. Curve is y = a sinh(x/a)
The equations of motion are
m.(d2s/dt2) = mv(dv/ds) = mg sin
and
m(v2/) = mg cos – R,
i.e.
v(dv/ds) = g sin,
(1)
and
(v2/) = - (R/m) + g cos.
(2)
Now
y = a sinh(x/a), so (dy/dx) = cosh(x/a) = tan,
i.e.
tan = cosh(x/a) = [1 + sinh2(x/a)]
=  [1 + (y2/a2)] =  [(a2 + y2)/a2].
Also
(d2y/dx2) = (1/a).sinh(x/a) = y/a2, so that
[1 + (dy/dx)2]3/2
[1 + cosh2(x/a)]3/2
(2a2 + y2)3/2
=
=
=
2
2
2
(d y/dx )
(y/a )
ay
At = , y = b, tan = [1 + (b2/a2)].
From (1)
v(dv/ds) = g sin = g.(dy/ds), so 2vdv = 2gdy
Integrating, we get v2 = 2gy + c
Initially when v = 0, y = b, then c = - 2gb,
so that
v2 = 2g.(y – b).
(3)
The particle leaves the curve when R = 0.
Therefore from (2) v2 = g cos.
(4)
From (3) and (4)
2g.(y – b) = g cos = g/(1 + tan2)
g[1 + cosh2(x/a)]3/2(a2/y)
g[2 + sinh2(x/a)]3/2(a2/y)
2g(y – b)=
2 2
=
(2 + y /a )
(2 + y2/a2)
= g.( 2 + y2/a2).(a2/y)
or
2(y – b) = (2a2/y) + y
or
y2 – 2yb + b2 = 2a2 + b2, i.e. (y – b)2 = b2 + 2a2
131
or
as
so
(y – b)2 = a2 + a2tan2 = a2sec2
tan2 = 1 + (b2/a2), so b2 = a2 tan2 – a2
y – b = a sec.
Ex.2 A particle descends a smooth curve under the action of gravity, describing equal
vertical distances in equal times, and starting in a vertical direction. Shew that the curve
is a semi-cubical parabola, the tangent at the cusp of which is vertical.
Sol.
Let the axis of x be horizontal and axis of y vertically downwards. „Describing
equal vertical distances in equal times‟ means velocity, i.e. (dy/dt) is constant, say k.
i.e.
(dy/dt) = k
(1)
Initially the particle starts in a vertical direction with velocity k.
Therefore v2 = k2 + 2gy;
[x = 0; u = y = k],
2
2
2
i.e.
(dx/dt) + (dy/dt) = k + 2gy; so (dx/dt) = (2gy) from (1)
Now (dx/dy) = [(dx/dt)/(dy/dt)] = (2gy)/k
or
dx = [(2gy)/k]dy, integrating x = [(2g)/k].(2/3).(y)3/2 + C
Initially when x = 0, y = 0; so C = 0.
Hence x = (2/3).[(2g)/k].y3/2
which represents a semi cubical parabola whose tangent at the cusp is vertical.
Ex.3 A particle is projected with velocity V from the cusp of a smooth inverted cycloid
down the arc; shew that the time of reaching the vertex is 2(a/g)tan-1[(4ag)/V].
Sol.
Equation of motion is
(d2s/dt2) = g.(s/4a) or [D2 + g/(4a)].s = 0, i.e. D = i(g/4a).
Its solution is s = A cos(g/4a).t + B sin(g/4a).t.
Initially when t = 0, s = 4a, (ds/dt) = - V; so A = 4a
and as (ds/dt) = - A(g/4a)sin(g/4a).t + B(g/4a) cos(g/4a).t
so
- V = B(g/4a); i.e. B = - V(4a/g).
Hence s = 4a cos(g/4a).t – V(4a/g) sin(g/4a).t
When it comes to the vertex, s = 0.
So
tan(g/4a).t = [4a/V(4a/g)] = (4ag)/V
Therefore t = 2(a/g) tan-1[(4ag)/V].
Ex.4 A particle slides down the arc of a smooth cycloid whose axis is vertical and lowest;
prove that the time occupied in falling down the first half of the vertical height is equal to
the time of falling down the second half.
Sol.
Let the equation of cycloid be s = 4asin, where s is measured from the vertex.
Also s = (8ay).
At t, let P be the position of the particle. Let P(x, y); at cusp A, s = 4a.
So vertical height of A = 2a.
Equation of motion is
m(d2s/dt2) = - mg sin
Y
or
(d2s/dt2) = - (g/4a).s
A
132
[as s = 4a sin.
A
2
R
Integrating (ds/dt) = - (g/4a).s + C.
Initially when s = 4a.(ds/dt) = 0;
P
2
so
C = (2/4a).16a = 4ag.
mg

Hence (ds/dt)2 = (g/4a).(16a2 – s2);
X
O
2
2
so
(ds/dt) = - (g/4a).(16a – s )
- ve sign being taken as the particle is moving in direction of s decreasing.
Therefore
t = - 2(a/g) [1/(16a2 – s2)].ds + D = 2(a/g) cos-1(s/4a) + D
Initially when s = 4a, t = 0; so D = 0.
So that
t = 2(a/g)cos-1[(8ay)/4a].
If t1, be the time taken in falling first half of the vertical distance and t2 that in
falling second half of vertical distance.
a
-1
So
t1 = 2(a/g).[cos {(8ay)/4a}] = 2(a/g)[/4 – 0] = 2(a/g).(/4)
2
2a
and
0
t2 = 2(a/g).[cos-1{(8ay)/4a}] = 2(a/g)[/2 – /4] = 2(a/g).(/4) = t1.
a
Ex.5 A particle is placed very close to the vertex of a smooth cycloid whose axis is
vertical and vertex upwards and is allowed to run down the curve. Shew that it leaves the
curve when it is moving in a direction making with the horizontal an angle of 450.
Sol.
Let at any time t the, the particle be at P. So the equations of motion are
m(d2s/dt2) = mg sin
(1)
2
and
m(v /) = mg cos - R
(2)
(1) may be written as
v(dv/ds) = g(s/4a) as s = 4a sin.
Integrating, v2 = g(s2/4a) + A.
Initially when s = 0, v = 0, A = 0, so that v2 = g(s2/4a)
(3)
From (2) and (3),
R = mg cos – [(mgs2)/(4a.4a cos)] as = (ds/d) = 4a cos
or
R = (mg/cos).(cos2 – sin2) = (mg/cos).cos2.
The particle will leave the curve when R = 0, i.e. when
cos2 = 0 or 2 = /2; so = /4.
O
T
X
R
P
mg
Motion on Rough Plane Curve:
ms
133
Let s be the arc measured from a fixed point A on the curve, in the direction of
motion, to the position of the particle P whose mass is m. Let R be the pressure measured
inwards, the coefficient of friction and T, N the components of the external force on the
particle along the tangent and the outward normal. Then the equations of motion are
mv(dv/ds) = T – R
(1)
2
and
mv / = R – N
(2)
In the case in which a heavy particle is sliding upwards on a curve in a vertical
plane, the equations are
mv.(dv/ds) = - R - mg sin
(3)
and
mv2/ = R - mg cos
(4)
where is the inclination of the tangent to the horizontal.
Eliminating R, we get
v(dv/ds) + (v2/ = - g(sincos),
or
dv2/d + 2v2 = - 2g (sincos),
(5)
From the intrinsic equation of the curve, say s = f(), we get = f  and the
equation (5) has an integrating factor e2.
y
mv2/
y
mv(dv/ds)
T
R
P
A
O
P
s
A

R
N
x
Note:
In applying the method of this article it should be noted that the friction must
always oppose the motion and that in the case of a two-sided constraint (i.e. bead on a
wire), if R changes sign during the motion, the motion will cease to be represented by
equation (5), because it will become necessary to change the sign of R in (4) but in (3).
Ex.1 The base of a rough cycloidal arc is horizontal and its vertex downwards; a bead
slides along it starting from rest at the cusp and coming to rest at the vertex. Show that
2e = 1.
Sol.
The intrinsic equation of cycloid is
s = 4a sin;
So
= ds/d = 4a cos
The equations of motion are
mv.(dv/ds) = R - mg sin
i.e.
½ m.(dv2/ds) = R - mg sin
(1)
2
and
(mv )/ = R - mg cos
(2)
Eliminating R between (1) and (2), we get
dv2/d - 2v2 = 2g(cos - sin)
= 8ga cos(cos - sin), since = 4a cos
134
The equation being linear in v2, the I.F. = e– 2 d = e– 2
and therefore the solution is
v2e– 28ga e– 2cos(cos - sin) d + A
ga e– 2cos2) – sin2] d + A
ga - ½ e– 2e– 2sin2) + 2(1 – 2)cos2 + A
4 + 42
At the cusp when = /2, v = 0.
So
A = - 4ga [- ½ e– + e– {- 2 (1 – 2 )}/(4 + 42)].
Also at the vertex when = 0, v = 0.
So
A = - 4ag [ - ½ + {2 (1 – 2 )}/( 4 + 42)].
Elimination of A between the last two equation gives
– 
2 (1 – 2 ) = - ½ + 2 (1 – 2 )
- ½ e–  - e
4 + 42
4 + 42
– 
2
or
e
 

.
=


1 + 2
1 + 2 
or
2e = 1.
Ex.2 A particle slides in a vertical plane down a rough cycloidal arc whose axis is
vertical and vertex downwards, starting from a point where the tangent makes an angle
with the horizon and coming to rest at the vertex. Shew that e = sin – cos.
Sol.
The intrinsic equation of cycloid is
s = 4a sin;
So
= ds/d = 4a cos
The equations of motion are
mv.(dv/ds) = R - mg sin
i.e.
½ m.(dv2/ds) = R - mg sin
(1)
2
and
(mv )/ = R - mg cos
(2)
Eliminating R between (1) and (2), we get
dv2/d - 2v2 = 2g(cos - sin)
= 8ga cos(cos - sin), since = 4a cos
The equation being linear in v2, the I.F. = e– 2 d = e– 2
and therefore the solution is
v2e– 28ga e– 2cos(cos - sin) d + A
ga e– 2cos2) – sin2] d + A
ga [- ½ e– 2e– 2sin2) + 2(1 – 2)cos2] + A
4 + 42
At the cusp when = , v = 0.
So
A = - 4ga [- ½ e– 2+ e– 2sin2) + 2(1 – 2)cos24 + 42)]
Also at the vertex when = 0, v = 0.
So
A = - 4ag [ - ½ + {2 (1 – 2 )}/( 4 + 42)].
Elimination of A between the last two equations gives
– 2
sin2) + 2(1 – 2)cos2 = - ½ + 2 (1 – 2 )
- ½ e– 2 - e
4 + 42
4 + 42
135
so that
e = sin – cos.
Ex.3 A bead moves along a rough curved wire, which is such that it changes its direction
of motion with constant angular velocity. Shew that a possible form of the wire is an
equiangular spiral.
Sol.
The equations of motion are
mv(dv/ds) = - R
(1)
and
mv2/ = R
(2)
Now we are given that
d/dt = (constant).
So
v = ds/dt = (ds/d).(d/dt) =  i.e. = v/
Thus from (2), v = R/m and therefore from (1), we have
v(dv/ds) = - v 
or dv = - ds.
Integrating
v = A - s.
so that v = = (ds/d) = A - s.
or
ds = d
A - s
Integrating further
log (A - s) = 
i.e.
(A - s) = e
or
s = (A  e)/() = a + b e (say)
which represents an equiangular spiral.
Ex.4 A rough cycloid has its plane vertical and the line joining its cusps horizontal. A
heavy particle slides down the curve from rest at a cusp and comes to rest again at the
point on the other side of the vertex where the tangent is inclined at 450 to the vertical.
Shew that the coefficient of friction satisfies the equation 3 + 4 log (1 + ) = 2 log2
Sol.
The intrinsic equation of cycloid is
s = 4a sin;
So
= ds/d = 4a cos
The equations of motion are
mv.(dv/ds) = R - mg sin
i.e.
½ m.(dv2/ds) = R - mg sin
(1)
2
and
(mv )/ = R - mg cos
(2)
Eliminating R between (1) and (2), we get
dv2/d - 2v2 = 2g(cos - sin)
= 8ga cos(cos - sin), since = 4a cos
The equation being linear in v2, the I.F. = e– 2 d = e– 2
and therefore the solution is
v2e– 28ga e– 2cos(cos - sin) d + A
ga e– 2cos2) – sin2] d + A
ga [- ½ e– 2e– 2sin2) + 2(1 – 2)cos2] + A
136
4 + 42
or
or
At the cusp when = /2, v = 0.
So
A = - 4ga [- ½ e– + e– {- 2 (1 – 2 )}/( 4 + 42)].
Also when = - /4, v = 0.
So
A = - 4ag e/ 2 [ - ½ - (4)/( 4 + 42)].
Elimination of A between the last two equations gives
– 
2 (1 – 2 )
4 e/ 2
/ 2
- ½ e–  - e
=
½
e
4 + 42
4 + 42
2
3/ 2
2 3/ 2
2 = (1 +  + 2) e
= (1 + ) e
Taking logarithm both sides, we get
log 2 = 2 log (1 + ) + (3)/2
3 + 4 log (1 + ) = 2 log2
Ex.5 A particle is projected along the inner surface of a rough sphere and is acted on by
no forces; show that it will return to the point of projection at the end of time
a(e2 – 1)/(V)
where a is the radius of the sphere, V is the velocity of projection and is the coefficient
of friction.
Sol.
Since there is no forces, the particle will continue to move in the plane section,
i.e. a circle in which it starts.
The equations of motion are
mv(dv/ds) = - R
(1)
2
and
mv /a = R
(2)
Eliminating of R between (1) and (2) gives
1 dv2 + v2
=0
2 ds
a
or
1 dv2 d +  v2 = 0
2 d ds
a
2
2
or
1 dv + 2 v
=0
a d
a
[since s = a, so that d/ds = 1/a]
or
or
dv2 + 2v2 = 0.
d
(3)
2
2 d
2 
The equation being linear in v , its integrating factor = e
=e
Thus the solution of the differential equation (3) is
v2 e2  = A, where A is the some constant.
Initially when = 0, v = V; so A = V2.
Hence v2 e2  = V2 or v e  = V
a d e = V,
dt
[since v = ds/dt = a (d/dt) from s = a]
So
=
2

a e
d = t
0
V
a [e ] 2
137
0
V
a [e2 - 1]
=
V
Ex.6 A bead of mass m slides on a rough circular wire of radius a in a vertical plane, the
coefficient of friction being 1/2. The bead is projected from the lowest point; show that
in order that it may just reach the highest point the velocity of projection must be
[(2/3)6gae2(/ 2 + )],
where is the acute angle tan –12.
Sol.
If v is the velocity when is the angular distance of the bead from its lowest
position, the equations of motion, in the early stages of the motion, are
mv.(dv/ds) = - mg sinR
(1)
2
and
mv /a = R - mg cos
(2)
leading to
dv2/d + 2v2 = - 2ga(sin cos),
(3)
which gives, on integration,
2
– 1) cos
v2e2 = C – 2gae2 3 sin + (2
2
4 + 1
(4)
For = 1/2, this becomes
v2e2 = v02 – 2gae2 sin
(5)
where v0 is the value of v when 

