Answer to Question #232999 in Mechanics | Relativity for Cee

Question

Three particles of masses 3.724kg, 3.236 kg, and 19.86 kg form an equilateral triangle of edge length 289 cm. If the 3.724kg mass is at the origin, and the 3.236kg mass is along the x- axis, and the 19.86kg mass is in the positive x, and positive y quadrant, the centre of mass of the whole system will be at

Answer 1

$\text{Let }A,B,C \text{ the points where three particles are located}$

$A(0,0);m_a= 3.724 kg$

$B(2.89,0);m_b=3.236kg$

$\text{Let's draw the height of the } CH \text{ to the } AB \text{ side}$

$CH \text{ is also the median since the triangle is equilateral}$

$AH = \frac{AB}{2}= \frac{2.89}{2}=1.445m$

$CH =\sqrt{AC^2-AH^2}=\sqrt{2.89^2-1.445^2}\approx2.5m$

$C(1.445,2.5);m_c=19.86$

$\text{Let point } O \text{ be the center of gravity of the particles}$

$x_o= \frac{x_am_a+x_bm_b+x_cm_c}{m_a+m_b+m_c}$

$x_o =\frac{0*3.724+2.89*3.236+1.445*19.86}{3.724+3.236+19.86}\approx1.42m$

$y_o= \frac{y_am_a+y_bm_b+y_cm_c}{m_a+m_b+m_c}$

$x_o =\frac{0*3.724+0*3.236+2.5*19.86}{3.724+3.236+19.86}\approx1.85m$

$O(1.42;1.85)$

$\text{Answer : }O(1.42;1.85) -\text{coordinates of the center of gravity in meters}$