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 coordinates \( (0, 50 \, \text{cm}) \)
- Sphere B: \( m_B = 35 \, \text{kg} \) at coordinates \( (0, 0) \)
- Sphere C: \( m_C = 200 \, \text{kg} \) at coordinates \( (-80 \, \text{cm}, 0) \)
- Sphere D: \( m_D = 50 \, \text{kg} \) at coordinates \( (40 \, \text{cm}, 0) \)
### Step 2: Calculate the gravitational force exerted on B by A
The gravitational force \( F_{AB} \) between two masses is given by the formula:
\[
F = G \frac{m_1 m_2}{r^2}
\]
where \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses.
The distance \( r_{AB} \) between A and B is:
\[
r_{AB} = \sqrt{(0 - 0)^2 + (50 - 0)^2} = 0.50 \, \text{m}
\]
Now, calculate the force:
\[
F_{AB} = G \frac{m_A m_B}{r_{AB}^2} = 6.67 \times 10^{-11} \frac{40 \times 35}{(0.50)^2}
\]
\[
F_{AB} = 6.67 \times 10^{-11} \frac{1400}{0.25} = 6.67 \times 10^{-11} \times 5600 = 3.74 \times 10^{-7} \, \text{N}
\]
The direction of this force is downward (negative y-direction), so:
\[
\vec{F}_{AB} = -3.74 \times 10^{-7} \hat{j} \, \text{N}
\]
### Step 3: Calculate the gravitational force exerted on B by C
The distance \( r_{BC} \) between B and C is:
\[
r_{BC} = \sqrt{(-80 - 0)^2 + (0 - 0)^2} = 0.80 \, \text{m}
\]
Now, calculate the force:
\[
F_{BC} = G \frac{m_B m_C}{r_{BC}^2} = 6.67 \times 10^{-11} \frac{35 \times 200}{(0.80)^2}
\]
\[
F_{BC} = 6.67 \times 10^{-11} \frac{7000}{0.64} = 6.67 \times 10^{-11} \times 10937.5 = 7.30 \times 10^{-7} \, \text{N}
\]
The direction of this force is to the left (negative x-direction), so:
\[
\vec{F}_{BC} = -7.30 \times 10^{-7} \hat{i} \, \text{N}
\]
### Step 4: Calculate the gravitational force exerted on B by D
The distance \( r_{BD} \) between B and D is:
\[
r_{BD} = \sqrt{(40 - 0)^2 + (0 - 0)^2} = 0.40 \, \text{m}
\]
Now, calculate the force:
\[
F_{BD} = G \frac{m_B m_D}{r_{BD}^2} = 6.67 \times 10^{-11} \frac{35 \times 50}{(0.40)^2}
\]
\[
F_{BD} = 6.67 \times 10^{-11} \frac{1750}{0.16} = 6.67 \times 10^{-11} \times 10937.5 = 7.30 \times 10^{-7} \, \text{N}
\]
The direction of this force is to the right (positive x-direction), so:
\[
\vec{F}_{BD} = 7.30 \times 10^{-7} \hat{i} \, \text{N}
\]
### Step 5: Calculate the net gravitational force on B
Now, we can find the net force \( \vec{F}_{net} \) on sphere B by summing the forces from A, C, and D:
\[
\vec{F}_{net} = \vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{BD}
\]
\[
\vec{F}_{net} = (-3.74 \times 10^{-7} \hat{j}) + (-7.30 \times 10^{-7} \hat{i}) + (7.30 \times 10^{-7} \hat{i})
\]
\[
\vec{F}_{net} = -3.74 \times 10^{-7} \hat{j} + 0 \hat{i}
\]
\[
\vec{F}_{net} = -3.74 \times 10^{-7} \hat{j} \, \text{N}
\]
### Final Answer
The net gravitational force on sphere B due to the other spheres is:
\[
\vec{F}_{net} = -3.74 \times 10^{-7} \hat{j} \, \text{N}
\]
