Page 154 - WCS - Electrical

WORKSHOP CALCULATION & SCIENCE - CITS

Arm of a Couple

The perpendicular distance between the lines of action of the two equal and opposite parallel forces, is known as
“arm of the couple” as shown in the figure
Moment of a Couple

The moment of a couple is the product of the force i.e., one of the forces of the two equal and opposite parallel
forces, and the arm of the couple.
Moment of a couple = Magnitude of the force x Arm of the couple
The couple may be clockwise or anti clockwise, but the algebraic sum of the forces constituting the couple is zero.
But it produces a motion of rotation in the body.
When two forces are acting simultaneously on a particle, be represented by the two adjacent sides of a
parallelogram in magnitude and direction, their resultant may be represented in magnitude and direction by the
diagonal of the parallelogram which passes through their point of intersection.
Let P and Q be two forces acting at a point O, shown in the figure and the angle between them is 0.
In the diagram, it is shown: Q.Sin θ
BC
P
=
OB represents force P OA represents force Q tan α = P + Q.cos θ P BC Q.Sin θ
Q
AC
=
Q
Draw parallel lines AC and BC which are parallel to OB and OA respectively. The parallel lines AC and BC tan α = P + Q.cos θ
AC
P
BC
intersect at point C. Join points O and C. Now the line OC which is known as diagonal of parallelogram represents
P× AC = Q× BC =
the resultant (R) of two forces P and Q. Distance covered BC P = BC
Q
AC

Q×
P×
AC =
Speed =
Let US find below the magnitudes and direction of resultant of two forces. Q AC Distance covered

QxBC
Time
P =
AC
In the figure drawn, it is shown that P and Q are two forces acting at an angle of theta in between them. A is the Speed = Time
QxBC

P =
resultant force of P and Q. a is the angle of resultant force R, with the force P i.e , α is the angle between R and
AC

Definite

Distance
direction
PxAC

Velocity =
P. Project the Line OB and drop perpendicular line CD on it so that the BD and CD meet at point D.
P =
BC Time PxAC Distance Definite direction

P =
Let line BC represent the force Q as BC is equal and perpendicular to OA In the triangle CBD: Velocity = Time
P
BC
BC

Q.Sin
θ
BC
P
=
CD
∴CD=Q.sin θ = tan α = tan α = Q.Sin θ
Q
AC
Sin θ = Q P AC P + Q.cos θ P + Q.cos θ
BC

Q P BC = a = change in velocity m/sec tan = Q.Sin θ

2 CD

θ
α
Q.Sin
tan
BD P BC Q = Q AC Time Sin θ = Q α = P + Q.cos θ change in velocity
AC
P× AC 2 Q 2 BC Q 2 AC P× AC = Q× BC P = BC P + Q.cos θ a = m/sec 2
OD = OD + CD =×=
Cos
θ =
Q
Q
AC
BD
Distance
covered

BC
P
covered

Distance
=(OB+BD) + CD 2 ∴CD=Q.sin θ Time
θ =
Cos
2
Speed =

P BC
BC =
AC =
Q×
P×
QxBC P× AC = Q× BC = AC 1 Time Speed =
Q
QxBC Q
2
Time
covered
Distance
AC
Q

P = =OB + 2.OB.BD+BD + CD = 2 Let us consider triangel COD

P
2
2 AC
2
Distance
R = P + Q + 2.P.Q cos θ AC 2 Speed = covered 1
QxBC
Speed =

Time

=P + 2.P.O Cos ϴ +O 2 As per pythagoras theorem
P =
2
QxBC
Time
2
2
P = PxAC P = PxAC Velocity = Distan R = De ce P + direction finite Q + 2.P.Q cos θ Definite direction 2
Distance
AC

Velocity =
tan α = =OB+P CDCD BC P = AC BC F = m  (v −u)  Time Time

=



direction
Distance
Definite

PxAC
OD
OB +
BD


CDocity =
CDnce
BD= O Cos ϴ P = PxAC  t tanVel Vel = Dista Definite direction  (v −u) 
P =
α =ocity =
Time
CD 2 2 BC BC OD OB + BD Time F = m   t  
CD
=BD + CD = O
θ
2
Sin = θ Q Sin θ = change in velocity
Q.Sin
= Q a = m/sec 2 change in velocity

P + =OC+R θ Q.cos CD Q.Sin θ m/sec 2
a =
Sin
θ
BD Si θ n = CD = BD Q Time = Time

Cos
θ =
=P + 2.P.O.Cos ϴ+O
Therefore R Cos θ = Q 2 P + Q.cos θ = change in velocity m/sec 2
2
2
change

velocity

in
a
Q
tan α = Q.Sin θ Q a = Time m/sec 2
BD
=P + O + 2.P.O.Cos ϴ
2
2
θ
P + Q.cos θ R Cos Cos BD = 2 1 Q.Sin θ Time
θ =
2 2 Q Q tan α = 1
R = P + Q + 2.P.Q cos R = P + Q + 2.P.Q cos θ 2 P + Q.cos θ 2 1
θ
2
2
2 2 2 2 1 2
2.P.Q
2 Q ++
cos
θ
R = P + Q + CD 2.P.Q CD R = R = 2 P Q + 2.P.Q cos θ cos θ 2
P +

Let = Angle between R and P CD CD F = m  (v −u)   (v −u) 
tan
=
α =

2 
= m

F
2
OD OB + BD tan α = OD = OB + BD  R = t P + Q + 2.P.Q cos θ t 

 (v
tan α CD = CD = CD CD F  (v  m  −u)  −u)   
=
Q.Sin θ tan α = OD = OD OB + BD F = m    t 
BD
OB +
= = Q.Sin θ  t 
P + Q.cos θ P + Q.cos θ
Q.Sin
Q.Sin θ θ
=
=
Q.cos
tan α = Q.Sin θ P + Q.cos θ θ θ
P +
Q.Sin
P + Q.cos θ tan α = P + Q.cos θ
θ
tan α = Q.Sin θ
Q.Sin
tan α = P + Q.cos θ
2 2 P + Q.cos θ 141
R = P + Q + 2.P.Q cos R = P + Q + CITS : WCS - Electrical - Exercise 13
θ
2
2
cos

θ
2.P.Q
R = 2 P 2 2 Q ++ 2 2.P.Q cos θ
R = P + Q + 2.P.Q cos θ