Question

find the position of centre of mass if two Kg 3 kg and 4 kg masses are kept at vertex of equilateral triangle of side 50 centimetre​

Answer 1

Answer:-  $(\frac{5}{18},\frac{\sqrt{3}}{9})$

Step by Step Explanation:-

Let take a triangle and suppose the position of masses are shown in figure.

Now,

AB = BC = AC= 50 cm = 1/2 m

A = (0,0)

B = (1/2 , 0)

C = ( 1/4 , √3/4)  [Explained later]

Draw a perpendicular from  point C on AB.

E is mid-point of AB, So coordinates of E is (1/4 , 0)

Now, in ΔAEC ,

tan60° = EC/AE

√3 = EC/(1/4))

√3 = 4EC

EC = √3/4

Hence,

Coordinates of C is  (1/4 , √3/4)

Now, again We define A , B and C

A = (x₁ , y₁ ) = (0 , 0)

B = (x₂ , y₂) = (1/2 , 0)

C = (x₃ , y₃) = (1/4 , √3/4)

m₁ = 2

m₂ = 3

m₃ = 4

We know  that,

$x_{COM}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}\\\;\\x_{COM}=\frac{(2\times0)+(3\times\frac{1}{2})+(4\times\frac{1}{4})}{2+3+4}\\\;\\x_{COM}=\frac{\frac{3}{2}+1}{9}\\\;\\x_{COM}=\frac{5}{18}$

Similarly,

$y_{COM}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}\\\;\\y_{COM}=\frac{(2\times0)+(3\times0)+(4\times\frac{\sqrt{3}}{4})}{2+3+4}\\\;\\y_{COM}=\frac{\sqrt{3}}{9}$

Hence coordinates of center of mass = $(\frac{5}{18},\frac{\sqrt{3}}{9})$