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Answer :
To find the magnitude of the net gravitational force exerted on the 61.2 kg mass by the two larger masses (193 kg and 348 kg), we can follow these steps:
1. Identify the Given Values:
- Mass of object 1 ([tex]\(m_1\)[/tex]) = 193 kg
- Mass of object 2 ([tex]\(m_2\)[/tex]) = 348 kg
- Mass of object 3 ([tex]\(m_3\)[/tex]) = 61.2 kg
- Distance between [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] = 0.328 m
- Universal gravitational constant ([tex]\(G\)[/tex]) = [tex]\(6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\)[/tex]
2. Determine the Position of [tex]\(m_3\)[/tex]:
- Since [tex]\(m_3\)[/tex] is placed midway between [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex], the distance from [tex]\(m_3\)[/tex] to both [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] is half of the distance between [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex].
- Distance from [tex]\(m_3\)[/tex] to [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] = 0.328 m / 2 = 0.164 m
3. Calculate the Gravitational Force Exerted by [tex]\(m_1\)[/tex] on [tex]\(m_3\)[/tex]:
[tex]\[
F_{m1\_m3} = \frac{G \cdot m_1 \cdot m_3}{(0.164)^2}
\][/tex]
[tex]\[
F_{m1\_m3} \approx 2.9300637715645445 \times 10^{-5} \, \text{N}
\][/tex]
4. Calculate the Gravitational Force Exerted by [tex]\(m_2\)[/tex] on [tex]\(m_3\)[/tex]:
[tex]\[
F_{m2\_m3} = \frac{G \cdot m_2 \cdot m_3}{(0.164)^2}
\][/tex]
[tex]\[
F_{m2\_m3} \approx 5.283223795359903 \times 10^{-5} \, \text{N}
\][/tex]
5. Determine the Net Gravitational Force:
- Since both forces act in opposite directions, the net gravitational force is the difference in magnitudes of these forces.
[tex]\[
F_{\text{net}} = |F_{m1\_m3} - F_{m2\_m3}|
\][/tex]
[tex]\[
F_{\text{net}} \approx 2.3531600237953585 \times 10^{-5} \, \text{N}
\][/tex]
Thus, the magnitude of the net gravitational force exerted by the two larger masses on the 61.2 kg mass is approximately [tex]\(2.3531 \times 10^{-5} \, \text{N}\)[/tex].
1. Identify the Given Values:
- Mass of object 1 ([tex]\(m_1\)[/tex]) = 193 kg
- Mass of object 2 ([tex]\(m_2\)[/tex]) = 348 kg
- Mass of object 3 ([tex]\(m_3\)[/tex]) = 61.2 kg
- Distance between [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] = 0.328 m
- Universal gravitational constant ([tex]\(G\)[/tex]) = [tex]\(6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\)[/tex]
2. Determine the Position of [tex]\(m_3\)[/tex]:
- Since [tex]\(m_3\)[/tex] is placed midway between [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex], the distance from [tex]\(m_3\)[/tex] to both [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] is half of the distance between [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex].
- Distance from [tex]\(m_3\)[/tex] to [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] = 0.328 m / 2 = 0.164 m
3. Calculate the Gravitational Force Exerted by [tex]\(m_1\)[/tex] on [tex]\(m_3\)[/tex]:
[tex]\[
F_{m1\_m3} = \frac{G \cdot m_1 \cdot m_3}{(0.164)^2}
\][/tex]
[tex]\[
F_{m1\_m3} \approx 2.9300637715645445 \times 10^{-5} \, \text{N}
\][/tex]
4. Calculate the Gravitational Force Exerted by [tex]\(m_2\)[/tex] on [tex]\(m_3\)[/tex]:
[tex]\[
F_{m2\_m3} = \frac{G \cdot m_2 \cdot m_3}{(0.164)^2}
\][/tex]
[tex]\[
F_{m2\_m3} \approx 5.283223795359903 \times 10^{-5} \, \text{N}
\][/tex]
5. Determine the Net Gravitational Force:
- Since both forces act in opposite directions, the net gravitational force is the difference in magnitudes of these forces.
[tex]\[
F_{\text{net}} = |F_{m1\_m3} - F_{m2\_m3}|
\][/tex]
[tex]\[
F_{\text{net}} \approx 2.3531600237953585 \times 10^{-5} \, \text{N}
\][/tex]
Thus, the magnitude of the net gravitational force exerted by the two larger masses on the 61.2 kg mass is approximately [tex]\(2.3531 \times 10^{-5} \, \text{N}\)[/tex].
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