Four uniform spheres, with masses `m_A=40 kg, m_B=35 kg , m_C=200 kg` , and `m_D` = 50 kg, have (x,y) coordinates of (0, 50 cm), (0, 0), (–80 cm, 0), and (40 cm, 0), respectively. In unit-vector notation, what is the net gravitational force on sphere B due to the other spheres?

Options

A. `(2.6 xx 10^(-5)N)hatj`
B. `(7.1xx10^(-3)N)hatj`
C. `(3.7xx10^(-7)N)hatj`
D. `(4.9xx10^(-4) N )hatj`

Explanation

To find the net gravitational force on sphere B due to the other spheres (A, C, and D), we will follow these steps:

### Step 1: Identify the positions and masses of the spheres
- Sphere A: \( m_A = 40 \, \text{kg} \) at \( (0, 50 \, \text{cm}) = (0, 0.5 \, \text{m}) \)
- Sphere B: \( m_B = 35 \, \text{kg} \) at \( (0, 0) \)
- Sphere C: \( m_C = 200 \, \text{kg} \) at \( (-80 \, \text{cm}, 0) = (-0.8 \, \text{m}, 0) \)
- Sphere D: \( m_D = 50 \, \text{kg} \) at \( (40 \, \text{cm}, 0) = (0.4 \, \text{m}, 0) \)

### Step 2: Calculate the gravitational force between B and A
The gravitational force between two masses is given by:
\[ F = G \frac{m_1 m_2}{r^2} \]
where \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \).

The distance \( r_{BA} \) between B and A is:
\[ r_{BA} = 0.5 \, \text{m} - 0 \, \text{m} = 0.5 \, \text{m} \]
Thus, the force \( F_{BA} \) is:
\[ F_{BA} = G \frac{m_B m_A}{r_{BA}^2} = 6.674 \times 10^{-11} \frac{35 \times 40}{(0.5)^2} \]
Calculating this gives:
\[ F_{BA} = 6.674 \times 10^{-11} \frac{1400}{0.25} = 6.674 \times 10^{-11} \times 5600 = 3.74 \times 10^{-7} \, \text{N} \]

### Step 3: Calculate the gravitational force between B and C
The distance \( r_{BC} \) between B and C is:
\[ r_{BC} = |0 - (-0.8)| = 0.8 \, \text{m} \]
Thus, the force \( F_{BC} \) is:
\[ F_{BC} = G \frac{m_B m_C}{r_{BC}^2} = 6.674 \times 10^{-11} \frac{35 \times 200}{(0.8)^2} \]
Calculating this gives:
\[ F_{BC} = 6.674 \times 10^{-11} \frac{7000}{0.64} = 6.674 \times 10^{-11} \times 10937.5 = 7.31 \times 10^{-7} \, \text{N} \]

### Step 4: Calculate the gravitational force between B and D
The distance \( r_{BD} \) between B and D is:
\[ r_{BD} = |0 - 0.4| = 0.4 \, \text{m} \]
Thus, the force \( F_{BD} \) is:
\[ F_{BD} = G \frac{m_B m_D}{r_{BD}^2} = 6.674 \times 10^{-11} \frac{35 \times 50}{(0.4)^2} \]
Calculating this gives:
\[ F_{BD} = 6.674 \times 10^{-11} \frac{1750}{0.16} = 6.674 \times 10^{-11} \times 10937.5 = 7.31 \times 10^{-7} \, \text{N} \]

### Step 5: Determine the direction of the forces
- The force \( F_{BA} \) acts upwards (positive y-direction).
- The force \( F_{BC} \) acts to the left (negative x-direction).
- The force \( F_{BD} \) acts to the right (positive x-direction).

### Step 6: Calculate the net gravitational force on B
Since \( F_{BC} \) and \( F_{BD} \) are equal in magnitude but opposite in direction, they will cancel each other out. Therefore, the net force on B will be solely due to A:
\[ \vec{F}_{net} = F_{BA} \hat{j} = 3.74 \times 10^{-7} \hat{j} \, \text{N} \]

### Final Answer
The net gravitational force on sphere B due to the other spheres is:
\[ \vec{F}_{net} = 3.74 \times 10^{-7} \hat{j} \, \text{N} \]