According to the data v decreases steadily and vanishes when so there will
be an angle between 0 and for which cosv2/(ag) and from (2) for this value of
the pressure R vanishes and changes sign. From this point onwards (3) ceases to
represent the motion, because (1) still holds good, but instead of (2) we have
mv2/a =  R – mg cos
(6)
which with (1) leads to
dv2/d – 2v2 = - 2ga(sin – cos),
(7)
giving, on integration,
2
v2e–2 = C – 2gae–2 – 3 sin 2+ (2 – 1) cos
4 + 1
(8)
For = 1/2, this becomes
v2e–2 = C + 2gae–2 sin
But v is to vanish when , therefore C = 0 and in the later stages of the
motion
v2 = 2ga sin
(9)
But the motion being continuous the values of v2 given by (5) and (9) are the
same at the point at which R vanishes, i.e. when v2 is also – ga cos. Substituting this
value in (9) gives
cos = 2 sin,
or
cos sin = 1.
=
- 2 1
3
so that = /2 + , where is the acute angle tan–12. Then from (5) and (9) we get
v02 = 22gae2 sin
138
= (2/3) 6gae2(/ 2 + 
2.4 Motion in Resisting Medium:
Particle falls under Gravity:
Theorem: A particle falls under gravity (supposed constant) in a resisting medium whose
resistance varies as the square of the velocity. Find the motion if the particle starts from
rest.
Proof:
Let v be the velocity when the particle has fallen a distance x in time t from rest.
The equation of motion is
d2x/dt2 = g – v2.
Let
= (g/k2), so that
d2x/dt2 = g (1 – v2/k2)
(1)
From (1) it follows that if v equaled k, the acceleration be zero; the motion would
then be unresisted and the velocity of the particle would continue to be k. For this reason
k is called the „terminal velocity‟.
From (1)
v (dv/dx) = g (1 – v2/k2),
so that (2g/k2).x = [2v/(k2 – v2)].dv = - log(k2 – v2) + A.
Since v and x are both zero initially, so A = log k2.
Therefore
k2 – v2 = k2exp(-2gx/k2).
Hence
v2 = k2[1 – exp(– 2gx/k2)]
(2)
It follows that x = when v = k. Hence the particle would not actually acquire the
„terminal velocity‟ until it had fallen an infinite distance.
Again (1) can be written
dv/dt = g(1 – v2/k2).
So
gt/k2 = [1/(k2 – v2)] dv = (1/2k) log [(k + v)/(k –v)] + B.
Since v and t were zero initially, so B = 0.
Hence [(k + v)/(k – v)] = e2gt/ k
So
v = k.[(e2gt/ k – 1) /(e2gt/ k + 1)] = k tanh(gt/k)
From (2) and (3), we have
exp(- 2gx/k2) = 1 – v2/k2 = 1 – tanh2(gt/k) = 1/cosh2(gt/k)
Therefore
exp(gx/k2) = cosh(gt/k) and x = (k2/g).log cosh(gt/k)
(3)
(4)
Particle Projected Upwards:
Theorem: If the particle were projected upwards instead of downwards. Find the motion.
Proof:
Let V be the velocity of projection.
The equation of motion now is
d2x/dt2 = - g – v2 = - g(1 + v2/k2)
(1)
where x is measured upwards.
139
Hence v(dv/dx) = - g(1 + v2/k2).
So
2g/k2 = - [2v/(v2 + k2)]dv = - log(v2 + k2) + A,
where
0 = - log(V2 + k2) + A.
Therefore 2gx/k2 = log[(V2 + k2)/(v2 + k2)].
(2)
Again (1) gives
dv/dt = - g(1 + v2/k2).
So
- gt/k2 = [1/(k2 + v2)].dv = (1/k).tan-1(v/k) + B,
where
0 = (1/k).tan-1(V/k) + B.
Hence gt/k = tan-1(V/k) – tan-1(v/k)
(3)
Equation (2) gives the velocity when the particle has described any distance and
(3) gives the velocity at the end of any time.
Ex.1 A heavy particle is projected vertically upwards with velocity u in a medium, the
resistance of which is gu2 tan2a times the square of the velocity, a being a constant.
Shew that the particle will return to the point of projection with velocity u cosa, after a
time
ug1cota[a + log{cosa/(1 – sina)}].
Sol.
Resistance = gu2tan2a.v2.
Consider the motion of the rising particle, the equation of motion is
v(dv/dx) = - g – gu-2tan2a.v2
or
0
x = - (u2/g) [v/(u2 + v2tan2a)].dv
u
Integrating the greatest height attained, will be given by
0
x = - [u2/(2gtan2a)].[log(u2 + v2tan2a)] = [(u2cot2a)/(2g)] log sec2a
u
(1)
When the particle falls, the equation of motion is
v(dv/dx) = g – v2gu2tan2a = (g/u2).[u2 – v2tan2a].
The velocity attained by the particle in falling through the distance given by (1),
will be obtained after integration as follows
i.e.
2
2
2
v
2
2
2
2 (u /2g) cot a log sec a
[v/(u – v tan a)]dv = (g/u ) 

