1. The problem statement, all variables and given/known data Three lead spheres, of mass 10.0 kg each, are located at three corners of a square of ice length 45.0 cm, as shown. A bead is released at the forth corner. By considering the gravitational forces among the four objects only, determine the magnitude and direction of the acceleration of the bead when released. 2. Relevant equations F=Gm1m2/r^2 F=ma 3. The attempt at a solution Each of the spheres applies a gravitation force on the particle in the y direction and/or x direction. F = G m1m2/r^2 where r is the distance between the bead and lead sphere. The distance between the bead and the sphere directly above it is 45 cm or 0.45 m (this is r) The sphere directly above applies a force only in the positive y direction. This force is... F = G m1m2/r^2 where r = 0.45 m and G = 6.67408 * 10 ^-11 and m1 = mass of bead and m2 is mass of the given sphere F = (6.67408 * 10^-11)(m1)(10)/0.45^2 F = 6.67408*10^-10 * m1 / 0.2025 --> However I do not know the mass of the bead, so I continued on anyways. The sphere directly to the right applies a force only in the positive x direction. This force is... F = (6.67408*10^-11)(m1)(10)/0.45^2 F = 6.67408*10^-10 m1 / 0.2025 The sphere diagonally across from the bead exerts a force. This force is... F = (6.67408*10^-11)(m1)(10)/r^2 --> the distance between them is not 45. I use a right triangle to find the hypotenuse if both sides are 0.45 m. Thus, the distance between them is: the square root of (0.45^2 + 0.45^2) = hypotenuse Hypotenuse is 0.63639 m Therefore the force is: F = (6.67408*10^-11)(m1)(10)/ (0.63639^2) F = (6.67408*10^-10)(m1)/0.405 After I get these three forces, one in the x direction, one in the y direction only, and one having both x and y direction, where do I go from here to find the direction of the acceleration of the bead and magnitude? I know F=ma. Please help, thank you!