A triangular sheet is removed from a rectangular sheet as shown in figure. The shift of CM is

Given that a rectangular lamina with dimensions l = 60 cm and b = 20 cm as shown below has C.M at centre O

i.e. point of symmetry of the lamina

Let mass per unit area of sheet or surface density $=\rho$

Now a triangular sheet ODC is removed.

Then we get C.M of the remaining lamina will be shifted to O₁ i.e. towards heavier part.

Let the removed triangular part has C.M at O₂

Now we get obviously the C.M of O₁ and O₂ at O upon joining them again.

Let the total mass m₁ At ${\text{O}}_{\text{1}}$ = surface density of sheet × area of remaining lamina

$=\rho \left[\text{lb}-\frac{1}{2}\text{b}\times \text{h}\right]$

${\text{m}}_{\text{1}}=\rho \left[20\times 60-\frac{1}{2}\times 20\times 30\right]$

$=\rho \left[1200-300\right]$

${\text{m}}_{\text{1}}=900\rho$

Let the total mass of triangular sheet m₂ at O₂ ,

${\text{m}}_2=\rho \times \text{area}$

$=\rho \times \frac{1}{2}\times 20\times 30=300\rho$

Consider origin at O and coordinate axes

We know for a triangular plate C.M lies on median such that

OO₂ : O₂E = 2 : 1

OE = 30cm

So OO₂ = 20cm, O₂E = 10cm

We have to obtain O₁O i.e. shift of C.M after removing triangular sheet.

Let O₁O = x, OO₂ = 20cm, ${\text{m}}_{\text{1}}=900\rho$  , ${\text{m}}_2=300\rho$

If we join them again we should get C.M at (0, 0)

We can have

$\text{O}=\frac{{\text{m}}_1{x}_1+{\text{m}}_2{x}_2}{{\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}},\text{O}=\frac{{\text{m}}_1{y}_1+{\text{m}}_2{y}_2}{{\text{m}}_{\text{1}}+{\text{m}}_{\text{2}}}$

$\Rightarrow \frac{900\rho x+300\rho \times 20}{900\rho +300\rho }=0$

$\Rightarrow 900x+6000=0$

$\therefore x=\frac{-6000}{900}=-6.67\text{cm}$

$\therefore$  Shift in C.M $=-6.67\text{cm}$ after removing triangular sheet.