Find out its centre of gravity:
A square plate of uniform thickness and density is bent along M1 M2 till corner C coincides along with centre C′. Find out its centre of gravity.
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) W1 = Weight corresponding to a square plate OACB
= (36 w) at location (3, 3)
(ii) W2 = Weight corresponding to overlapped portion M1 C ′ M2
= (4.5 w) at location (4, 4)
(iii) (- W3) = Portion (M1 C M2) which is eliminated
= (- 4.5 w) at location (5, 5)
∴ Resultant W = ∑ Wi = (W1 + W2 - W3 ) = 36 w .
∴ 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) W1′ = weight of rectangular plate (BM2 M3 O)
18 w with its mass-centre at (1.5, 3)
(ii) W2′ = weight of square plate (C′′ M1 AM3)
= 9 w with its mass-centre at (4.5, 1.5)
(iii) W3′ = weight of two triangular plates (M1 M2 C′′)
= 9 w with its mass-centre at (4, 4)
Note that resultant weight W = ∑ wi = (18 + 9 + 9) w = 36 w , as before.