CoG center

Find the position of the center of gravity of a system of four mass points having masses, m₁, m₂ = 2 m1, m₃ = 3 m1, and m₄ = 4 m₁, if they lie at the vertices of an isosceles tetrahedron. (in all cases, between adjacent material points, the distance is a=1. )

Correct answer:

x = -0.05
y = -0.0816
z = 0.7348

Step-by-step explanation:

m_{1} = 1 m_{2} = 2 m_{3} = 3 m_{4} = 4 a = 1 h_{33} = 3 / 2 \cdot a h = a \cdot 6 / 3 = 1 \cdot 6 / 3 ≐ 0.8165 h_{1} = \frac{2}{3} \cdot h = \frac{2}{3} \cdot 0.8165 ≐ 0.5443 h_{2} = h - h_{1} = 0.8165 - 0.5443 ≐ 0.2722 A = (0, 0, 0) B = (0, h_{1}, h) = (0, 0.54433105395182, 0.81649658092773) C = (a / 2, - h_{2}, h) = (0.5, - 0.27216552697591, 0.81649658092773) D = (- a / 2, - h_{2}, h) = (- 0.5, - 0.27216552697591, 0.81649658092773) m = m_{1} + m_{2} + m_{3} + m_{4} = 1 + 2 + 3 + 4 = 10 x = \frac{m _{1} \cdot A _{x} + m _{2} \cdot B _{x} + m _{3} \cdot C _{x} + m _{4} \cdot D _{x}}{m} = \frac{1 \cdot 0 + 2 \cdot 0 + 3 \cdot 0 . 5 + 4 \cdot ( - 0 . 5 )}{1 0} = - \frac{1}{2 0} = - 0.05

y = \frac{m _{1} \cdot A _{y} + m _{2} \cdot B _{y} + m _{3} \cdot C _{y} + m _{4} \cdot D _{y}}{m} = \frac{1 \cdot 0 + 2 \cdot 0 . 5 4 4 3 + 3 \cdot ( - 0 . 2 7 2 2 ) + 4 \cdot ( - 0 . 2 7 2 2 )}{1 0} = - 0.0816

z = \frac{m _{1} \cdot A _{z} + m _{2} \cdot B _{z} + m _{3} \cdot C _{z} + m _{4} \cdot D _{z}}{m} = \frac{1 \cdot 0 + 2 \cdot 0 . 8 1 6 5 + 3 \cdot 0 . 8 1 6 5 + 4 \cdot 0 . 8 1 6 5}{1 0} = 0.7348

Did you find an error or inaccuracy? Feel free to write us. Thank you!

Tips for related online calculators

Looking for a statistical calculator?
Our vector sum calculator can add two vectors given by their magnitudes and by included angle.
The Pythagorean theorem is the base for the right triangle calculator.
Calculation of an isosceles triangle.
See also our trigonometric triangle calculator.