rigid bodies equilibrium-engg mechanics

1. RIGID BODIES EQUILIBRIUM
 A rigid body is an idealization of a solid body where the deformations occurring on
the body are neglected.
 In other words, the distance between any two given points of a rigid body remains
constant regardless of the external force acting upon it.
 Principle of force transmissibility will be applied, which is
The principle of transmissibility states that the point of application of a force
can be moved anywhere along its line of action without changing the external reaction
forces on a rigid body.
OR, The principle of transmissibility states that the conditions of equilibrium or
motion of a rigid body will remain unchanged if a force F acting at a given point of the
rigid body is replaced by a force F’ of the same magnitude and same direction, but
acting at a different point, provided that the two forces have the same line of action.

3. A.B = ABcos𝜃
If A & B are solved as per their rectangular components, then,
A.B = 𝐴𝑥𝑖 + 𝐴𝑦𝑗 + 𝐴𝑧𝑘 . (𝐵𝑥𝑖 + 𝐵𝑦𝑗 + 𝐵𝑍𝑘)
= 𝐴𝑥𝐵𝑥 + 𝐴𝑦𝐵𝑦+𝐴𝑧𝐵𝑍
∵ 𝑖. 𝑖 = 𝑗. 𝑗 = 𝑘. 𝑘 = 1 &
𝑖. 𝑗 = 𝑗. 𝑘 = 𝑘. 𝑖 = 0

4. A× 𝐵 = 𝐶 = 𝐴𝐵𝑠𝑖𝑛𝜃
𝐴 × 𝐵 = 𝐴𝑥𝑖 + 𝐴𝑦𝑗 + 𝐴𝑧𝑘 × 𝐵𝑥𝑖 + 𝐵𝑦𝑗 + 𝐵𝑍𝑘
A× 𝐵 = - (B× 𝐴)
; A× 𝐵 ≠ B × 𝐴

6. Moment of a Force about a point

7. Or,

8. 𝑴𝒐 = 𝒓 × 𝑭
𝐹 = 𝐹𝑥𝑖 + 𝐹𝑦𝑗 + 𝐹𝑧𝑘
r = dx𝑖 + 𝑑𝑦𝑗 + 𝑑𝑧𝑘 𝑠𝑖𝑛𝑐𝑒, 𝑑𝑥 = 𝑥 − 0 = 𝑥;
𝑑𝑦 = 𝑦 − 0 = 𝑦;
𝑑𝑧 = 𝑧 − 0 = 𝑧
‘r’ will be,
r = x𝑖 + 𝑦𝑗 + 𝑧𝑘
Hence, 𝑴𝒐 is
𝑴𝒐 =
𝑖 𝑗 𝑘
𝑥 𝑦 𝑧
𝐹𝑥 𝐹𝑦 𝐹𝑧
= 𝑦𝐹𝑧 − 𝑧𝐹𝑦 𝑖 + 𝑧𝐹𝑥 − 𝑥𝐹𝑧 𝑗 + 𝑥𝐹𝑦 − 𝑦𝐹𝑥 𝑘
And also, 𝑴𝒐 = 𝑀𝑥𝑖 + 𝑀𝑦𝑗 + 𝑀𝑧𝑘
(0,0,0)

9. 𝑴𝑩 = 𝒓𝑨𝑩 × 𝑭
𝒓𝑨𝑩 = (𝑥𝐴 − 𝑥𝐵)𝑖 + (𝑦𝐴 − 𝑦𝐵)𝑗 + (𝑧𝐴 − 𝑧𝐵)𝑘
𝐹 = 𝐹𝑥𝑖 + 𝐹𝑦𝑗 + 𝐹𝑧𝑘
Hence, 𝑴𝑩 is
𝑴𝑩 =
𝑖 𝑗 𝑘
(𝑥𝐴−𝑥𝐵) (𝑦𝐴 − 𝑦𝐵) (𝑧𝐴 − 𝑧𝐵)
𝐹𝑥 𝐹𝑦 𝐹𝑧
And also, 𝑴𝑩= 𝑀𝑥𝑖 + 𝑀𝑦𝑗 + 𝑀𝑧𝑘

10. 𝑴𝒐 = 𝒓 × 𝑭
𝑴𝑶𝑳 = 𝝀. 𝑴𝒐 = 𝝀. (𝒓 × 𝑭)

19. B
= =
= =
B
Actual system
Actual system
Actual system
Equivalent system
Equivalent system
Equivalent system