Step-by-step explanation:
Let the small mass be m and position on x axis be y.
A 20 kg sphere is at the origin.
Distance to 20 kg mass = y
A 10 kg sphere is at x = 20 cm
Distance to 10 kg sphere = 20 - y
We have forces between them are equal
We have gravitational force
[tex]F=\frac{GMm}{r^2}[/tex]
Where G = 6.67 x 10⁻¹¹ N m²/kg²
M = Mass of body 1
M = Mass of body 2
r = Distance between them
Here we have
[tex]\frac{G\times 20\times m}{y^2}=\frac{G\times 10\times m}{(20-y)^2}\\\\800-80y+2y^2=y^2\\\\y^2-80y+800=0\\\\y=68.28cm\texttt{ or }y=11.72cm[/tex]
So the small mass should be placed at x = 11.72 cm or x = 68.28 cm
The positions on the x-axis obtained using quadratic equation such that net gravitational force is zero are 68.28 and 11.72 cm respectively.
Recall the force of universal gravitational attraction :
Sphere 1 :
Sphere 2 :
Gravitational force ; sphere 1 :
Gravitational force ; sphere 2 :
Equate (1) and (2)
[tex] \frac{G \times 20 \times m_{2}}{d^{2}} = \frac{G \times 10 \times m_{2}}{(20 - d)^{2}} [/tex]
[tex] 20(20 - d)^{2} = 10 \times d^{2} [tex]
20(400 - 40d + d²) = 10d²
8000 - 800d + 20d² = 10d²
10d² - 800d + 8000 = 0
Divide through by 10
d² - 80d + 800 = 0
Using a quadratic equation solver :
d = 68.28 cm or d = 11.72 cm
The possible positions are 68.28 cm and 11.72 cm
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