Find out its centre of gravity:
A square plate of uniform thickness and density is bent along M_{1} M_{2} till corner C coincides along with centre C′. Find out its centre of gravity.
Solution
Consider w be the uniform weight of the plate per unit area. The whole plate after it is
bent can be considered to be made up of three parts.
(i) W_{1} = Weight corresponding to a square plate OACB
= (36 w) at location (3, 3)
(ii) W_{2} = Weight corresponding to overlapped portion M_{1} C ′ M_{2}
= (4.5 w) at location (4, 4)
(iii) (- W_{3}) = Portion (M_{1} C M_{2}) which is eliminated
= (- 4.5 w) at location (5, 5)
∴ Resultant W = ∑ W_{i} = (W_{1 }+ W_{2} - W_{3} ) = 36 w .
Table
∴ x¯ =103.5/36 =2.88
and, also y = 2.88 m .
The example may also be solved with other choice way by considering overall plate as follows:
(i) W_{1}′ = weight of rectangular plate (BM2 M3 O)
18 w with its mass-centre at (1.5, 3)
(ii) W_{2}′ = weight of square plate (C′′ M_{1} AM_{3})
= 9 w with its mass-centre at (4.5, 1.5)
(iii) W_{3}′ = weight of two triangular plates (M_{1} M_{2} C′′)
= 9 w with its mass-centre at (4, 4)
Note that resultant weight W = ∑ w_{i} = (18 + 9 + 9) w = 36 w , as before.