Locate the centre of mass of a system of bodies of masses
m 1 = 1 kg, m 2 = 2 kg and m 3 = 3 kg situated at the corners of
an equilateral triangle of side 1 m.
Answers
Answered by
1
let one side of triangle on X-axis
co ordinate of each point of triangle is
A = (0,0)
B = (1,0)
C = (1/2,√3/2)
X1 = (0*1+1*2+3*1/2)/(1+2+3)
= 7/12
Y1 = (0*1+0*2+3*√3/2)/(1+2+3)
= 3√3/12
= √3/4
hence centre of mass = (X1,Y1) = (7/12,√3/4)
co ordinate of each point of triangle is
A = (0,0)
B = (1,0)
C = (1/2,√3/2)
X1 = (0*1+1*2+3*1/2)/(1+2+3)
= 7/12
Y1 = (0*1+0*2+3*√3/2)/(1+2+3)
= 3√3/12
= √3/4
hence centre of mass = (X1,Y1) = (7/12,√3/4)
Similar questions