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

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

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

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

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

 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