Answer :
Sure! To find the magnitude of the net gravitational force exerted by the two larger masses on the 38.4 kg mass, we begin by using Newton's law of universal gravitation, which states:
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( G = 6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \)[/tex] is the universal gravitational constant,
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses interacting, and
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
In this problem:
- Mass 1 ([tex]\( m_1 \)[/tex]) is 138 kg,
- Mass 2 ([tex]\( m_2 \)[/tex]) is 688 kg,
- The third mass ([tex]\( m_3 \)[/tex]) is 38.4 kg, placed midway between the other two masses,
- The total distance between Mass 1 and Mass 2 is 0.387 m, so the distance from Mass 3 to each of the other two masses is half of this, i.e., 0.387 m / 2 = 0.1935 m.
### Step 1: Calculate the gravitational force exerted by Mass 1 on Mass 3.
Using the formula:
[tex]\[ F_1 = \frac{G \cdot m_1 \cdot m_3}{r_1^2} \][/tex]
Substitute the known values:
[tex]\[ F_1 = \frac{6.672 \times 10^{-11} \cdot 138 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_1 = 9.44 \times 10^{-6} \, \text{N} \][/tex]
### Step 2: Calculate the gravitational force exerted by Mass 2 on Mass 3.
Using the formula:
[tex]\[ F_2 = \frac{G \cdot m_2 \cdot m_3}{r_2^2} \][/tex]
Substitute the known values:
[tex]\[ F_2 = \frac{6.672 \times 10^{-11} \cdot 688 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_2 = 4.71 \times 10^{-5} \, \text{N} \][/tex]
### Step 3: Determine the net gravitational force on Mass 3.
Since the forces exerted by the two larger masses on the 38.4 kg mass are opposite in direction, the net force is the difference between the two forces:
[tex]\[ F_{\text{net}} = F_2 - F_1 \][/tex]
So:
[tex]\[ F_{\text{net}} = 4.71 \times 10^{-5} \, \text{N} - 9.44 \times 10^{-6} \, \text{N} \][/tex]
This results in:
[tex]\[ F_{\text{net}} = 3.76 \times 10^{-5} \, \text{N} \][/tex]
Thus, the magnitude of the net gravitational force exerted on the 38.4 kg mass is approximately [tex]\( 3.76 \times 10^{-5} \, \text{N} \)[/tex].
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( G = 6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \)[/tex] is the universal gravitational constant,
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses interacting, and
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
In this problem:
- Mass 1 ([tex]\( m_1 \)[/tex]) is 138 kg,
- Mass 2 ([tex]\( m_2 \)[/tex]) is 688 kg,
- The third mass ([tex]\( m_3 \)[/tex]) is 38.4 kg, placed midway between the other two masses,
- The total distance between Mass 1 and Mass 2 is 0.387 m, so the distance from Mass 3 to each of the other two masses is half of this, i.e., 0.387 m / 2 = 0.1935 m.
### Step 1: Calculate the gravitational force exerted by Mass 1 on Mass 3.
Using the formula:
[tex]\[ F_1 = \frac{G \cdot m_1 \cdot m_3}{r_1^2} \][/tex]
Substitute the known values:
[tex]\[ F_1 = \frac{6.672 \times 10^{-11} \cdot 138 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_1 = 9.44 \times 10^{-6} \, \text{N} \][/tex]
### Step 2: Calculate the gravitational force exerted by Mass 2 on Mass 3.
Using the formula:
[tex]\[ F_2 = \frac{G \cdot m_2 \cdot m_3}{r_2^2} \][/tex]
Substitute the known values:
[tex]\[ F_2 = \frac{6.672 \times 10^{-11} \cdot 688 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_2 = 4.71 \times 10^{-5} \, \text{N} \][/tex]
### Step 3: Determine the net gravitational force on Mass 3.
Since the forces exerted by the two larger masses on the 38.4 kg mass are opposite in direction, the net force is the difference between the two forces:
[tex]\[ F_{\text{net}} = F_2 - F_1 \][/tex]
So:
[tex]\[ F_{\text{net}} = 4.71 \times 10^{-5} \, \text{N} - 9.44 \times 10^{-6} \, \text{N} \][/tex]
This results in:
[tex]\[ F_{\text{net}} = 3.76 \times 10^{-5} \, \text{N} \][/tex]
Thus, the magnitude of the net gravitational force exerted on the 38.4 kg mass is approximately [tex]\( 3.76 \times 10^{-5} \, \text{N} \)[/tex].