dx
0
0
2
(u /2g) cot2a log sec2a
v
2
2
2
2
2
½ cot a.[log(u – v tan a)] = (g/u ).[x]
0
0
or
½ cot2a.log[(u2 – v2tan2a)/u2] = (g/u2).[(u2cot2a)/2g].log sec2a
or
[(u2 – v2tan2a)/u2] = cos2a
or
v2tan2a = u2(1 – cos2a) = u2sin2a.
or
v2 = u2 cos2a.
which gives the velocity attained v = u cosa i,e, the particle reaches the point of
projection with velocity u cosa.
Now to find the time, we have
dv. = d2x
g 2
2
2
2 = – 2(u + v tan a)
dt
dt
u
for upward motion.
140
gt =  0
dv
2
u
2
u
u + v2 tan2a
tan– 1 v tana 0 = – a
= 1
u tana
u
u
u tana
i.e. the time taken in reaching the greatest height say t1 is given by t1 = (au cota)/g.
Again to find the time of falling to the point of projection we have
dv/dt = (g/u2)[u2 – v2 tan2a].
So the time from the greatest height to the point of projection, say t2 is given by
2
u cosa
dv
t2 = u 0
2
g
u  v2 tan2a
cosa
= u cota log
g
1  sina
Hence the total time taken by the particle in returning to the point of projection is
cosa
au cota
t1 + t2 = u cota log
+
g
1  sina
g
1
= ug cota[a + log{cosa/(1 – sina)}].
Ex.2 A particle of mass m is projected vertically under gravity, the resistance of the air
being mk times the velocity; shew that the greatest height attained by the particle is
(V2/g).[ - log(1 + )], where V is the terminal velocity of the particle and V is its initial
vertical velocity.
Sol.
Resistance = mkv
The equation of motion for falling particle is
m(d2x/dt2) = mg – mkv
In case the particle falls under gravity,
(d2x/dt2) = 0.
So
0 = mg – mkv or v = g/k = V (given as terminal velocity)
Thus V = g/k
(1)
When the particle is projected upwards, the equation of motion is
m(d2x/dt2) =  mg – mkv
or
d2x/dt2 = – g – (gv/V) from (1)
or
v(dv/dx) = – (g/V).(V + v)
So
– (g/V).dx = [v/(V + v)].dv = [1 – V/(V + v)].dv
Integrating, (– g /V).x = v – V log(V + v) + A
Initially when x = 0, v = V; A = V log(V + V) – V.
Thus we have (- g/V).x = v – V log(V + v) + V log(V + V) – V.
The particle will be at the greatest height when v = 0, therefore putting v = 0 in
the above equation, we have
(- g/V).x = V + Vlog(1 + )
i.e.
x = (V2/g).[ log(1 + )]
Ex.3 A person falls by means of a parachute from a height of 800 yards in 2½ minutes.
Assuming the resistance to vary as the square of the velocity, shew that in a second and a
half his velocity differs by less than one percent from its value when he reaches the
ground and find an approximate value for the limiting velocity.
141
Sol.
When the parachute has fallen a space x in time t, we have, if
= g/k2,
v2 = k2[1 – exp(– 2gx/k2)]
(1)
v = k tanh(gt/k)
(2)
2
and
x = (k /g)log cosh(gt/k)
(3)
Here 2400(g/k2) = log cosh(150g/k)
So
exp(2400g/k2) = [e150g / k + e– 150g / k]/2
(4)
The second term on the right hand is very small, since k is positive.
Hence (4) is approximate equivalent to exp(2400g/k2) = ½ e150g / k.
So 2400g/k2 = 150g/k – log2 = 150g/k, nearly.
Hence k = 16 is a first approximation.
Putting k = 16(1 + y), (4) gives, for a second approximation,
300(1 – 2y) = ½ [e300(1 – y) + e– 300 (1 – y)] = ½ e300 (1 – y), very approx.
So
e-300 y = ½
Therefore y = (1/300)log2 = 0.693/300 = 0.0023.
Hence a second approx. is k = 16(1 + 0.0023), giving the terminal velocity.
Also the velocity v1, when the particle reaches the ground, is, by (1), given by
v12 = k2[1 – exp(- 2.32 2400/162)] = k2[1 – e6 0 0] = k2, for all practical
purpose.
When v is 99% of the terminal velocity, (2) gives
tanh (gt/k) = 99/100 = 0.99.
2gt/ k
So e
= 199 = e5 . 2, from the tables.
Therefore t = k/(2g) 5.3 = (16/64) 5.3 = 1.235 approx., i.e. t is less than 1½ secs.
Ex.4 A particle of mass m is falling under the influence of gravity through a medium
whose resistance equals times the velocity. If the particle be released from rest, show
that the distance faller through in time t is g(m2/2).[e(t/ m) – 1 +(t/m)].
Sol.
Resistance = v for the mass m.
The equation of motion is
m(d2x/dt2) = mg – v
or
d2x/dt2 = g – (/m).(dx/dt), so v = dx/dt
i.e.
d2x/dt2 + (/m).(dx/dt) = g
This is a linear differential equation in dx/dt and hence its
I.F. = e(/ m)dt = et/ m.
Therefore the solution is
(dx/dt).et/m = g et/ m dt + A = (mg/).et/ m + A
Initially when t = 0, dx/dt = 0. So A = - mg/
so that et/ m.(dx/dt) = (mg/).[et/m – 1]
or
dx = (mg/).[1 – e– t/ m ]dt.
Integrating, x = (mg/).[t + (m/).e– t/ m] + B.
Initially when t = 0, x = 0; so B = - m2g/2.
Hence x = g.(m2/2).[e– t/ m – 1 + t/m].
142
Ex.5 If the resistance vary as the fourth power of the velocity, the energy of m lbs. At a
depth x below the highest point when moving in a vertical line under gravity will be
E tan(mgx/E) when rising and E tanh(mgx/E) when falling, where E is the terminal
energy in the medium.
Sol.
Resistance = v4.
When the particle is falling, equation of motion is
d2x/dt2 = g – v4.
The acceleration is zero if g – v4 = 0
i.e.
v = (g/)1/4 = K the terminal velocity (say). So = g /K4.
If K is the terminal velocity then terminal energy E = ½ mK2.
Now the equation of motion when the particle is rising, is
d2y/dt2 = - g – (g/K4).v4, i.e. vdv/dy = - (g/K4).[K4 + v4]
or
[2v/(K4 + v4)].dv = - (2g/K4)dy.
Integrating, (1/K2) tan– 1(v2/K2) = - (2gy/K4) + A.
Initially when y = 0, let v = u; so A = (1/K2).tan– 1(u2/K2).
Thus 2gy = K2 tan– 1(u2/K2) – K2 tan– 1(v2/K2).
(1)
This gives the velocity of the particle at a height y from the ground.
If h be the greatest height to which the particle rises then putting v = 0 in (1), we
get
2gh = K2 tan– 1(u2/K2).
(2)
The height which is x feet below the heightest point is
= h – x = (K2/2g).tan– 1(u2/K2) – x.
The velocity of the particle at this height is obtained by substituting
(K2/2g). tan– 1(u2/K2) – x for y in (1), whence we get
2
2g[(K /2g).tan– 1(u2/K2) – x] = K2 tan– 1(u2/K2) – K2 tan– 1(v2/K2).
or
2gx = K2 tan– 1(v2/K2), i.e. v2 = K2 tan (2gx/K2).
Putting K2 = 2E/m, the energy at this point is given by
½ mv2 = E tan (mgx/E).
Further when the particle is falling the equation of motion is
v(dv/dx) = (g/K4).(K4 – v4), i.e., [v/(K4 – v4)].dv = (g/K4).dx
or
(1/2K2).[{v/(K2 + v2)} + {v/(K2 – v2)}] dv = (g/K4)dx.
Integrating, (1/4K2).log[(K2 + v2)/(K2 – v2)] = (g/K2).x + B.
Initially when x = 0, v = 0, so B = 0.
Thus log[(K2 + v2)/(K2 – v2)] = 4gx/K2,
i.e.
(K2 + v2)/(K2 – v2) = exp(4gx/K2).
Applying componendo and dividendo,
v2/K2 = [exp(4gx/K2) – 1]/[exp(4gx/K2) + 1]
= [exp(2gx/K2) – exp(– 2gx/K2)]/[exp(2gx/K2) + exp(– 2gx/K2)]
= tanh (2gx/K2), so that v2 = K2 tanh (2gx/K2).
Putting K2 = 2E/m, the energy at the point is
½ mv2 = E tanh (mgx/E).
Ex.6 A heavy particle is projected vertically upwards in a medium the resistance of
which varies as the square of the velocity. It has a kinetic energy K in its upwards path at
a given point; when it passes the same point on the way down, show that its loss of
143
energy is [K2/(K + K)], where K is the limit to which its energy approaches in its
downwards course.
Sol.
Resistance = v2.
For the falling particle, equation of motion is
d2x/dt2 = g – v2.
Acceleration is zero if g – v2 = 0
i.e.
v = (g/) = k the terminal velocity; so = g/k2.
If k be terminal velocity, then terminal energy = ½ mk2 = K (given).
When the particle is going upwards, equation of motion is
v(dv/dx) = - (g/k2).(k2 + v2), since d2x/dt2 = - g – v2
or
[2v/(k2 + v2)]dv = - 2g/k2.
Integrating log(k2 + v2) = - (2g/k2).x + A.
Initially when x = 0, let v = V; so A = log(k2 + V2).
Thus 2gx/k2 = log[(k2 + V2)/(k2 + v2)],
(1)
This gives velocity of the particle at a certain height x. If h be the greatest height
attained by the particle, then by putting v = 0 in (1), we have
2gh/k2 = log[(k2 + V2)/k2]
(2)
When the particle falls, the equation of motion is
v(dv/dy) = (g/k2).(k2 – v2) or – [2v/(k2 – v2)].dv = - (2g/k2)dy.
Integrating log(k2 – v2) = - (2gy/k2) + B.
Initially when y = 0, v = 0; so B = logk2.
Thus 2gy/k2 = log[k2/(k2 – v2)]
(3)
This gives velocity of particle after falling a distance y. If V2 be the velocity of
the falling particle when distant x1, from the ground i.e. (h – x1) from the highest point,
therefore putting y = h – x1 in (3), we get
(2g/k2).(h – x1) = log[k2/(k2 – V22)]
(4)
If V1 be the velocity of the rising particle at the same height x 1 then from (1), we
get
(2g/k2).x1 = log[(k2 + V2)/(k2 + V12)]
(5)
Adding (4) and (5), we get
2gh/k2 = log[{k2(k2 + V2)}/{(k2 + V12).(k2 – V22)}]
(6)
Comparing (2) and (6), we have
log[(k2 + V2)/k2] = log[{k2(k2 + V2)}/{(k2 + V12).(k2 – V22)}]
i.e.
(k2 + V2)/k2 = k2(k2 + V2)/[(k2 + V12).(k2 – V22)]
or
- k2V22 – V12V22 + k2V12 = 0, i.e. V22 = k2V12/(k2 + V12)
Kinetic energy at P when passing down
= ½ mV22 = ½ (mk2.V12)/(k2 + V12) = (½ mk2. ½ mV12)/( ½ mk2 + ½ mV12)
= (K.K)/(K + K)
and kinetic energy at P when going up
= ½ mV12 = K.
Loss of energy = V – VV/(V + V) = V2/(V + V)
Ex.7 A particle is projected in a resisting medium whose resistance varies as (velocity)n
and it comes to rest after describing a distance s in time t. Find the values of s and t and
144
show that s is finite if n < 2, but infinite if n = or > 2, whilst t is finite if n < 1, but infinite
if n = or > 1.
Sol.
Resistance = vn
The equation of motion is
d2x/dt2 = v(dv/dx) = - vn or dx = - dv/vn – 1
so that x = - (1/vn – 1)dv = (1/n – 2).(1/vn – 2) + A
Initially when x = 0, let v = V; so A = – (1/n – 2).(1/Vn – 2)
Thus x = (1/n – 2).[(1/vn – 2) – (1/Vn – 2)]
If n > 2 and v = 0, x = .
If n = 2, then x = - (1/v) dv = log(V/v)
Constant being logV as v = V, when x = 0, so that x = when v = 0.
If n < 2 let n = 2 – , then
x = - (1/v1 – )dv = (1/).(V – v) as v = V when x = 0.
Hence s = [x] (at v = 0) = (1/)V = a finite quantity.
Again dv/dt = - vn, so that t = - dv/vn
= (1/n – 1).[(1/vn – 1) – (1/Vn – 1)] as v = V when x = 0.
Let n < 1 and = 1 – p, then
t = (1/p).(Vp – vp) as v = V when t = 0 so that when v = 0,
t = (1/p)Vp = (1/1 – n)V1 – n = a finite quantity.
Let n > 1 and = 1 + p, then t = (1/p).[(1/vp) – (1/Vp)] (as v = 0 when t = 0).
Hence when v = 0, t = .
Let n = 1, then t = - dv/v = - logv + logV (as v = V when x = 0)
= log(V/v).
Hence when v = 0, t = .
Conclusively
s is finite if n < 2 but infinite if n = 0or > 2
and
t is finite if n < 1 but infinite if n = 0 or > 1
Ex.8 In the previous question if the resistance be k (velocity) and the initial velocity be
V, show that v = Vek t and s = (V/k).(1 – ek t).
Sol.
Resistance = kv.
The equation of motion is d2x/dt2 = dv/dt =  kv.
So
kt =  dv/v =  logv + A.
Initially when t = 0, v = V; so A = logV,
So that
kt = log(V/v); so v = Ve  kt.
Further,
v = ds/dt = Ve kt
Therefore
s = Vektdt =  (V/k).ekt + B.
Initially when s = 0; so B = V/k.
Hence
s = (V/k).[1 – ekt].
Ex.9 A particle falls from rest at a distance a from the centre of the Earth towards the
Earth, the motion meeting with a small resistance proportional to the square of the
velocity v and the retardation being for unit velocity; show that the kinetic energy at
145
distance x from the centre is mgr2[(1/x) – (1/a) + 2(1 – x/a) – log(a/x)], the square of
being neglected and r being radius of the Earth.
Sol.
When the particle moves towards the centre of the earth and is at a distance x
from the centre of the earth; attraction of the earth = /x2 towards the centre and
resistance = v2 away from the centre (opposite to the motion).
Therefore the equation of motion of the particle is
v(dv/dx) = v2 – (/x2) = v2 – (gr2/x2).
So on earth‟s surface g = (/r2) or = gr2
and acceleration at a distance x = (/x2) = gr2/x2
i.e.
dv2/dx – 2v2 = - 2gr2/x2.
Being linear differential equation in v2, its I.F. = e– 2dx = e – 2x and therefore its
solution is
v2.e– 2x = - 2gr2 (1/x2).e – 2x dx + A
or
v2(1 – 2x) = - 2gr2 [(1 – 2x)/x2]dx + A (expanding exponential terms
and neglecting squares of )
2
= - 2gr [ - (1/x) – 2log x] + A
= 2gr2[(1/x) + logx2] + A.
Initially when x = a, v = 0; so A = - 2gr2.[(1/a) + log a2],
so that v2(1 – 2x) = [(1/x) – (1/a) + log (x2/a2)] 2gr2
i.e.
v2 = 2gr2(1 – 2x)-1[(1/x) – (1/a) + log (x2/a2)]
= 2gr2 (1 + 2x)[(1/x) – (1/a) + 2log (x/a)] (expending binomially and
neglecting 2 etc.)
2
= 2gr [(1/x) – (1/a) + 2(1 – x/a) + 2log(x/a)] (neglecting 2 etc.)
So K.E. = ½ mv2 = mgr2 [(1/x) – (1/a) + 2(1 – x/a) – 2log(a/x)]
Ex.10 An attracting force, varying as the distance, acts on a particle initially at rest at a
distance a. Show that, if V be the velocity when the particle is at a distance x and V the
velocity of the same particle when the resistance of the air is taken into account, then
V = V[1 – (1/3).k.{(2a + x).(a – x)/(a + x)}]
nearly, the resistance of the air being given to be k times the square of the velocity per
unit of mass, when k is very small.
Sol.
The equation of motion is
v(dv/dx) =  x + kv2 or dv2/dx – 2kv2 =  2x.
Being linear differential equation in v2, its I.F. = e2kdx = e2 k x and therefore its
solution is
x
2k x
(v2/).e2k x =  2xe
dx
a
x 2k x
x
=  2[{(xe2kx)/( 2k)} + (1/2k) e
dx]
a
a
or
= (xe2kx)/k – (ae2a k)/k + (1/2k2)[e2k x – e-- 2ak]
= [(1 + 2kx)/2k2].e2k x – [(1 + 2ak)/2k2].e 2ak
v2/ = [(1 + 2kx)/2k2] – [(1 + 2ak)/2k2].e2k(x – a)
146
Since at a distance x, v = V (taking resistance of air into account)
V2
1 + 2kx
1 + 2ka [1 + 2k(x –a) + 4k2(x – a)2 8k3(x – a)3
+ .]
+

