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Answer :
To solve the problem of finding the magnitude of the net gravitational force exerted by the two larger masses on the 37.9 kg mass, we can follow these steps:
1. Understand the Problem: We have three masses:
- Mass 1 (m₁) = 224 kg
- Mass 2 (m₂) = 781 kg
- Mass 3 (m₃) = 37.9 kg
The distance between Mass 1 and Mass 2 is 0.32 m, and Mass 3 is placed exactly midway between them.
2. Determine Distances: Since Mass 3 is in the middle, the distance from Mass 3 to each of the other two masses is half of 0.32 m:
[tex]\[
\text{Distance from } m₃ \text{ to } m₁ \text{ and to } m₂ = \frac{0.32}{2} = 0.16 \text{ meters}
\][/tex]
3. Calculate Gravitational Forces:
The gravitational force between two masses is given by Newton's law of universal gravitation:
[tex]\[
F = G \frac{m_1 \cdot m_2}{r^2}
\][/tex]
where [tex]\( G = 6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \)[/tex].
- Force exerted by Mass 1 on Mass 3:
[tex]\[
F_{1,3} = G \frac{(224 \, \text{kg}) \cdot (37.9 \, \text{kg})}{(0.16 \, \text{m})^2} \approx 2.2126 \times 10^{-5} \, \text{N}
\][/tex]
- Force exerted by Mass 2 on Mass 3:
[tex]\[
F_{2,3} = G \frac{(781 \, \text{kg}) \cdot (37.9 \, \text{kg})}{(0.16 \, \text{m})^2} \approx 7.7145 \times 10^{-5} \, \text{N}
\][/tex]
4. Determine the Net Gravitational Force:
Since the forces are exerted in opposite directions (Mass 1 pulls towards itself and Mass 2 pulls towards itself as well), we subtract the smaller force from the larger force:
[tex]\[
F_{\text{net}} = F_{2,3} - F_{1,3} \approx 7.7145 \times 10^{-5} \, \text{N} - 2.2126 \times 10^{-5} \, \text{N} = 5.5019 \times 10^{-5} \, \text{N}
\][/tex]
The magnitude of the net gravitational force exerted by the two larger masses on the 37.9 kg mass is approximately [tex]\(5.5019 \times 10^{-5}\)[/tex] N.
1. Understand the Problem: We have three masses:
- Mass 1 (m₁) = 224 kg
- Mass 2 (m₂) = 781 kg
- Mass 3 (m₃) = 37.9 kg
The distance between Mass 1 and Mass 2 is 0.32 m, and Mass 3 is placed exactly midway between them.
2. Determine Distances: Since Mass 3 is in the middle, the distance from Mass 3 to each of the other two masses is half of 0.32 m:
[tex]\[
\text{Distance from } m₃ \text{ to } m₁ \text{ and to } m₂ = \frac{0.32}{2} = 0.16 \text{ meters}
\][/tex]
3. Calculate Gravitational Forces:
The gravitational force between two masses is given by Newton's law of universal gravitation:
[tex]\[
F = G \frac{m_1 \cdot m_2}{r^2}
\][/tex]
where [tex]\( G = 6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \)[/tex].
- Force exerted by Mass 1 on Mass 3:
[tex]\[
F_{1,3} = G \frac{(224 \, \text{kg}) \cdot (37.9 \, \text{kg})}{(0.16 \, \text{m})^2} \approx 2.2126 \times 10^{-5} \, \text{N}
\][/tex]
- Force exerted by Mass 2 on Mass 3:
[tex]\[
F_{2,3} = G \frac{(781 \, \text{kg}) \cdot (37.9 \, \text{kg})}{(0.16 \, \text{m})^2} \approx 7.7145 \times 10^{-5} \, \text{N}
\][/tex]
4. Determine the Net Gravitational Force:
Since the forces are exerted in opposite directions (Mass 1 pulls towards itself and Mass 2 pulls towards itself as well), we subtract the smaller force from the larger force:
[tex]\[
F_{\text{net}} = F_{2,3} - F_{1,3} \approx 7.7145 \times 10^{-5} \, \text{N} - 2.2126 \times 10^{-5} \, \text{N} = 5.5019 \times 10^{-5} \, \text{N}
\][/tex]
The magnitude of the net gravitational force exerted by the two larger masses on the 37.9 kg mass is approximately [tex]\(5.5019 \times 10^{-5}\)[/tex] N.
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