(^52) „„„„„ A Textbook of Engineering Mechanics
Equating the moments (iii) and (iv),
PQxPa Qb^22 +×=×()–()×
∴
22
()–()Pa Qb
x
PQ
××

• Ans.
  Example 4.9. Three forces, acting on a rigid body, are represented in magnitude, direction
  and line of action by the three sides of a triangle taken in order as shown in Fig.4.15
  Fig. 4.15.
  Prove that the forces are equivalent to a couple whose moment is equal to twice the area of the
  triangle.
  Solution. The system of forces on the triangle ABC is shown in Fig. 4.16. Now complete the
  figure as discussed below :

1. Through A draw a line EF parallel to BC.


2. Extend CA to D, such that AD is equal to Q (i.e. CA).


3. Now apply two equal and opposite forces (P) at A repre-
   sented by AE and AF.


4. Complete the parallelogram ABED with adjacent sides AB
   and AD.
   We know that the diagonal AE represents in magnitude and
   direction of the resultant of the two forces R and Q.
   Thus the force AF (equal to P) will set the forces Q and R in
   equilibrium. Thus we are left with only two forces BC (equal to P)
   and AE (equal to P) which will constitute a couple. Now from A, draw AH perpendicular to BC. Let
   this perpendicular be equal to h.
   We know that moment of the couple,
   M = P × a = P × h ...(i)

and area of triangle

11

Base Height

22

=× × =××Ph ...(ii)

From equations (i) and (ii), we find that moment of the couple

= Twice the area of triangle. Ans.

Example 4.10. A machine component of length 2.5 metres and height 1 metre is carried

upstairs by two men, who hold it by the front and back edges of its lower face.

If the machine component is inclined at 30° to the horizontal and weighs 100 N, find how

much of the weight each man supports?

Solution. Given : Length of machine component = 2.5 m; Height of the component = 1 m ;

Inclination = 30° and weight of component = 100 N

Fig. 4.16.