Question

Masses each 1 kg are placed at the vertices of an isosceles triangle ABC in which AC is equals to BC is equals to 5 cm and a b is equals to 8 cm the distance of centre of mass of the system from the vertex c is

Answer 1

The distance of centre of mass of the system from the vertex C is 5.58 cm.

Explanation:

The masses on each vertex of the isosceles triangle are equal to m1=m2=m3=1 kg.

Side AC = side BC = 5 cm

Side AB = 8 cm  

So, referring to the figure attached below,

Let’s assume the vertex C as the point of origin with coordinates = (x1,y1) = (0,0).

Then the coordinates of vertex A = (x2,y2) = (8,5) and the coordinates of vertex B (x3,y3) = (8,0).

Now,

The centre of mass at x-axis,  

Xcom = $\frac{m1x1 + m2x2 + m3x3}{m1+m2+m3}$ = $\frac{1*0 + 1*8 + 1*8}{1+1+1}$ = $\frac{16}{3}$

And,

The centre of mass at y-axis,  

Ycom = $\frac{m1y1 + m2y2 + m3y3}{m1+m2+m3}$ = $\frac{1*0 + 1*5 + 1*0}{1+1+1}$ = $\frac{5}{3}$  

∴ The coordinates of Centre of Mass of the system = (X,Y) = ($\frac{16}{3}$ ,$\frac{5}{3}$ )

Thus,  

The distance of the centre of mass of the system  from the vertex C (origin) is,

= √[(X-0)² + (Y-0)²]

= √[($\frac{16}{3}$ )² + ($\frac{5}{3}$ )²]

= √[$\frac{256}{9}$ + $\frac{25}{9}$]

= √[$\frac{281}{9}$ ]

= √[31.22]

= 5.58 cm

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