In deep space, three masses are held in fixed positions (by rods light compared

Question

In deep space, three masses are held in fixed positions (by rods light compared to the masses) at three corners of a square of side length as shown. A fourth mass is placed at point A, and released from rest. By symmetry it accelerates straight down and reaches point B (the midpoint between the two 3.0 Times 104 kg masses). How fast is the fourth mass moving at point B? (Note: the acceleration is not constant.) If the fourth mass is 70 kg, what is the magnitude of the net force acting on it at point A?

Explanation / Answer

part a: easiest to handle this as an energy problem.

Before you begin, note that the distances are easy because A to B to either of the 3 * 109 masses is a 3-4-5-Pythagorean triangle. So initial distances from A are 5,8, and 5 m; final distances are 3, 3, and 3 m.

U = -GMm/r   [ formula for gravitational potential energy ]

So initial U = -(6.67 * 10-11)(3 * 109 kg)(70 kg)/(5 m) = -2.803 J   for each of the two masses on the side

U = -(6.67 * 10-11)(5.64 * 109 kg)(70 kg)/(8 m) = -3.2935 J for the other mass

Total initial U is -8.8995 J

Final U is calculated by repeating the above calculation with all distances changed to 3 m.

Final U is -18.126 J. The 70 kg mass has converted 9.2265 J of potential energy into kinetic energy. Solving for v using (1/2)mv2, we get v = 0.5134 m/s.

part b: force from bottom mass: GMm/r2 = (6.67 * 10-11)(5.64 * 109 kg)(70 kg)/(8 m)2 = 0.412 N

which is directed all vertically.

force from each side mass: (6.67 * 10-11)(3 * 109 kg)(70 kg)/(5 m)2 = 0.561 N but this force is not directed vertically; vertical component is F cos 36.87 = F * .8 so each side mass contributes: 0.448 N

Total force: 1.31 N, all vertical

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.