Three uniform spheres are located at the corners of an equilateral triangle. Eac

Question

Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 1.19 m. Two of the spheres have a mass of 2.66 kg each. The third sphere (mass unknown) is released from rest. Considering only the gravitational forces that the spheres exert on each other, what is the magnitude of the initial acceleration of the third sphere?

Explanation / Answer

let's imagine the triangle is oriented with the unknown mass at the top, and the vertex lies along the y axis...this is not necessary, but it will make it easier for me to describe the subsequent motion the two 2.66 kg spheres will exert both x and y forces on the sphere; but the x forces will cancel out, so we need to consider only the y forces acting on the sphere the component of each sphere's force on M is Fy = Ftotal sin(theta) = G M(2.66 kg)/(1.19m)^2 x sin(theta) where theta = 60 deg (find theta by solving for the height of the triangle...use the pythagorean theorem and realize that the height of the triangle divides the base into two sections of 0.3m, then the height = Sqrt[1.19^2-0.595^2] = 1.062 and sin theta = 1.062/1.19 => theta = 63.19 therefore, Fy = G M(2.66kg)/1.19^2 x sin 63.19= 2.24x10^-10 x M there are two spheres acting on M, so the total force is 4.48x10^-10N x M now, this force on M equals M a, so we have M a = 4.48 x 10^-10 x M a = 4.48 x 10^-10 m/s/s

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