Engineering Mechanics

Chapter 3 : Moments and Their Applications „„„„„ 35

Solution. The system of given forces is shown in Fig. 3.12.
Magnitude of the resultant force
Resolving all the forces horizontally,
∑H = P – 3P = – 2P ...(i)
and now resolving all forces vertically,
∑V = 2P – 4P = – 2P ...(ii)
We know that magnitude of the resultant forces,

R= (∑HV)^22 +( )∑ =+(–2)P P2 2(–2)

= 22 P Ans.

Direction of the resultant force
Let θ = Angle, which the resultant makes with the horizontal.

∴

–2

tan 1

–2

VP

HP

∑

θ= = =

∑

or θ = 45°

Since ∑H as well as ∑V are –ve, therefore resultant lies between 180°and 270°. Thus actual
angle of the resultant force = 180° + 45° = 225° Ans.
Position of the resultant force
Let x= Perpendicular distance between A and the line of action of the resultant
force.
Now taking moments of the resultant force about A and equating the same,
22 Px Pa×= × + × = ×(2 ) (3Pa Pa) 5

∴

5

22

a

x= Ans.

Note. The moment of the forces P and 4P about the point A will be zero, as they pass
through it.

Example 3.7. ABCD is a square, each side being 20 cm and E is the middle point of AB.
Forces of 7, 8 , 12 , 5 , 9 and 6 kN act on the lines of directions AB, EC, BC, BD, CA and DE
respectively. Find the magnitude, direction and position of the resultant force.
Solution. The system of the given forces is shown in Fig. 3.13
Magnitude of resultant force

Let ∠BEC = α

We know that

∴

20

tan 2

10

α= =

2

sin 0.894

5

α= =

and

1

cos 0.447

5

α= =

Resolving all the forces horizontally,

ΣH = 8 sin α + 12 + 5 sin 45° – 9 sin 45° – 6 sin α

= (8 × 0.894) + (12) + (5 × 0.707) – (9 × 0.707) – (6 × 0.894) kN

= 10.96 kN ...(i)

Fig. 3.13.

Fig. 3.12.