2k2
2k2
2!
3!
1 + 2kx
1 + 2ak
(1 + 2ka).(x – a)
2k2
2k2
k
2
 (1 + 2ka)(x – a) – (2k/3)(1 + 2ka)(x – a)3 + ……
= (1/2k2) + x/k – (2/2k2) – a/k – (x – a)/k – 2a(x – k) – (x – a)2
 2ak (x – a)2 – (2k/3)(x – a)3, neglecting k2 etc.
=  2ax + 2a2 – x2 + 2ax – a2 – (2k/3)(x – a)2(3a + x – a)
= a2 – x2 – k(x – a)2. (2/3).(x + 2a).
(1)
But as the velocity is V when there is no resistance of air, i.e. k = 0.
Therefore
V2/ = a2 – x2
(2)
x
2
2
2
2
[or thus v(dv/dx) =  x; so v =  2 xdx
giving K / = a – x ].
a
Dividing (1) by (2), we get
K2/K2 = 1 – (2k/3).[(a – x)(2a + x)/(a + x)]
K = K[1 – (2k/3).{(a – x)(2a + x)/(a + x)}]1/2
= K[1 – ½ (2k/3).{(a – x)(2a + x)/(a + x)} + …… ]
= K[1 – (k/3).{(a – x)(2a + x)/(a + x)}], neglecting k2 etc.
So
or
Particle Projected under Gravity:
Theorem: A particle is projected under gravity and a resistance equal to mk (velocity)
with a velocity u at an angle to the horizon. Find the motion.
Proof:
Let the axes of x and y be respectively horizontal and vertical and the origin at the
point of projection. Then the equations of motion are
d2x/dt2 =  k(ds/dt).(dx/ds) =  k(dx/dt),
and
d2y/dt2 =  k(ds/dt).(dy/ds) – g =  k(dy/dt) – g.
Integrating, we have
log (dx/dt) =  kt + const. =  kt + log (u cos),
and
log {k(dy/dt) + g} =  kt + const. =  kt + log(ku sin + g);
So dx/dt = u cos ekt
(1)
and
k (dy/dt) + g = (ku sin + g) ekt
(2)
kt
kt
Therefore
x =  (u cos/k)e + const. = (u cos/k)(1 – e )
(3)
kt
and
ky = + gt =  [(ku sin + g)/k].e + const.
= [(ku sin + g)/k](1 – ekt)
(4)
Eliminating t, we have
y = (g/k2).log[1 – kx/(u cos)] + [x/(u cos)](u sin + g/k)
(5)
which is the equation to the path.
The greatest height is attained when dy/dt = 0, i.e. when
ekt = g/[ku sin + g], i.e. at time (1/k) log(1 + ku sin/g),
and then
y = u sin/k – (g/k2)log(1 + ku sin/g).
147
It is clear from equation (3) and (4) that when t = , x = (u cos/k and y =  .
Hence the path has a vertical asymptote at a horizontal distance u cos/k from the point
of projection. Also, then, dx/dt = 0 and dy/dt =  g/k i.e. the particle will then have just
attained the limiting velocity.
Cor: If the right-hand side of (5) be expanded in powers of k, it becomes
kx
k2x2
k3x3
-…]
y = (g/k2) [ –
u cos
2u2 cos2
3u3 cos3
+ (x/ucos)(u sin + g/k),
gx2
gkx3
gk2x4
….
y = x tan
2u2 cos2
3u3 cos3
4u4 cos4
On putting k = 0, we have the ordinary equation to the trajectory for unresisted
motion.
Particle Moving under Gravity:
Theorem: A particle is moving under gravity in a medium whose
resistance = m(velocity)2. Find the motion.
Proof:
When the particle has described a distance s, let its tangent make an angle with
the upward drawn vertical and let v be its velocity.
The equation of motions are then
v(dv/ds) =  g cos – v2
(1)
2
and
v / = g sin
(2)
2
(1) gives
dv d
= - 2g cos – v2
d ds
i.e. from (2), 1 d(sin)
= - 2 cos – 2sin
d
So
(1/)(d/d).sin + 3 cos =  2sin,
Therefore
d (1 ) 1
3 cos 1
2
3
4
d  sin 
sin   sin3
Hence 1/(sin3) = 21/sin3d
=  (cos/sin2) – log[(1 + cos)/sin] + A
(3)
(2) then gives
v2[A – (cos/sin2) – log {(1 + cos)/sin}] = g/sin2
Equation (3) gives the intrinsic equation of the path, but cannot integrated further.
Bead Moves on a Smooth Wire:
Theorem: A bead moves on a smooth wire in a vertical plane under a
resistance = k (velocity)2. Find the motion.
Proof:
When the bead has described an arcual distance s, let the velocity be v at an angle
to the horizon and let the reaction of the wire be R.
The equations of motion are
148
v(dv/ds) = g sin – kv2
and
v2/ = g cos – R
Let the curve be s = f().
Then (1) gives
d ( v2) = f ()[g sin – kv2]
d2
i.e.
d(v2)
+ 2kf ().v2 = 2g sin.f ()
d
a linear equation to give v2.
(1)
(2)
Cor: Let the curve be a circle so that s = a, if s and be measured from the highest
point.
(1) then gives
d(v2)
+ 2akv2 = 2ag sin
d
So
v2e2ak = 2ag sin.e2akd = [2ag/(1 + 4a2k2)].e2ak(2ak sin – cos) + C.
Therefore v2 = [2ag/(1 + 4a2k2)].(2ak sin – cos) + Ce-2ak.
Ex.1 A particle, of mass m, is projected in a medium whose resistance is mk (velocity)
and is acted on by a force to a fixed point (= m..distance). Find the equation to the path,
in the case when 2k2 = 9, show that it is a parabola and that the particle would
ultimately come to rest at the origin, but that the time taken would be infinite.
Sol.
The equations of motion are
d2x = - x - k dx
dt2
dt
(1)
and
d2y = - y - k dy
dt2
dt
(2)
2
(1) may be written as (D + kD + )x = 0, where D = d/dt, which gives
2
D = k (k – 4)
2
Hence putting
2
p = k + (k – 4)
2
and
2
q = - k - (k – 4)
2
the solution of (1) is x = Aept + Beqt
Similarly the solution of (2) is
y = Cept + Deqt,
The initial conditions determine the constants A, B, C and D and we thus have the
path.
Now if 2k2 = 9, then p =  (½) and q = 2(½); hence by putting (½) =
p1, we have
x = Aep1t + Be2 p1t and y = Cep1t + De2p1t
149
Consider
(Dx – By)2 = (ADep1t + BDe2p1t – BCep1t – BDe2p1t)2
= (AD – BC)2e 2p1t
But also as
Ay – Cx = ACep1t + ADep1t – ACep1t – BCe2p1t
= (AD – BC)e2p1t
Hence (Dx – By)2 = (AD – BC).(AD – BC)e2p1t
= (AD – BC)(Ay – Cx),
which represents a parabola.
Again the particle will be at the rest at the origin at time t given by
0 = ep1t[A + Bep1t] and 0 = ep1t[A + 2Bep1t]
(on putting x = 0 and dx/dt = 0)
0 = ep1t[C + Dep1t] and 0 = ep1t[C + 2Dep1t]
(on putting y = 0 and dy/dt = 0)
All the four equations are satisfied by
ep1t = 0 or ep1t =, i.e. t = 
Hence the time taken would be infinite.
Ex.2 A particle of unit mass is projected with velocity u at an inclination  above the
horizon in a medium whose resistance is k times the velocity. Show that its direction will
again make an angle  with the horizon after a time (1/k) log [1 + (2ku/g).sin].
Sol.
Resistance = kv = k(ds/dt)
The equation of motions are
d2x = - k ds cos = - k dx
dt2
dt
dt
(1)
and
d2y = - g - k ds sin = - g - k dy
dt2
dt
dt
(2)
Integrating (1), log(dx/dt) =  kt + A.
Initially when t = 0, dx/dt = u cos; so A = log(u cos), so that
log(dx/dt) =  kt + log(u cos), i.e. (dx/dt) = u cos.ek t
(3)
Integrating (2),
log[k(dy/dt) + g] = kt + log(g + ku sin),
so when t = 0, dy/dt = u sin
or
k(dy/dt) + g = (g + ku sin)ekt
or
dy/dt =  (g/k) + (1/k)(g + ku sin)ekt
(4)
Dividing (4) by (3)
dy = g/k + (1/k)(g + kusin)ekt= gekt + g + ku sin
dx
u cos.e-kt
ku cos
ku cos
If the particle again makes an angle with the horizon after time t, then
dy
gekt
+ g + ku sin
= tan( ) =  tan =
dx
ku cos
ku cos
or
gekt
sin
g
+
ku
sin
=
+
ku cos
cos
ku cos
kt
or
e = (2ku sin)/g + 1
150
So
t = (1/k) log[1 + (2ku/g) sin]
Ex3. If a high throw is made with a diabolo spool the vertical resistance may be
neglected, but the spin and the vertical motion together account for a horizontal drifting
force which may be taken as proportional to the vertical velocity. Show that if the spool
is thrown so as to rise to the height h and return to the point of projection, the spool is at
its greatest distance c from the vertical through that point when it is at a height 2h/3 and
show that the equation to the trajectory is of the form 4h3x2 = 27c2y2(h – y).
Sol.
Neglecting the vertical resistance, equation of motion is
d2y/dt2 =  g.
Integrating w.r.t. t, we get
dy/dt =  gt + A
Initially when t = 0, dy/dt = v (say); so A = v.
Thus dy/dt = v – gt or dy = (v – gt)dt
Integrating again y = vt – ½ gt2 + B
Initially when y = 0, t = 0; so B = 0
so that y = vt – ½ gt2
(1)
Also when the particle reaches the highest height h,
0 = v2 – 2gh, i.e. v2 = 2gh
(2)
Now the horizontal drifting is given by
d2x/dt2 =  (dy/dt)
Integrating, dx/dt =  y + C
Initially when y = 0, dx/dt = u (say); so C = u,
so that dx/dt = – vt + ½ gt2 + u, from (1)
Integrating it again,
x = ut – v(t2/2) + (g/6)t3 + D
Initially when t = 0, x = 0; so D = 0
Thus x = ut – ½ vt2 + (1/6)gt3
(3)
Now the spool goes to height h and again comes to the position y = 0 in time
given by
0 = vt – ½ gt2
[on putting y = 0 in (1)]
whence t = (2v/g) and then x is given to be zero, so that from (3)
0 = u(2v/g) – (v/2)(2v/g)2 + (g/6)(2v/g)3
i.e.
u = (v2/g) – (v2/3g) = (1/3)(v2/g)
Substituting this value of u in (3), we get
x = (g/6)[t3 – (3vt2/g) + (2v2/g2).t] = (gt/6)[t2 – 3vt/g + (2v2/g2)]
= (gt/6)(v/g – t)(2v/g – t) = (/3)(v/g – t)(vt – ½ gt2)
= (/3)[(2/g)(h – y)]y, {as h – y = (v2/2g) – vt + ½ gt2 = (g/2)(v/g – t)2
So v/g – t = [(2/g)(h – y)]}
Squaring both sides,
9gx2 = 22y2(h – y)
Now x = c when y = 2h/3
Therefore 9gc2 = 22(4h2/9)(h/3) = 82h3/27 i.e. 2 = (279gc2)/8h3
Hence 9gx2 = 2 (279gc2)y2(h – y)/(8h3)
151
or
4h3x2 = 27c2y2(h – y)
Ex.4 If the resistance vary as the velocity and the range on the horizontal plane through
the point of projection is a maximum, show that the angle  which the direction of
projection makes with the vertical is given by (1 + cos)/(cos + ) = log[1 + sec],
where is the ratio of the velocity of projection to the terminal velocity.
Sol.
Resistance varies the velocity. Let be the angle which the direction of projection
makes with the horizontal, so that = /2  .
The equations of motion are
d2x = - kv cos = -k ds dx = - k dx
dt2
dt ds
dt
(1)
2
and
d y = - kv sin - g = -k ds dy - g = - k dy - g
dt2
dt ds
dt
(2)
.. .
(1) may be written as x/x =  k
Integrating, log(dx/dt) =  kt + A
Initially when t = 0, dx/dt = u cos; A = log u cos,
so that log(dx/dt) =  kt + log u cos, i.e. x. = u cos.ekt
(3)
..
.
and (2) may be written as ky/(g + ky) = - k
.
Integrating, log(ky + g) =  kt + B
.
Initially when t = 0, y = u sin; so B = log(g + ku sin),
.
so that log(ky + g) =  kt + log(g + ku sin),
i.e.
k(dy/dt) + g = (g + ku sin)ekt
(4)
Again integrating (3), we have
x =  (u cos.ekt)/k + C
Initially when t = 0, x = 0; so C = (1/k) u cos
Therefore x = (u cos/k)(1 – e-kt)
(5)
Similarly integrating (4), we get
y = [(g + ku sin)/k2](1 – ekt) – gt/k
(6)
When the particle strikes the horizontal plane, y = 0 and therefore the time of
flight is obtained by putting y = 0 in (6),
i.e.
kgt = (g + ku sin)(1 – ekt)
(7)
Differentiating (7) w.r.t., we get
kg(dt/d) = ku cos(1 – ekt) + k(g + ku sin)ekt(dt/d)
or
dt/d = u cos(1 – ekt)/[g(1 – ekt) – ku sin ekt]
(8)
From (5) the range x = (u cos/k)(1 – ekt), where t is the time of flight given by
(7).
So
dx/d =  (u sin/k)(1 – ekt) + u cos ekt(dt/d)
For maximum x,
dx/d = 0, i.e. dt/d = [u sin(1 – ekt)]/[ku cos ekt]
(9)
Equating the two values of dt/d from (8) and (9), we have
cos
sin
=
kt
kt
g – ge – ku sin e
ku cos ekt
or
ku cos2 = g sin ekt – g sin – ku sin2
152
ekt = [ku + g sin]/(g sin), i.e. kt = log[(ku + g sin)/(g sin)]
(10)
Eliminating t between (7) and (10), we get
g log[(ku + g sin)/(g sin)] = (g + ku sin)[1 – (g sin)/(ku + g sin)]
= (g + ku sin)[ku/(ku + g sin)]
(11)
In order that acceleration is zero at any time, dx/dt = 0, y = 0, i.e. dx/dt = 0 and
d2y/dt2 = 0/k, i.e., the terminal velocity = g/k.
Now the given ratio = u/(terminal velocity) = , i.e., ku = g.
Putting this value of ku in (11), we have
log[(+ sin)/sin] = (1 + sin)/(+ sin)
Putting = /2 - , log[(+ cos)/cos] = (1 + cos)/(+ cos)
or
log(1 + sec) = (1 + cos)/(+ cos)
or
Ex.5 If a particle be moving in a medium whose resistance varies as the velocity of the
particle, show that the equation of the trajectory can, by a proper choice of axes, be put
into the form y + ax = b logx.
Sol.
Resistance = kv = k(ds/dt)
Assuming that the tangent to the curve at any point of the trajectory makes an
angle with the horizon, the equations of motion of the particle in horizontal and
vertical direction are
d2x = - k(ds/dt) cos = -k ds dx = - k dx
dt2
dt ds
dt
(1)
2
and
d y = - k(ds/dt) sin - g = -k ds dy - g = - k dy - g
dt2
dt ds
dt
(2)
Integrating and applying the initial conditions that when t = 0, dx/dt = u cos,
dy/dt = u sin, we get
dx/dt = u cos ekt,
(3)
kt
and
k(dy/dt) + g = (ku sin + g)e
(4)
Integrating again and applying the initial conditions that when t = 0, x = 0, we get
x = [(u cos)/k](1 – ekt)
(5)
kt
and
ky + gt = [(ku sin + g)/k](1 – e )
(6)
where u is the velocity and the angle of projection.
Eliminating t between (5) and (6), we have
y = (g/k2).log[1 – (kx)/(u cos)] + [x/(u cos)].[u sin + g/k]
g
k ( u cos - x)} + x
(ku sin + g)
= 2 log {
k
u cos k
ku cos
g
k
g
u cos - x) + x
(ku sin + g)
= 2 log
+ log (
k
u cos k2
k
ku cos
g
k
g
u cos - x) – u cos ku sin + g
= 2 log
+ 2 log (
k
u cos k
k
k
ku cos
u
cos
ku
sin
+
g
- x)
+(
k
ku cos
or
k )]
[ y - g2( ku sin + g + log
k
g
u cos
+ ( u cos - x) ku sin + g = g log ( u cos - x)
153
k
ku cos
k2
k
(7)
Putting
k
[ y - g2( ku sin + g + log
)] = Y
k
g
u cos
and
( u cos - x) = X
k
the equation (7) becomes
Y + [(ku sin + g)/(ku cos)].X = (g/k2).logX
which is of the form Y + aX = b logX,
where
a = [ku sin + g]/(ku cos) and b = g/k2
Generalising, we get y + ax = b logx
Ex.6 A particle moves in a resisting medium with a given central acceleration P; the path
of the particle being given, show that the resistance is
- 1 2 d ( p3 dr P )
2p ds
dp
Sol.
Let Q be the resistance.
Also let be the angle which the radius vector makes with the tangent, so that p =
r sin
Acceleration along the tangent is given by
v(dv/ds) =  Q – P cos =  P(dr/ds) – Q
(1)
2
and normal acceleration is v / = P sin = P.(p/r)
or
v2 = P.(p/r). = P.(p/r).r(dr/dp) = P.p(dr/dp)
(2)
From (1), we have
Q =  P(dr/ds) – v(dv/ds) =  P(dr/ds) – (1/2).(dv2/ds),
or
 2Q = 2P(dr/ds) + dv2/ds
i.e.
 2Qp2 = 2Pp2(dr/ds) + p2(dv2/ds)
= 2Pp2(dr/ds) + p2 d ( Pp dr )
[from (2)]
ds
dp
= 2p(dp/ds)[Pp(dr/dp)] + p2 d ( Pp dr )
since dr/ds = (dr/dp).(dp/ds)
ds
dp
2
d
p
Pp
dr )
(
=
ds
dp
3 dr
so
1
d
(p
Q= - 2
P)
2p ds
dp
Ex.7 If a point travel on an equiangular spiral towards the pole with uniform angular
velocity about the pole, show that the projection of the point on a straight line represents
a resisted simple vibration.
Sol.
154
Taking the given straight line as initial line, let the equation of the equiangular
spiral be r = ae and the angular velocity d/dt = - (- ve sign being taken as the point
moves in the direction of decreasing).
Let be the projection of r, so that
=
= ae.cos
. r cos

