1. Four particles with the same mass of 0.2kg
Four particles with the same mass of 0.2kg are placed on the corners of a square with sides of 1m. The fifth particle with the same mass is placed in the center of the square. Find the force that acts on that.
Break it down:
Four particles with the same mass of 0.2kg
m1 = 0.2 kg m2 = 0.2 kg
m3 = 0.2 kg m4 = 0.2 kg
are placed on the corners of a square with sides of 1m.
L = 1m the side of a square. We use this value to calculate the distance between each mass and the fifth mass in the center, r, which is half of the length of diagonal of the square.
The fifth particle with the same mass
m5 = 0.2 kg
is placed in the center of the square. Find the force that acts on that.
The force that acts on the fifth particle is the resultant of the individual forces acting on that from each mass.
Physics.com

2. Four particles with the same mass of 0
Four particles with the same mass of 0.2kg are placed on the corners of a square with sides of 1m. The fifth particle with the same mass is placed in the center of the square. Find the force that acts on that.
Solution:
Draw a diagram m4 m1 r1 r4 r3 m5 r2 m3 m2
L = 1m
Physics.com

3. Draw free body diagram (FBD) for m5
Four particles with the same mass of 0.2kg are placed on the corners of a square with sides of 1m. The fifth particle with the same mass is placed in the center of the square. Find the force that acts on that.
Solution:
Draw free body diagram (FBD) for m5
F1 is the gravitational force between m1, and m5
F1 = G (m1m5)/ (r12) F1 F4
Calculate r1 m5
We know that diagonals of a square are perpendicular. Then we use trigonometry: r12 = (L/2)2 + (L/2)2
 r1 = (0.5)2 + (0.5)2  r1 = 0.7m F3 F2 r1
Which is half of the length of the diagonal. Then we can write:
r1 = r2 = r3 = r4 = 0.7m
L/2 = 0.5m
L/2 = 0.5m
Physics.com

4. F1 = G (m1m5)/ (r12)  F1 = G (0.2) (0.2)/ (0.7)2 = 5.4*10-12N
Four particles with the same mass of 0.2kg are placed on the corners of a square with sides of 1m. The fifth particle with the same mass is placed in the center of the square. Find the force that acts on that.
Solution:
F1 = G (m1m5)/ (r12)  F1 = G (0.2) (0.2)/ (0.7)2 = 5.4*10-12N
Because all four particles have the same mass and the distance from them to the fifth particle is the same, it is obvious that magnitude of individual gravitational forces are the same.
r1 = r2 = r3 = r4 and m1 = m2 = m3 = m4  F1 = F2 = F3 = F4
F1 = 5.4*10-12N
F2 = 5.4*10-12N
F3 = 5.4*10-12N
F4 = 5.4*10-12N
Physics.com

5. Calculate the component of the resultant force on fifth particle
Four particles with the same mass of 0.2kg are placed on the corners of a square with sides of 1m. The fifth particle with the same mass is placed in the center of the square. Find the force that acts on that.
Solution:
Calculate the component of the resultant force on fifth particle
Fx = F1x + F2x + F3x + F4x
Fy = F1y + F2y + F3y + F4y
The angle between the force F1 and axes x and y is 45 degrees. Actually this is the angle between diagonal and chosen axes in this diagram. And gravitational force F1, acts on the line between m1 and m5 which is diagonal of the square.
F1x = F1 Cos45° F2x = F2 Cos45° F3x = F3 Cos45° F4x = F4 Cos45°
F1y = F1 Sin45° F2y = F2 Sin45° F3y = F3 Sin45° F4y = F4 Sin45°
Physics.com

6.  Fx = F1 Cos45° + F2 Cos45° + (-F3 Cos45°) + (-F4 Cos45°)
Four particles with the same mass of 0.2kg are placed on the corners of a square with sides of 1m. The fifth particle with the same mass is placed in the center of the square. Find the force that acts on that.
Solution:
Fx = F1x + F2x + F3x + F4x
 Fx = F1 Cos45° + F2 Cos45° + (-F3 Cos45°) + (-F4 Cos45°)
 Fx = Cos45° (F1+ F2 - F3 - F4) = 0
Fy = F1y + F2y + F3y + F4y
 Fy = F1 Sin45° + (- F2 Sin45°) + F3 Sin45° + (-F4 Sin45°)
 Fy = Sin45° (F1- F2 + F3 - F4) = 0
Magnitude of the resultant force is zero. Because the masses
of particles are the same as well as their distances from m5,
it will stay in an equilibrium state, with no movement.
F = (Fx2 + Fy2)1/2  F = (0 + 0)1/2
 F = 0
Physics.com