A particle of mass 3 m is located 1.40m from a particle of mass m . Part A Where

Question

A particle of mass 3 m is located 1.40m from a particle of mass m.

Part A

Where should you put a third mass M so that the net gravitational force on M due to the two masses is exactly zero?

Part B

Is the equilibrium of M at this point stable or unstable for points along the line connecting m and 3m?

Is the equilibrium of at this point stable or unstable for points along the line connecting and 3?

SubmitMy AnswersGive Up

Part C

Is the equilibrium of M at this point stable or unstable for points along the line passing through M and perpendicular to the line connecting m and 3m?

Is the equilibrium of at this point stable or unstable for points along the line passing through  and perpendicular to the line connecting and 3 ?

Explanation / Answer

The first thing you should recognize is that the position of the third object must be on the center line and between the other two masses. That's the only general location where the sum of forces will equal zero.

For example, if the third object were not on the center line, there would be vertical forces from both masses, pulling the third object back towards the center line on the X axis. So the real issue boils down to where should the third object be placed on the X axis as that is the center line between the two masses m and M.

To zero out, the two masses M = 21 kg and m = 12 kg must exert equal but opposite forces on the third object. This will happen when the respective gravity fields g are equal.

Thus g = GM/R^2 = Gm/r^2 = g must be true (these are the respective gravity fields g). Thus, M/m = (R/r)^2 and R^2 = (M/m)r^2 so that R = r sqrt(M/m) = r sqrt(21/12).

As the distance between the two masses is D = r + R = 18 cm = r (1 + sqrt(7/4)), we have r = D/(1 + sqrt(7/4)) = 18/(1 + sqrt(1.75)) = 7.75 cm then R = 18 - 7.75 = 10.25 cm

Then the point from M at the origin is p(7.75,0) AN

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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