Answer:
[tex]1.32391\times 10^{-7}\ N[/tex]
2.69 m
Explanation:
M = Mass in the middle = 35 kg
[tex]m_1[/tex] = 600 kg
[tex]m_2[/tex] = 300 kg
r = Distance
The net force between the objects would be
[tex]F=\dfrac{GMm_1}{r^2}-\dfrac{GMm_2}{r^2}\\\Rightarrow F=\dfrac{GM(m_1-m_2)}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 35(600-300)}{2.3^2}\\\Rightarrow F=1.32391\times 10^{-7}\ N[/tex]
The magnitude of the net gravitational force exerted by these objects is [tex]1.32391\times 10^{-7}\ N[/tex]
Now,
[tex]\dfrac{GMm_1}{r^2}=\dfrac{GMm_2}{(4.6-r)^2}\\\Rightarrow \dfrac{m_1}{r^2}=\dfrac{m_2}{(4.6-r)^2}\\\Rightarrow r^2-9.2r+21.16=\dfrac{m_2}{m_1}r^2\\\Rightarrow r^2-9.2r+21.16=\dfrac{300}{600}r^2\\\Rightarrow r^2-9.2r+21.16=0.5r^2\\\Rightarrow 0.5r^2-9.2r+21.16=0\\\Rightarrow 50r^2-920r+2116=0[/tex]
Solving the above equation we get
[tex]r=\frac{-\left(-920\right)+\sqrt{\left(-920\right)^2-4\cdot \:50\cdot \:2116}}{2\cdot \:50}, \frac{-\left(-920\right)-\sqrt{\left(-920\right)^2-4\cdot \:50\cdot \:2116}}{2\cdot \:50}\\\Rightarrow r=15.7, 2.69[/tex]
So, the distance at which the force will ben zero is 2.69 m
Answer:
Explanation:
m1 = 300 kg
m2 = 600 kg
d = 4.6 m
(a) m = 35 kg
Let the force between m1 and m is F1. Use the Newton's gravitation law
[tex]F_{1}=\frac{Gm_{1}m}{\left (0.5d \right )^{2}}[/tex]
[tex]F_{1}=\frac{6.67\times 10^{-11}\times 300\times 35}{2.3\times 2.3}[/tex]
F1 = 1.324 x 10^-7 N (towards m1, i.e., leftwards)
Let the force between m2 and m is F2. Use the Newton's gravitation law
[tex]F_{1}=\frac{Gm_{2}m}{\left (0.5d \right )^{2}}[/tex]
[tex]F_{1}=\frac{6.67\times 10^{-11}\times 600\times 35}{2.3\times 2.3}[/tex]
F1 = 2.65 x 10^-7 N (towards m2, i.e., rightwards)
Net force on m is
F = F2 - F1
F = (2.65 - 1.324) x 10^-7
F = 1.33 x 10^-7 N towards right
(b) Let it is placed at a distance r from m1 so that the net force is zero.
[tex]\frac{Gm_{1}m}{r^{2}}=\frac{Gm_{2}m}{\left ( d-r \right )^{2}}[/tex]
[tex]\frac{300}{r^{2}}=\frac{600}{\left ( d-r \right )^{2}}[/tex]
1.414 r = 4.6 - r
2.4 r = 4.6
r = 1.92 m
Thus, the net force on 35 kg is zero at a distance of 1.92 m from 300 kg.
