Engineering Mechanics

(^18) „„„„„ A Textbook of Engineering Mechanics
Example 2.4. A machine component 1.5 m long and weight 1000 N is supported by two
ropes AB and CD as shown in Fig. 2.2 given below.
Fig. 2.2.
Calculate the tensions T 1 and T 2 in the ropes AB and CD.
Solution. Given : Weight of the component = 1000 N
Resolving the forces horizontally (i.e., along BC) and equating the same,
T 1 cos 60° = T 2 cos 45°
∴ 1222 cos 45 0.707 1.414
cos 60 0.5
TTTT
°
=×=×=
°
...(i)
and now resolving the forces vertically,
T 1 sin 60° + T 2 sin 45° = 1000
(1.414 T 2 ) 0.866 + T 2 × 0.707 = 1000
1.93 T 2 = 1000
∴ (^2)
1000
518.1 N
1.93
T == Ans.
and T 1 = 1.414 × 518.1 = 732.6 N Ans.
2.14.METHOD OF RESOLUTION FOR THE RESULTANT FORCE

1. Resolve all the forces horizontally and find the algebraic sum of all the horizontal
   components (i.e., ∑H).


2. Resolve all the forces vertically and
   find the algebraic sum of all the
   vertical components (i.e., ∑V).


3. The resultant R of the given forces will be
   given by the equation :

(^) R= ()()∑HV^22 + ∑

4. The resultant force will be inclined at an
   angle θ, with the horizontal, such that

tan

V

H

∑

θ=

∑

Notes : The value of the angle θ will vary depending upon the values of ∑V and ∑H as

discussed below :

1. When ∑V is +ve, the resultant makes an angle between 0° and 180°. But when ∑V is –ve,
   the resultant makes an angle between 180° and 360°.


2. When ∑H is +ve, the resultant makes an angle between 0° to 90° or 270° to 360°. But
   when ∑H is –ve, the resultant makes an angle between 90° to 270°.

Component Force Vector

Resultant

Component Force Vector