So
= ae .cos.(d/dt) – ae.sin.(d/dt)

– sin],
since d/dt = 
..= ae [cos

and
= ae (d/dt)[cos – sin] + ae [- sin – cos].(d/dt)
= a2e[( 2 - 1)cos – 2 sin]
since d/dt = 
= – (1 – 2)2 + 2.ae .[cos – sin] – 22a2e cos
= – (1 – 2)2 + 2 – 
= – (1 + 2) 2 + 2
which is of the form
d2/dt2 + (d/dt) + n2 = 0
and therefore it represents a resisted simple harmonic motion.
Ex.8 A particle acted on by gravity is projected in a medium of which the resistance
varies as the velocity. Show that tits acceleration retains a fixed direction and diminishes
without limit to zero.
Sol.
Resistance = kv.
The equations of motion are
..
x = - k(dx/dt)
(1)
..
and
y = - k(dy/dt) – g
(2)
.
Integrating (1) and (2) and applying the initial conditions that when t = 0, x = u
.
cos and y = u sin, we get
dx/dt = u cos.ekt
(3)
kt
and
k(dy/dt) + g = (ku sin + g)e
(4)
Direction of acceleration
..
k(dy/dt) + g
= ..y
=
x
k(dx/dt)
[from (1) and (2)]
kt
= (ku sin + g)e
= ku sin + g
ku cos ekt
ku cos
[from (3) and (4)]
which is independent of t; hence the acceleration relations a fixed condition.
Now substituting from (3) and (4) to (1) and (2), we have
Total acceleration = [(d2x/dt2)2 + (d2y/dt2)2]1/2
= [{ k (dx/dt)}2 + {k (dy/dt) + g}2]1/2
[from (1) and (2)]
kt 2
kt 2
= [(ku cos e ) + {(ku sin + g)e } ] [from (3) and (4)]
= ekt(k2u2 + 2kug sin + g2),
which decreases as t increases and is ultimately zero.
Ex.9 A heavy bead, of mass m, slides on a smooth wire in the shape of a cycloid, whose
axis is vertical and vertex upwards, in a medium whose resistance is m(v2/2c) and the
distance of the starting point from the vertex is c; show that the time of descent to the
cusp is [8a(4a – c)/gc], where 2a is the length of the axis of the cycloid.
155
Sol.
Resistance = (mv2)/(2c)
The intrinsic equation of the cycloid is s = 4a sin
(1)
Now the equation of motion is
v(dv/ds) = g sin – (v2)/(2c) = g.(s/4a) – (v2)/(2c) [from (1)]
or
2v(dv/ds) = (gs)/(2a) – v2/c
or
dv2 = g.s - v2
ds
2a c
or
dv2 + v2
gs
ds
c
2a
Being linear differential equation in v2, its I.F. = e(1/c)ds = es/c and therefore its
solution is
v2. es/c = [g/(2a)] s. es/c ds + A
= [g/(2a)][s.c es/c – c2 es/c ] + A
Initially when s = c, v = 0; so A = 0
Hence v2. es/c = [g/(2a)].c(s – c) es/c, i.e.
v2 = (ds/dt)2 = [(gc)/(2a)](s – c)
or
ds/dt = [(gc)/(2a)](s – c)
4a
So
t = [(2a)/(gc)] [1/(s
– c)]ds = [(2a)/(gc)][2(s – c)] 4a
c
c
= [(2a)/(gc)].2(4a – c) = [8a(4a – c)/(gc)]
Ex.10 A particle, moving in a resisting medium, is acted on by a central force (/rn); if
the path be equiangular spiral of angle , whose pole is at the centre of force, show that
the resistance is [(n – 3)/2].(cos/rn).
Sol.
Let P be the resistance per unit mass; then the equations of motion are
v(dv/ds) = – (/rn)cos – P
(1)
2
n
and
v / = (/r )sin
(2)
cot 
We know that in an equiangular spiral r = ae
, we always have
= (constant). So p = r sin = r sin
and therefore = r(dr/dp) = r cosecHence from (2),
v2 = (/rn)sin = /rn – 1
since = r cosec
Differentiating it w.r.t. s, we get
2v(dv/ds) = – (n – 1) (/rn)(dr/ds)
v(dv/ds) = – [(n – 1)/2].(/rn)cos since dr/ds = cos = cos
Substituting this value of v(dv/ds) in (1), we have
P = – (/rn) cos + ½ [(n – 1)/rn].cos= [(n – 3)/2].[cos/rn]
2.5 Motion of Particles of Varying Mass:
The equation P = mf is only true when the mass m is constant. Newton‟s second
law in its more fundamental form is
P = d(mv)
dt
(1)
156
Suppose that a particle gains in time t an increment m of mass and that this
increment m was moving with a velocity u.
Then in time t the increment in the momentum of the particle
= m.v + m(v + v – u)
and the impulse of the force in this time is Pt.
Equating these we have, on proceeding to the limit.
m(dv/dt) + v(dm/dt) – u(dm/dt) = P
i.e.
d (mv) = P + u dm
dt
dt
(2)
When u = 0,we have the result (1)
Ex.1 A falling raindrop has its uniformly increased by access of moisture. If it has given
to it a horizontal velocity, show that it will then describe a hyperbola, one of whose
asymptotes is vertical.
Sol.
Initially taking the radius of the raindrop as unity, its mass
= (4/3).l2 = 4/3 = m (say).
(1)
Let the increase of volume at an instant t, when its radius is r, be times its
surface at that instant, so that (dM/dt) = 4r2., where M is the mass at time t, i.e. M =
(4/3)r3.
So
dM
d{(4/3)r3.)
=
dt
dt
2
= 4r . where dr/dt = .
Integrating, r = t + A.
Initially when t = 0, r = 1; so A = 1 and thus r = 1 + t.
(2)
3
3
As such
M = (4/3)r = m(1 + t) .
(3)
Since there is no horizontal force,
d {m (1 + t)3 dx }= 0
dt
dt
Integrating, (1 + t)3(dx/dt) = B.
Initially when t = 0, the horizontal velocity
dx/dt = 2a (say); so B = 2a,
so that (1 + t)3.(dx/dt) = 2a or dx = [2a/(1 + t)3]dt
Integrating again, x = - a/(1 + t)2 + C
Initially when t = 0, x = 0; so C = a.
Hence x = a[1 – 1/(1 + t)2]
Again considering the vertical motion, we have
d {m (1 + t)3 dy
}= m (1 + t)3.g,
dt
dt
Integrating,
(1 + t)3(dy/dt) = (g/4).(1 + t)4 + D
Initially when t = 0, dy/dt = 0; so D = - g/4
Thus (1 + t)3(dy/dt) = (g/4)[(1 + t)4 – 1]
(4)
157
dy/dt = (g/4)[(1 + t) – 1/(1 + t)3]
Integrating again
2
1 ]+E
y = g [ (1 + t) +
4
2
2 (1 + t)2
Initially when t = 0, y = 0; so E = - g/(42)
Hence y = [g/(82)].[(1 + t)2 + 1/(1 + t)2 – 2]
= [g/(82)].[1 + t – 1/(1 + t)]2
(5)
In order to find the equation of the trajectory we have to eliminate t between (4)
and (5) and for this let us put (1 + t)2 = a/(a – x) from (4) to (5), whence we get
x2
y = g 2[ a + a – x - 2 ] = g 2
8 a – x
a
8 a (a – x)
2
2 2
2
2
or
8
ay
8
a
y
8
axy
x =
(a – x) =
g
g
g
or
x2 + (82a/g).xy = (82a2/g).y,
which represents a hyperbola as
(42a/g) > 
Note:
Condition for ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 to represent a hyperbola is
2
h > ab.
Its asymptote parallel to y-axis is
(82a/g).x – (82a2/g) = 0, i.e. x = a,
which is obviously vertical.
or
Ex.2 A chain, of great length a, is suspended from the top of a tower so that its lower end
touches the earth; if it be then let fall, show that the square of its velocity, when its upper
end has fallen a distance x, is 2grlog[(a + r)/(a + r –x)], where r is the radius of the earth.
Sol.
Let m be the mass per unit length of the chain. Also let at an instant of time t, the
chain have fallen a distance x, so that the mass of the chain deposited per unit of time
d(mx) = m.dx
=
dt
dt
Now d [m (a – x) dx ]
dt
dt
= change in the momentum per unit of time
= the difference of the momentum of the deposited mass and the mass of
the vertical portion of the chain.
dx
 E.m.dy
= - (m. ) dx +  a – x
dt
dt
0
(r + a – x – y)2
where is constant of gravitation, E the mass of the earth and y an element of the
vertical portion of the chain at a height y from the surface.
or
m (a – x)(d2x/dt2) – m (dx/dt)2 = – m(dx/dt)2
a–x
dy
.
+ gr2m 
0
(r + a – x –y)2
[since g = (E.1)/r2]
2
or
1
(a – x) d 2x = gr2 [
]a – x
dt
r+a–x–y 0
158
or
= gr2[1/r – 1/(r + a – x)] = [gr/(r + a – x)].[r + a – x –r]
= gr(a – x)/(r + a – x)
= d2x/dt2 = gr/(r + a – x).
Multiplying both sides by 2 (dx/dt)dt and integrating.
(dx/dt)2 = – 2gr log(r + a – x) + A
Initially when x = 0, dx/dt = 0; so A = 2gr log(r + a)
Hence (dx/dt)2 = 2gr log[(a + r)/(a + r – x)]
Ex.3 A spherical raindrop of radius a cms. Falls from rest through a vertical height h,
receiving throughout the motion an accumulation of condensed vapour at the rate of k
grammes per square cm. per second, no vertical force but gravity acting; show that when
it reaches the ground its radius will be
k(2h/g)[1 + {1 + (ga2)/(2hk2)}]
Sol.
Let the radius and the mass of the raindrop be respectively r and M when it has
fallen through a distance x in time t. Then using the relation P = d(mv) , we have
d ( M dx ) = Mg
dt
dt
dt
(1)
3
Now M = (4/3).r ., where is the density/c.c.
and
dM/dt = 4r2.k as the vapour accumulates at the rate of k gm./sp.cm./sec.
Hence from (1),
d [(4/3)r3. dx ] = .(4/3)r3g
dt
dt
or
d [r3 dx ] = r3g
dt dt
(2)
Again dM = d .(4/3)r3] = .4r2k
dt
dt
or
r2(dr/dt) = r2k or dr/dt = k
(3)
Integration of (3) gives r = kt + A whilst initially when t = 0, r = a; so A = a, so
that
r = kt + a
(4)
With the help of (3), (2) may be written as
d (r3 dx ) = g r3 dr
dt
dt
k dt
[since dr/dt = k from (3)].
Integrating,
r3(dx/dt) = (gr4)/(4k) + B,
while initially when r = a, dx/dt = 0; so B = – (ga4)/(4k),
so that
r3(dx/dt) = g(r4 – a4)/(4k)
or
dx = g (r – a4 ) dr
dt
4k2
r3 dt
from (3)
Integrating again,
x = [g/(4k2)].[r2/2 + a4/(2r2)] + C
Initially when r = a, x = 0; so C = - (ga2)/(4k2)
Thus,
g (r2 + a4 - 2a2) = g ( r2 – a2 )2
x = 8k2
r2
8k2
r
159
or
or
When x = h, we have [(r2 – a2)/r]2 = 8k2h/g
(r2 – a2)/r = 2k(2h/g)
r2 – 2k(2h/g).r – a2 = 0, which gives
2
2
r = 2k(2h/g) [(8k h)/g + 4a ]
2
= k(2h/g)[1 + {1 + (ga2)/(2hk2)}] taking + ve sign.
Ex.4 A particle of mass M is at rest and begins to move under the action of a constant
force F in a fixed direction. It encounters the resistance of a stream of fine dust moving in
the opposite direction with velocity V, which deposits matter on it at a constant rate .
Show that its mass will be m when it has traveled a distance
(k/2)[m – M{1 + log(m/M}]
where k = F – V.
Sol.
When the mass of the particle is m, let us suppose that it gains in time t an
increment m of mass and this increment m was moving with a velocity V in the
direction opposite to that of the mass. Hence
mv + m[v + v – (– V)] = increases in the momentum in time t, i.e.
mv + m(v + v + V) = Ft
Dividing throughout by t and proceeding to the limit t 0, we get
m(dv/dt) + (dm/dt).(v + V) = F,
So
Lim v. m = 0
t 0 t
where m = M + t, since the deposition of matter in time t is t. So
(M + .t)dv + (v + V) d (M + .t) = F
dt
dt
(M + .t)dv + v d (M + .t) + V d (M + .t) = F
dt
dt
dt
or
d {(M + .t)v} = F – V d (M + .t)
dt
dt
= F – V[0 + ], since dM/dt = 0, M being a constant quantity
= F – V = k as given, F – V = k.
Integrating with regard to t,
(M + t)v = k dt = kt + A.
Initially when t = 0, v = 0; so A = 0, so that
(M + t)v = kt
or
v = ds/dt = (kt)/(M + t), i.e. ds = (kt dt)/(M + t)
Integration again,
kt dt
=  k (t  + M – M) dt
s = 
M + t

