Respuesta :
Answer:
[tex]x_4=-4.1031\ m[/tex]
[tex]y_4=-1.7618\ m[/tex]
Explanation:
Given:
coordinates of center of mass of the system, [tex](x_c,y_c)\equiv (0,0)[/tex]
The system comprises of the following masses with their position:
[tex]m_1=7.2\ kg\ at\ (4\ m,1\ m)[/tex]
[tex]m_2=2.3\ kg\ at\ (0\ m,4.3\ m)[/tex]
[tex]m_3=2.5\ kg\ at\ (4.4\ m, 0\ m)[/tex]
[tex]m_4=9.7\ kg\ at\ (x_4,y_4)[/tex]
We know the center of mass is given as:
[tex]x_c=\frac{m_1.x_1+m_2.x_2+m_3.x_3+m_4.x_4}{m_1+m_2+m_3+m_4}[/tex]
[tex]0=\frac{7.2\times 4+2.3\times 0+2.5\times 4.4+9.7\times x_4}{7.2+2.5+2.3+9.7}[/tex]
[tex]x_4=-4.1031\ m[/tex]
&
[tex]y_c=\frac{m_1.y_1+m_2.y_2+m_3.y_3+m_4.y_4}{m_1+m_2+m_3+m_4}[/tex]
[tex]0=\frac{7.2\times 1+2.3\times 4.3+2.5\times 0+9.7\times y_4}{7.2+2.5+2.3+9.7}[/tex]
[tex]y_4=-1.7618\ m[/tex]
Answer:
Explanation:
m1 = 7.2 kg at (4m, 1m)
m2 = 2.3 kg at (0 m, 4.3 m )
m3 = 2.5 kg at (4.4 m, 0 m)
Let the fourth mass, m4 = 9.7 kg is placed at (x' y') so that the centre of mass is at (0, 0) of the system of masses.
use the formula of centre of mass
[tex]x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}+m_{4}x_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}[/tex]
[tex]y_{cm}=\frac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}+m_{4}y_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}[/tex]
Here, xcm, ycm = (0, 0)
[tex]0=\frac{7.2\times 4+ 2.3\times 0 + 2.5 \times 4.4+ 9.7 \times x'}{7.2 + 2.3 + 2.5 + 9.7}[/tex]
21.7 = 28.8 + 11 + 9.7 x'
- 18.1 = 9.7 x'
x' = - 1.87 m
[tex]0=\frac{7.2\times 1+ 2.3\times 4.3 + 2.5 \times 0+ 9.7 \times y'}{7.2 + 2.3 + 2.5 + 9.7}[/tex]
21.7 = 7.2 + 9.89 + 9.7 y'
4.61 = 9.7 y'
y' = 0.475 m
