1. The problem statement, all variables and given/known dat Three solid spheres of lead, each of mass 9.8 kg, are located at three corners of a square with side lengths of 50 cm. A small object is released at the forth corner. Considering only the gravitational forces among the four objects, determine the magnitude and direction of the acceleration of the smaller object when it is released. 2. Relevant equations F = Gm1m2/r^2 a = F/m mo = mass of object 3. The attempt at a solution Ok so I drew a square and added the three 9.8 kg balls. Top right of the square is Ball #1, Top left of the square is ball #2, bottom left is Ball #3. The smaller object is in the bottom right corner of the square. ∑F = F1 Cos 45 + F2 + F3 Cos 45 Substituting into this the formula for force one gets ΣF = G*mo*m1 / r^2 * Cos 45 + G*mo*m2 / 2r^2 + G*mo*m3 / r^2 Cos 45 This simplifies to: ΣF = G*mo /r^2 ( Cos 45 * m1 + 0.5m2 + Cos 45*m3) If we then equate this to F = mo *a, you get the mo to cancel and are left with: a = G / r^2 (Cos 45*(9.8kg) + 0.5(9.8kg) + Cos 45 * (9.8kg)) substitue in G = 6.67x10^-11 and 0.05m for r and you get an a of: 5.0 x 10^-7 m/s^2 in the direction of the centre of the square. Have I done this right?