M + t 
= k  M . ] dt

(M + t)
= (k/)[t – (M/).log(M + t)] + B
Initially when t = 0, s = 0.
160
So
B = (Mk/2)logM
Hence s = (k/)[t – (M/).log(M + t)] + (Mk/2).logM
= (k/2)[t – M.logm] + (Mk/2).logM, since M + t = m
= (k/2)[m – M – M log(m/M)], since t = m – M
= (k/2)[m – M{1 + log(m/M)}] where k = F – V.
Ex.5 Show slides off a roof clearing away a part of uniform breadth; show that, if it all
slide at once, the time in which the roof will be cleared is
1/ 2
6a (2/3)
g sin (1/6)
but that, if the top move first and gradually set the rest in motion, the acceleration is
(1/3)gsin and the time will be [6a/(gsin)], where  is the inclination of the roof and a
the length originally covered with snow.
Sol.
Let b be the breadth of the roof and the surface density of snow. Then if y be the
length on the roof at time t, the mass of snow = y.b.
Using the relation
d(mv) = P, we have
dt
d [yb (- dy )] = yb.g sin
dt
dt
Since the roof is inclined at an angle  to the horizon
or
d [y (- dy )] = yg sin
dt
dt
or
dy
d
2y
[y dy ] = - 2y2g sin
dt dy dt
or
d [y dy ]2 = - 2y2.g sin
dy dt
Integrating with regard to y,
[y dy]2 = – (2/3)y3.g sin + A
dt
Initially when y = a, dy/dt = 0; so A = (2/3)g sin a3.
Thus
y2 ( dy )2 = (2/3).g sin (a3 – y3)
dt
or
dy/dt = (1/y).[(2gsin)/3].(a3 – y3)
So
2g sin T
a
y dy
(
). dt = 
3
0
0 (a3 – y3)
[put y = az1/3, so dy = (1/3)az2/3dz]
or
az1/3
a dz
( 2g sin).T = 1 3/2
3
0 a (1 – z) 3z2/3
–1/3
(1 – z) –1/2 dz
= a 1 z
3 0
2/3 – 1
(1 – z)1/2 – 1 dz
= a 1 z
3 0
161
= a  (2/3, 1/2)
3
= a (2/3) (1/2)
3 (2/3 + 1/2)
Hence

T = (1/3) ( 3a ) (2/3)

2g sin (1/6)(1/6)

=  ( 6a ) (2/3)

g sin (1/6)
For the second part, supposing a length x to be in motion, we have
d [bxdx ] = bx.g sin
dt
dt
or
or
or
or
so
or
or
or
d [x dx ] = x.g sin
dt dt
2xdx d [x dx ] = 2x2.g sin
dt dx dt
d [x dx ]2 = 2x2.g sin
dx
dt
Integrating,
[x dx ]2 = (2/3)x3.g sin + C
dt
Initially when x = 0, dx/dt = 0; so C = 0, so that
[x dx ]2 = (2/3)x3.g sin 
dt
dx/dt = [(2gsin)/3].(x)
2g sin T
a
dx
(
). dt = 
3
0
0 (x)
2g sin ).T = [ x ]a = 2a)
(
3
(1/2) 0
(1)
(2)
3
.

T = 2a) (
)
2g sin
=  ( 6a .)  
g sin
Further to find the acceleration, we have
d [x dx ] = x.g sin
dt dt
from (1)
(dx/dt)2 + x(d2x/dt2) = gx sin
x(d2x/dt2) = gx sin – (2/3)gx sin, [from (2), (dx/dt)2 = (2/3)gx sin
= (1/3)xg sin
So acceleration d2x/dt2 = (1/3)g sin
162
Ex.6 A spherical raindrop, falling freely, receive in each instant an increase of volume
equal to time its surface at that instant; find the velocity at the end of time t and the
distance fallen through in that time.
Sol.
When the raindrop has fallen through a distance x in time t, let its radius be r and
its mass M. Then
d [M dx ] = Mg 
dt
dt
(1)
Now M = (4/3)r3, so that 4 r2(dr/dt) = dM/dt = 4 r2, by the question.
So
dr/dt = , and r = a + t,
where a is the initial radius.
Hence (1) gives
d [(a + t)3 dx ] = (a + t)3g 
dt
dt
4
4
3 dx
so
=  (a + t) g – a g 
(a + t)
dt
4  
since the velocity was zero to start with.
so
dx
g
a4
]
=  [a + t – 
dt
4 
(a + t)3
2
4
3
and
x = g 2  (a + t) +  a
– g .2 a 
2
4
2 
2(a + t) 
4

since x and t vanish together, so
4
] 
x = g 2 [(a + t)2 – 2a2 +  a
2
8

2(a
+
t)

2
x = g 2 [(a + t) –a
]2
8
(a + t)


2
and
gt
2a
+
t
2
x=
[
]
8
a + t


Ex.7 If a rocket, originally of mass M, throw off every unit of time a mass eM with
relative velocity V and if M be the mass of the case etc., show that it cannot rise at once
unless eV > g, nor at all unless eMV/ M > g. If it just rises vertically at once, show that
its greatest velocity is
V log(M/M) – (g/e).(1 – M/M),
and that the greatest height it reaches is
(V2/2g).[log(M/M)]2 + (V/e).[1 – (M/M) – log(M/M)].
Sol.
Let u, u be the velocities communicated to the rocket and the mass thrown off in
unit time by the explosion; then the relative velocity of the mass V = u – (– u) = u + u.
When a mass eM is thrown off the remaining mass of the rocket = M – eM = (1 – e)M.
At the start of the explosion:
Momentum of rocket = momentum of mass thrown off, i.e.
(1 – e)Mu = eMu = eM(V – u), since V = u + u.
So
u = eV
(1)
163
But we know that the gravity in unit time gives a velocity g. Hence the rocket will
not rise at once unless
g < u, i.e. g < eV.
Now when all the powder is burnt (except the last element), the law of
conservation of momentum gives
Mu = eMu, where M is the mass of the case
or
Mu = eM(V – u), i.e. u = (eMV)/(M + eM), as u = V – u (2)
Again the velocity imparted by the gravity in unit time being g, it will not start at
all unless
g < (eMV)/(M + eM), i.e. eMV > (M + eM)g > Mg
or unless
(eMV)/M > g
Further when the rocket has risen a distance x in time t, then the mass of the
rocket
= initial mass – mass thrown off in time t.
= M – eM.t = (1 – e.t)M
Thus
d [(1 – et)M dx ]
dt
dt
= change in the momentum
= – eM.(dx/dt – V) – (1 – e.t)M.g
2
or
(1 – et)d 2x - e dx
dt
dt
= – e(dx/dt) + eV – g + etg
2
or
(1 – et)d 2x
dt
= eV – g(1 – et);
or
d2x = eV . – g
dt2
1 - et
Integrating with respect to t,
dx/dt = - V log(1 – et) – gt + A
Initially when t = 0, dx/dt = 0; so A = 0.
Thus dx/dt = - V log(1 – et) – gt
(3)
Integrating again,
x = - V log(1 – et).1 dt – ½ gt2 + B
2
= - V[log(1 – et)t -  - e.t dt] – gt + B
1 – et
2
2
1
gt
)dt] –
= - V[t log(1 – et) - 
+B
1 – et
2
= – V[t log(1 – et) – t – (1/e) log(1 – et)] – gt2/2 + B
= (V/e).[(1 – et).{log(1 – et)} + et] – gt2/2 + B
Initially when t = 0, x = 0; so B = 0.
Hence x = (V/e)[(1 – et) log(1 – et) + et] – gt2/2
= (V/e)[(1 – et) log(1 – et) – (1 – et) + 1] – gt2/2
= (V/e)[(1 – et){log(1 – et) – 1}] – gt2/2 + V/e
(4)
164
Now the greatest velocity is attained when the powder is all burnt, i.e. when mass
of the rocket = mass of the case, i.e. when (1 – et)M = M or 1 – et = (M/M) and then
from (3) and (4), we have
V1 = dx/dt = - V log(M/M) – (g/e).(1 – M/M), since t = (1/e)(1 – M/M),
(5)
2
2
and
x1 = x = (V/e).[(M/M){log(M/M) – 1}] – (g/2).(1/e ).(1 – M/M) + V/e
= (V/e).(1 – M/M) – [(MV)/(eM)].log(M/M) – [g/(2e2)].(1 – M/M)2
[since log(M/M) = - log(M/M)]
So total height attained
= x1 + V12/(2g)
= (V/e)(1 – M/M) – [(MV)/(eM)].log(M/M) – [g/(2e2)].(1 – M/M)2
+ [1/(2g)].[V log(M/M) + (g/e).(1 – M/M)]2
= (V/e)(1 – M/M) – [(MV)/(eM)].log(M/M) – [g/(2e2)].(1 – M/M)2
+ [V2/(2g2)].[log(M/M)]2 – (V/e).(1 – M/M).log(M/M)
+ [g/(2e2)].(1 – M/M)2
{since log(M/M) = - log(M/M) and [log(M/M)]2 = [log(M/M)]2}
= [V2/(2g)].[log(M/M)]2
+ (V/e).[1 – M/M – (M/M).log(M/M) – log(M/M) + (M/M).log(M/M)]
= [V2/(2a)].[log(M/M)]2 + (V/e).[1 – M/M – log(M/M)]
Ex.8 A uniform string, whose length is l and whose weight is W, rests over a small
smooth pulley with its end and just reaching to a horizontal plane; if the string be slightly
displaced, show that when a length x has been deposited on the plane the pressure on it is
W[2log{l/(l – x)} – (x/l)],
and that the resultant pressure on the pulley is W(l – 2x)/(l – x).
Sol.
Initially for equilibrium the length ½ l of the string hangs on either side of the
pulley. But when a length x has been deposited on the plane in an interval of time t (say),
then the length (½ l – x) of the string hangs on one side and ½ l hangs vertically on the
other side. Let T be the tension of the string and d2x/dt2 the acceleration, then equations
of motion are
m.(l/2).g – T = m.(l/2).(d2x/dt2)
(1)
where m is the mass per unit length
and
T – m(l/2 – x)g = m (l/2 – x)(d2x/dt2)
(2)
Adding,
[l/2 – (l/2 – x)]g = [l/2 + (l/2 – x)](d2x/dt2)
d2x/dt2 = xg/(l – x)
(3)
d2x/dt2 = – g[(l – x – l)/(l – x)] = g [l/(l – x) – 1]
Multiplying both side by 2.(dx/dt).dt and integrating, we get
(dx/dt)2 = 2g.[ – l log(l – x) – x] + A.
Initially when x = 0, dx/dt = 0; so A = 2gl logl.
Thus (dx/dt)2 = 2g[l log{l/(l – x)} – x]
(4)
Now if M be the mass of the whole string, then its mass per unit length = M/l = m
and its weight W = Mg.
giving
or
165
Again if R be the part of reaction due to the impact, then R.t = m.x.(dx/dt).
Dividing by t and proceeding to the limit, we get
R = m.(dx/dt)2
Total pressure = R + mg.x = m[(dx/dt)2 + gx]
Mg [ l {2g(l log l . - x)} + x]
=
l
g
l–x
from (4) and also m = M/l
= W[2log{l/(l – x)} – x/l]
Further to find the pressure on the pulley, which is = 2T, let us subtract (2) from
(1) after multiplying (1) by (l/2 – x) and (2) by l/2, whence get
– T[l/2 – x + l/2] + mg[2.(l/2).(l/2 – x)] = 0
or
T = mgl (l – 2x)
2 (l – x)
hence the pressure on the pulley = 2T
= mgl (l – 2x)
(l – x)
Mgl
(l
– 2x)
=
l
(l – x)
(l
–
2x)
=W
(l – x)
Ex.9 A weightless string passes over a smooth pulley. One end is attached to a coil of
chain lying on a horizontal table and the other to a length l of the same chain hanging
vertically with its lower end just touching the table. Show that after motion ensues the
system will first be at rest when a length x of chain has been lifted from the table, such
that (l – x)e(2x/ l) = l. Why cannot the Principle of energy be directly applied to find the
motion of such a system?
Sol.
Let m be the mass per unit of length, so that a mass ml is always in motion and
the mass deposited per unit of time
mx = m dx
= Lim
t 0
t
dt
So
d [ml dx ] 
dt
dt
= change in the momentum per unit of time
= – m(dx/dt).(dx/dt) + m(l – x).g – m.xg,
[since the gravity in unit time gives a velocity g]
or
l(d2x/dt2) = – (dx/dt)2 + (l – 2x)g
or
l(d2x/dt2) + (dx/dt)2 = (l – 2x)g
(1)
or
d
dx
dx
2
dx
2
2
2 ( )
+ (
) = (l – 2x)g
dx dt dt
l dt
l
or
d ( dx )2 + 2 ( dx )2 = 2 (l – 2x)g
dx dt
l dt
l
(2)
which is linear differential equation in (dx/dt)2.
So its I.F. = e(2/ l)dx = e2x/ l
Solution is
(dx/dt)2e2x/ l = (2g/l) (l – 2x)e2x/ ldx + A
166
= (2g/l)[(l – 2x).(l/2).e2x/ l + 2.(l/2) e2x/ ldx ] + A
= (2g/l)[(l/2).(l – 2x)e2x/ l + (l2/2)e2x/ l] + A
= g[l – 2x + l]e2x/ l + A = 2g(l – x)e2x/ l + A
Initially when x = 0, dx/dt = 0; so A = -2gl.
Hence
(dx/dt)2e2x/ l = 2g(l – x)e2x/ l – 2gl.
Again the string comes to rest when dx/dt = 0,
i.e.
2gl(l – x)e2x/ l – 2gl = 0
or
(l – x)e2x/ l = l.
Further since energy is lost by the deposit of the portions of the chain on the table,
the principle of energy cannot be directly applied.
Ex.10 A mass M is fastened to a chain of mass m per unit length coiled up on a rough
horizontal plane (coefficient of friction = ). The mass is projected from the coil with
velocity V; show that it will be brought to rest in a distance
M [(1 + 3mV2 )1/3 – 1]
m
2Mg
Sol.
Let a length x of the coiled up chain lifts in time t, then equation of motion is
d [(M +mx) dx
]
dt
dt
= – (M + mx)g, since P = – R = – (M + x)g
or
2(M + mx)dx d [(M + mx)dx ]
dt dt
dt
2
= – 2(M + mx) .g
or
d [(M + mx)dx 2
]
dt
dt
= – 2(M + mx)2.g
Integrating with regard to x, we get
(M + mx)2.(dx/dt)2 = – [2/(3m)].g(M + mx)3 + A
Initially when x = 0, dx/dt = V; so A = M2V2 + (2M3g)/(3m)
Hence,
(M + mx)2.(dx/dt)2 = – [2/(3m)].g(M + mx)3 + M2V2 + (2M3g)/(3m)
The mass will be brought to rest when dx/dt = 0
i.e. when
0 = – (2g)/(3m).(M + mx)3 + M2V2 + (2M3g)/(3m)
or
(M + mx)3 = M3 + (3M2mV2)/(2g)
= M3[1 + (3mV2)/(2Mg)]
Extracting cube root of either side,
M + mx = M[1 + (3mV2)/(2Mg)]1/3
so that
2
x = M [(1 + 3mV )1/3 – 1]
m
2Mg
Ex.11 A mass in the form of a solid cylinder, the area of whose cross-section is A, moves
parallel to its axis, being acted on by a constant force F, through a uniform cloud of fine
167
dust of volume density which is moving in a direction opposite to that of the cylinder
with constant velocity V. If all the dust meets the cylinder clings to it, find the velocity
and distance described in any time t, the cylinder being originally at rest and its initial
mass m.
Sol.
Let M be the mass at time t and v the velocity. Then
M.v + M(v + v + V) = increase in the momentum in time t = Ft.
So
M(dv/dt) + v(dM/dt) + V(dM/dt) = F
(1)
in the limit.
Also dM/dt = A(v + V)
(2)
(1) gives Mv + MV = Ft + constant = Ft + mV.
Therefore (2) gives
M(dM/dt) = A(Ft + mV)
So
M2 = A(Ft2 + 2mVt) + m2
Therefore (2) gives
Ft + mV
v=-V+
M
Ft + mV
.
=-V+
2
2
(m + 2mAVt + AFt )
(3)
Also if the hinder end of the cylinder has described a distance x from rest, so that
v = dx/dt, then x = - Vt + [1/(A)].(m2 + 2mAVt + AFt2) – [m/(A)]
From (3) we have that the acceleration
dv =
m2(F – AV2)
.
2
2 3/2
dt
(m + 2mAVt + AFt )
so that the motion is always in the direction of the force, or opposite, according as F > or
< AV2.
Ex.12 A machine gun, of mass M, stands on a horizontal plane and contains shot, of mass
M. The shot is fired at the rate of mass m per unit of time with velocity u relative to the
ground. If the coefficient of sliding friction between the gun and the plane is , show that
the velocity of the gun backward by the time the mass M is fired is
M u – (M + M)2 – M2 g
M
2mM
Sol.
Let the gun move backward a distance x in time t, then its velocity is dx/dt.
The mass of the shot fired out in time t = mt.
Then the mass of the machine gun with shot = M + M – mt.
Now the change in momentum per unit of time
= impulsive in one second = impulsive force 1 sec.
i.e.
d [(M + M – mt)dx ] – mu
dt
dt
= – (M + M – mt)g
or
d [(M + M – mt)dx ]
dt
dt
= – (M + M – mt)g + mu
168
Integrating with regard to t, we get
[(M + M – mt)dx ]
dt
= [(g)/(2m)].(M + M – mt)2 + mut + A
Initially when t = 0, dx/dt = 0; so A = – [(g)/(2m)].(M + M)2
Thus
[(M + M – mt)dx ]
dt
= [(g)/(2m)].(M + M – mt)2 + mut – [(g)/(2m)].(M + M)2
= [(g)/(2m)].[(M + M – mt)2 – (M + M)2] + mut
= [(g)/(2m)].[(M + M)2 – 2m(M + M)t + m2t2 – (M + M)2] + mut
= [(g)/(2m)].[– 2(M + M) + mt]mt + mut
(1)
Now when whole of the mass M (of shot) is fired, then
M = mt.
Hence from (1), we have
[(M + M – M) dx ]
dt
= [(g)/(2m)].[– 2(M + M) + M]M + Mu
= [(g)/(2m)].[– M2 – 2MM– M + M2] + Mu
= [(g)/(2m)].[(M + M)2 – M2] + Mu
dx = M u – (M + M)2 – M2 g
dt
M
2mM
Ex.13 A uniform cord, of length l, hangs over a smooth pulley and a monkey, whose
weight is that of the length k of the cord, clings to one end and that system remains in
equilibrium. If he start suddenly and continue to climb with uniform relative velocity
along the cord, show that he will cease to ascend in space at the end of time
[(l + k)/(2g)]1/2 cosh1(1 + l/k)
Sol.
Let m be the mass per unit length of the cord; then the mass of monkey = mk.
If V is the given relative velocity and V1 the starting velocity of the cord, then
Mk(V – V1) = pull of the monkey on the cord
= mlV1
so that
V1 = (kV)/(k + l)
(1)
Initially the lengths of the cord on the two sides of the pulley are
(l – k)/2 and (l + k)/2
Let at time t, a length x of the cord have gone over the pulley and the monkey
have ascended a distance y along the cord so that dy/dt = V and d2y/dt2 = 0, V being
constant.
Then if P be the weight of the pulley, we have
d [{( l – k + x) + ( l + k – x)}m.dx ]
dt
2
2
dt
= m[(l – k)/2 + x]g – m[(l + k)/2 – x]g + P,
i.e.
ml(d2x/dt2) = (2x – k)mg + P
(2)
d [mk( dx – dy )]
169
dt
dt dt
= mkg – P,
2
2
i.e.
[mk( d 2x – d 2y )]
dt
dt
= mkg – P,
or
mk(d2x/dt2) = mkg – P, since d2y/dt2 = 0
(3)
Adding (2) and (3), we get
m(l + k).(d2x/dt2) = 2mgx, i.e. d2x/dt2 = (2gx)/(k + l)
Multiplying both side by 2(dx/dt).dt and integrating, we get
(dx/dt)2 = [2g/(k + l)].x2 + A
Initially when x = 0, dx/dt = V1 = kV/(k + l) from (1)
So
A = k2V2/(k + l)2
2 2
2
so that
( dx )2 = 2g x + k V 2
dt
(k + l)
(k + l)
2 2
2
2
or
]
( dx ) = 2g [x + k V
dt
(k + l)
2g(k + l)
(4)
or
dx = [ 2g ] [x2 + k2V2 ]
dt
(k + l)
2g(k + l)
or
2g
[
].t
(k + l)
dx
.
= 
2 2
[x2 + k V
]
2g(k + l)
= sinh-1[ x{2g(k + l)} ] + B
kV
Initially when t = 0, x = 0, so B = 0.
Hence
or
[ 2g ].t
(k + l)
= sinh1[x{2g(k + l)} ]
kV
(5)
Now the monkey stops ascending in space when dx/dt = V, i.e. when from (4),
2 2
2
V2 = 2g [x + k V ]
k+l
2g(k + l)
or
2gx2 = (k + l)V2 – (k2V2)/(k + l) = [V2/(k + l)].(k2 + 2kl + l2 – k2)
=V2.[(l2 + 2kl)/(k + l)],
which gives
2
x = V(l + 2kl)
{2g(k + l)}
Substituting this value of x in (5), we get
[ 2g ].t
(k + l)
{l2 + 2kl}
= sinh1[
]
k
so
sinh [t ( 2g . )]
(k + l)
170
2
1/ 2
=  (l + 22kl)
k
Hence
cosh2 [t ( 2g . )]
(k + l)
so
= 1 + sinh2 [t ( 2g )]
(k + l)
= 1 + (2l/k) + l2 /k2 = (1 + l/k)2,
which gives
t = [(l + k)/(2k)]1/2cosh–1(1 + l/k)
Ex.14 A heavy chain, of length l, is held by its upper end so that its lower end is at a
height l above a horizontal plane; if the upper end is let go, show that at the instant when
half the chain is coiled up on the plane is to the weight of the chain in the ratio of 7:2.
Sol.
Let at time t, a length x of the chain be coiled up on the plane. Then this length x
has to move a distance (l + h) as the lower end is already at a height l above the plane.
Since the moving part of the chain falls with acceleration g, using the relation v2 = u2 +
2gh, where the initial velocity u = 0, the moving part is given by
v2 = 2g(l + x)
(1)
The part of the reaction say R due to the impact is given by
Impulse = the change in momentum
i.e.
R.t = m x.v (supposing that x length of chain coils up in time t)
Dividing by t and proceeding to the limit.
R = m.(dx/dt).v = mv2 = 2mg(l + x), from (1).
Hence the total pressure P
= R + weight of the length x of the chain
= R + mgx where m is the mass per unit length when half the chain is
coiled up i.e. x = l/2, we have
Pat x = l/ 2 = 2mg(l + l/2) + mg.(l/2), since R = 2mg(l + x)
= (7/2).mlg
which shows that the pressure on the plane when half the chain is coiled up, is to the
weight of the chain in the ratio 7:2.
Ex.15 A uniform chain, of length l and mass ml, is coiled on the floor and a mass mc is
attached to one end and projected vertically upwards with velocity (2gh). Show that,
according as the chain does or does not completely leave the floor, the velocity of the
mass on finally reaching the floor again is the velocity due to a fall through a height
(1/3)[2l – c + a3/(l + c)2] or a – c,
3
2
where a = c (c + 3h).
Sol.
The equation of motion is
d {(mc + mx) dx }= - (mc + mx)g
dt
dt
Supposing that a length x of the coil lifts in time t
or 2(c + x) dx d {(c + x) dx }= - 2(c + x)2g
171
dt dx
dt
or
{(c + x) dx }2 = – 2g(x2 + 2cx + c2)
dt
Integrating with respect to x, we get
or
{(c + x) dx }2 = – 2g(x3/3 + cx2 + c2x) + A
dt
Initially when x = 0, dx/dt = (2gh); since A = c2.2gh
Hence
(x + c)2.(dx/dt)2 = – (2g/3)[ x3 + 3cx2 + 3c2x] + c22gh
(1)
Case I. Let us assume that the chain leaves the floor. Then putting x = l in (1) and
supposing that the velocity then is V, we have
(l + c)2V2 = – (2g/3).[l3 + 3cl2 + 3c2l] + c2.2gh
= (2g/3).[3c2h + c3 – {l3 + 3l2c + 3lc2 + c3}]
= (2g/3).[a3 – (l + c)3], since a3 = 3c2h + c3 = c2(c + 3h) (given)
or
V2 = (2g/3).[a3/(l + c)2 – (l + c)]
Using the relation v2 = u2 – 2H, where u = V, the mass will rise till v = 0, i.e. the
height attained H = V2/(2g)
Hence the total height = l + H = l + V2/(2g)
= l + (1/3).[a3/(l + c)2 – l – c]
= (1/3).[2l – c + a2/(l + c)2]
Case II. If we assume that the chain does not leave the floor, then the mass will come to
rest when dx/dt = 0, i.e. when from (1),
0 = – (2g/3).(x3 + 3cx2 + 3c2x) + c2.2gh
or
3c2h + c2 – (x3 + 3cx2 + 3c2x + c3) = 0
or
a3 = (x + c)3, i.e. when x + c = a or x = a – c.
Ex.16 A uniform chain is coiled up on a horizontal plane and one end passes over a small
light pulley at a height a above the plane; initially a length b, > a, hangs freely on the
other side; find the motion.
Sol.
When the length b has increased to x, let v be the velocity; then in the time t next
ensuing the momentum of the part (x + a) has increased by m(x + a).v, where m is the
mass per unit length. Also a length m.x has been jerked into motion and given a velocity
v + u. Hence
m(x + a)v + mx(v + v) = change in the momentum
= impulse of the acting force = mg(x – a).t
Hence, dividing by t and proceeding to the limit, we have
(x + a).(dv/dt) + v2 = (x – a)g
So
v(dv/dx).(x + a) + v2 = (x – a)g
x
So
v (x + a) = 2(x2 – a2)g dx = 2[(x3 – b3)/3 – a2(x – b)].g
2
2
b
So that
2
2
2
v2 = 2g (x – b)(x + bx + b2 – 3a )
3
(x + a)
This equation cannot be integrated further.
172
In the particular case when b = 2a, this gives v2 = (2g/3).(x – b), so that the end
descends with constant acceleration g/3.
This tension T of the coil is clearly given by Tt = mx.v, so that T = mv2.
Ex.17 A ball of mass m, is moving under gravity in a medium which deposits matter on
the ball at a uniform rate . Show that the equation to the trajectory, referred to horizontal
and vertical axes through a point on itself, may be written in the form
k2uy = kx(g + kv) + gu(1 – ekx/ u),
where u, v are the horizontal and vertical velocities at the origin and mk = 2.
Sol.
The mass of the ball at time t = m + t. Since there is no horizontal force, using
the relation,
d(mv) = P
dt
we have,
d [(m + t) dx
]=0
dt
dt
Integrating,
[(m + t) dx ] = A
dt
Initially when t = 0, dx/dt = u, so A = mu, so that
[(m + t) dx ] = mu
dt
(1)
or
dx = [mu/(m + t)].dt = (mu/).[/(m + t)].dt
Integrating, x = (mu/) log(m + t) + B
Initially when t = 0, x = 0; so B = – (mu/) logm.
Then x = (mu/).log[(m + t)/m], i.e. m + t = me x/m u
(2)
Substituting the value of m + t from (2) in (1), we get
dx/dt = ue– (x/m u)
(3)
Now considering the vertical motion, we have
d [(m + t) dx
] = - (m + t)g
dt
dt
or
d [(m + t) dy dx ] dx = - (m + t)g
dx
dx dt dt
x/m
u
or
d [me
dy ue–(x/m u)] ue– (x/m u)
= – mex/m u g
dx
dx
[from (1) and (2)]
or
d [mu dy ] ue– (x/m u)
x/m u
= – me
g
dx
dt
or
u2(d2y/dx2) = – ge2x/m u = – gk x/ u
as mk = 2
Integrating,
u2(dy/dx) = – g.(u/k)ek x/u + C
Initially when x = 0, dy/dx = (dy/dt)/(dx/dt) = v/u, so that
C = u2v/u + (g/k)u
Thus u(dy/dx) = – (g/k)ek x/u + (v + g/k)
Integrating again,
173
or
uy = – (g/k2).uek x/u + (v + g/k)x + D
Initially when x = 0, y = 0; so D = gu2/k2
Hence uy = – (gu/k2)ek x/u + (v + g/k)x + (g/k2).u2
k2uy = kx(g + kv) + gu(1 – ek x/u)
Ex.18 A mass of in the form of a solid cylinder, of radius c, acted upon by no forces,
moves to its axis through a uniform cloud of fine dust which of volume density , which
is at rest. If the particles of dust which meet the mass adhere to it, and if M and u be the
mass and velocity at the beginning of the motion, prove that the distance x traversed in
time t is given by the equation (M + c2x)2 = M2 + 2uc2Mt.
Sol.
When the mass M has fallen through a distance x in time t, its mass
= M + mass adhered to it = M + c2.x.
Since there is no force i.e. P = 0, using the relation
d(mv) = P
dt
we have,
d [(M + c2x.) dx
]=0
dt
dt
Integrating,
[(M + c2x) dx ] = A
dt
Initially when t = 0, dx/dt = u, so A = mu, so that
[(M + c2x) dx ] = Mu
dt
or
2c2(M + c2x)(dx/dt) = 2Mu.c2
Integrating again, we get
(M + c2x)2 = 2Mc2ut + B
Initially when x = 0, t = 0; so B = M2
Hence (M + c2x)2 = 2Mc2ut + M2
Ex.19 A chain, of length l, is coiled at the edge of a table. One end is fastened to a
particle, whose mass is equal to that of the whole chain and the other end is put over the
edge. Show that, immediately after leaving the table, the particle is moving with velocity
½ (5gl/6).
Sol.
Let M be the total mass of the chain, so that the mass per unit length is M/l. When
a length x of the chain has moved over the edge in time t, we have
d [(M/l).x dx
] = (M/l).xg
dt
dt
[since mass of the chain of the length x = (M/l).x]
or
d
dx
dx
2x [x ]
= 2x2g
dt
dt dt
or
d [x dx ]2 = 2x2g
dx
dt
Integrating,
174
or
or
[x dx ]2 = (2/3)x3g + A
dt
Initially when x = 0, dx/dt = 0; so A = 0, so that
[x dx ]2 = (2/3)x3g
dt
[ dx ]2 = (2/3)xg
dt
giving
[ dx ] = (2/3)xg
dt
(1)
Hence, just before the mass is jerked, i.e. the chain has run over a length l, the
velocity of the chain say V = [(2/3).gl], by putting x = l in (1).
When the mass is just to run off the plane, let the horizontal velocity
communicated to the mass be V1 and as the mass is just off the plane, let the jerk give it a
vertical velocity V2. Then the law of conservation of momentum gives
(M + M)V1 = MV
(2)
and
(M + M)V2 = MV1
(3)
(1) gives V1 = V/2 and then (2) gives V2 = V/2
Therefore, total velocity of the particle, then = (V12 + V22)
=  (V2/4 + V2/16) = (V/4) 5 = [(2/3).gl] ( 5/4) = ½ (5gl/6)
Ex.20 A uniform chain, of mass M and length l, is coiled up at the top of a rough plane
inclined at an angle  to the horizon and has a mass M fastened to one end. This is
projected down the plane with velocity V. If the system comes to rest when the whole of
the chain is just straight, show that V2 = (14gl/3) sec sin(– ), where is the angle of
friction.
Sol.
The mass per unit length of the chain = M/l. Let a length x be uncoiled up the
chain at time t. considering the equilibrium of mass M and a mass of length x of the
chain, we have
The reaction R = [Mg + (M/l).xg] cos
and so R = [Mg + (M/l).xg].cos, where = tan
Resolved part of the weight [Mg + (M/l).xg] along the plane is
[M + (M/l)x]g sin
Therefore the equation of motion is
d [{M + (M/l).x} dx
] = [M + (M/l).x]g sin - R
dt
dt
= [M + (M/l)x]g (sin – cos)
or
d [(x + l) dx
] = (l + x)g (sin - cos)
dt
dt
or 2(x + l)dx d [(x + l) dx
] = 2(l + x)2g (sin - cos)
dt dx
dt
or
d [(x + l) dx 2
] = 2g(sin - tancos).(x2 + 2xl + l2)
dx
dt
Integrating,
or
[(x + l) dx ]2
175
dt
= (2g/cos[cos sin – sin cos].[x3/3 + lx2 + l2x] + A
Initially when x = 0, dx/dt = V; so A = l2V2
Hence
[(x + l) dx ]2
dt
= (2g secsin( – ).[x3/3 + lx2 + l2x] + l2V2
The system again comes to rest when the whole of the chain is just straight, when
x = l, dx/dt = 0. Thus we have
0 = 2g sec sin(- ).[l3/3 + l3 + l3] + l2V2
or
V2 = (14gl/3) sec sin( - )
2.6 Unit Summary:
1. For upwards projection
v2 = u2 – 2gh
and for downward projection
v2 = u2 + 2g
where h is the vertical distance between the two points.
2. In the case in which a heavy particle is sliding upwards on a curve in a vertical plane,
the equations are
mv.(dv/ds) = - R - mg sin
and
mv2/ = R - mg cos
where is the inclination of the tangent to the horizontal.
3. Suppose that a particle gains in time t an increment m of mass and that this increment
m was moving with a velocity u.
Then in time t the increment in the momentum of the particle
= m.v + m(v + v – u)
and the impulse of the force in this time is Pt.
Equating these we have, on proceeding to the limit.
m(dv/dt) + v(dm/dt) – u(dm/dt) = P
i.e.
d (mv) = P + u dm
dt
dt
2.7 Assignments:
1. A particle moves in a smooth tube in the form of a catenary, being attracted to the
directrix by a force proportional to the distance from it. Shew that the motion is simple
harmonic.
2. A smooth parabolic tube is placed, vertex downwards, in a vertical plane; a particle
slides down the tube from rest under the influence of gravity; prove that in any position
the reaction of the tube is 2w.[(h + a)/], where w is the weight of the particle, the
radius of curvature, 4a the latus rectum and h the original vertical height of the particle
above the vertex.
176
3. A curve in a vertical plane is such that the time of describing any arc, measured from a
fixed point O, is equal to the time of sliding down the chord of the arc; shew that the
curve is a lemniscate of Bernouilli, whose node is at O and whose axis is inclined at 450
to the vertical.
4. A particle, of mass m, moves in a smooth circular tube, of radius a, under the action of
a force, equal to m distance, to a point inside the tube at a distance c from its centre; if
the particle be placed very nearly at its greatest distance from the centre of force, show
that it will describe the quadrant ending at it least distance in time
(a/c) log(2 + 1)
5. A bead is constrained to move on a smooth wire in the form of an equiangular spiral. It
is attracted to the pole of the spiral by a force, =m(distance) 2 and starts from rest at a
distance b from the pole. Shew that, if the equation to the spiral be r = aecot, the time of
arriving at the pole is (/2).(b3/2).sec. Find also the reaction of the curve at any
instant.
6. A particle is projected horizontally from the lowest point of a rough sphere of radius a.
After describing an arc less than a quadrant it returns and comes to rest at the lowest
point. Shew that the initial velocity must be sin[2ga(1 + 2)/(1 – 22)], where is the
coefficient of friction and a is the arc through which the particle moves.

7. A bead slides down a rough circular wire, which is in a vertical plane, starting from
rest at the end of a horizontal diameter. When it has described an angle about the
centre, show that the square of its angular velocity is
2g
[(1 – 22) sin + 3 (cos – e– 2)],
2
a(1 + 4 )
where is the coefficient of friction and a the radius of the rod.
8. A ring which can slide on a rough circular wire in a vertical plane is projected from the
lowest point with such velocity as will just take it to the horizontal diameter; if the ring
returns to the lowest point, show that its velocity on arrival is to its velocity on departure
as
(1 – 22 – 3e–)1/ 2 : (1 – 22 + 3e)1/ 2,
where is the coefficient of friction.
9. A heavy particle slides, from cusp, down the arc of a rough cycloid, the axis of which
is vertical. Prove that its velocity at the vertex is to its velocity at the same point when the
cycloid is smooth as (e– – 2) : (1 + 2), where is the coefficient of friction.
10. A particle is projected with velocity V along a smooth horizontal plane in a medium
whose resistance per unit mass in times the cube of the velocity. Shew that the distance
it has described in time t is
(1/V)[(1 + 2V2t) – 1]
and that its velocity then is
V/(1 + 2V2t).
11. A heavy particle is projected vertically upwards with a velocity u in a medium the
resistance of which varies as the cube of the particles velocity. Determine the height to
which the particle will ascend.
12. If the resistance to the motion of a railway train vary as its mass and the square of the
velocity and the engine work at constant H.P., show that full speed will never be attained
and that the distance traveled from rest when half-speed is attained is (1/3) log(8/7),
177
where is the resistance per unit mass per unit velocity. Find also the time of describing
this distance.
13. A ship, with engines stopped, is gradually brought to rest by the resistance of the
water. At one instant the velocity is 10 ft. per sec. and one minute later the speed has
fallen to 6 ft. per sec. For speeds below 2 ft. per sec. the resistance may be taken to vary
as the speed and for higher speeds to vary as the square of the speed. Show that, before
coming to rest, the ship will move through 900[1 + log5] feet, from the point when the
first velocity was observed.
14. A particle moves from rest at a distance a from a fixed point O under the action of a
force to O equal to times the distance per unit of mass; if the resistance of the medium
in which it moves be k times the square of the velocity per unit of mass, show that the
square of the velocity when it is at a distance x from O is
(x/k) – (a/k).e2k(x – a) + (/2k2)[1 – e2k(x – a)].
Show also that when it first comes to rest it will be at a distance b given by
(1 – 2kb)e2kb = (1 + 2ka)e-2ak.
15. Show that in the motion of a heavy particle in a medium, the resistance of which
varies as the velocity, the greatest height above the level of the point of projection is
reached in less than half the total time of the flight above that level.
16. If the resistance of the air to a particle‟s motion be n times its weight and the particle
be projected horizontally with velocity V, show that the velocity of the particle, when it is
moving at an inclination to the horizontal, is V(1 – sin)(n – 2)/2(1 + sin) - (n + 1)/2.
17. A heavy bead slides down a smooth wire in the form of a cycloid, whose axis is
vertical and vertex downwards, from rest at a cusp and is acted on besides its weight by a
tangential resistance proportional to the square of the velocity. Determine the velocity
after a fall through the height x.
18. If a body move under a central force in a medium which exerts a resistance equal to k
times the velocity per unit of mass, prove that
d2u + u =
P e2kt
2
2 2
d
hu
where h is twice the initial momentum of momentum about the centre of force.
19. A particle moves with a central acceleration P in a medium of which the resistance is
k (velocity)2; show that the equation to its path is
d2u + u =
P e2ks
2
2 2
d
hu
where s is the length of the arc described and h is twice the initial moment of momentum
about the centre of force.
20. A uniform chain is partly coiled on a table, one end of it being just carried over a
smooth pulley at a height h immediately above the coil and attached there to a weight
equal to that of a length 2h of the chain. Show that until the weight strikes the table, the
chain uncoils with uniform acceleration g/3 and that, after it strikes the table, the velocity
at any moment is (2gh/3)e(x – h)/2h, where x is the length of the chain uncoiled.
21. A chain, of mass m and length 2l, hangs in equilibrium over a smooth pulley when an
insect of mass M alights gently at one end and begins crawling up with uniform velocity
V relative to the chain; Show that the velocity with which the chain leaves the pulley will
be
178
2
2M + m gl]1/2
2
[M
2 V +
(M + m)
M+m
22. One end of a heavy uniform chain, of length 5a and mass 5ma, is fixed at a point O
and the other passes over a small smooth peg at a distance a above O; the whole hangs in
equilibrium with the free end at a depth 2a below the peg. The free end is slightly
displaced downwards; prove that its velocity V, when the length of the free portion is x,
is given by
2
V2 = 2(x – 2a) (x + 10a)g
(x + 6a)2
and find the impulsive tension at O at the instant when the part of the chain between O
and the peg becomes tight.
23. A spherical raindrop, whose radius is 0.04 inches, begins to fall from a height of 6400
feet and during the fall its radius grows, by precipitation of moisture, at the rate of 10 -4
inches per second. If its motion be unresisted, show that its radius when it reaches the
ground is 0.0420 inches and that it will have taken about 20 seconds to fall.
24. A string of length l, hangs over a smooth peg so as to be at rest. One end is ignited
and burns away at a uniform rate v. Show that the other end will at time t be at a depth x
below the peg, where x is given by the equation (l – vt).(d2x/dt2 + g) – v(dx/dt) – 2gx = 0.
2.8 References:
1. Loney, S.L.:
An Elementary Treatise on the Dynamics of a Particle and of Rigid
Bodies, S.Chand & Company (Pvt.) Ltd, Ram Nagar, New Delhi,
1952.
2. Lamb, Horace: Dynamics: Cambridge University Press, 1961
3. Ramsey, A.S.: Dynamics Part I & II, : Cambridge University Press,1954