Question

13.9 A particle of mass 3m is located 2.00 m from a particle of mass m. (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? (b) Is the equilibrium of M at this point stable or unstable (i) for points along the line connect- ing m and 3m, and (ii) for points along the line passing through M and perpendicular to the line connecting m and 3m?

Answer 1

The third mass M should be placed at a point along the line connecting the two masses, closer to the larger mass, to have no net gravitational force acting on it. The equilibrium of M is stable for points along the line connecting the two masses, and unstable for points along the line passing through M and perpendicular to the line connecting the two masses.

Step-by-step explanation:

a) To find the position of the third mass M where the net gravitational force is zero, we can use the concept of the center of mass. The center of mass is the point where the masses are evenly distributed. For two masses, the center of mass lies along the line connecting the two masses and it is closer to the larger mass. In this case, the center of mass is located at a point 3m/4 from the mass 3m and 1m/4 from the mass m. Therefore, the third mass M should be placed at this point to have no net gravitational force acting on it.

b) (i) For points along the line connecting m and 3m, the equilibrium of M is stable. If M is displaced slightly from this line, it will experience a gravitational force trying to restore it back to the line. The equilibrium is stable because the net force that M experiences is a restoring force that brings it back to the equilibrium position.
(ii) For points along the line passing through M and perpendicular to the line connecting m and 3m, the equilibrium of M is unstable. If M is displaced slightly from this line, it will experience a gravitational force pushing it further away from the line. The equilibrium is unstable because the net force that M experiences is a destabilizing force that moves it further away from the equilibrium position.

Answer 2

The question seeks the position of a third mass where gravitational forces from two other masses are balanced, and whether this equilibrium is stable. The equilibrium along the line connecting the two fixed masses is unstable, but it's stable on a perpendicular line. The total force calculations must consider both Newton's law of gravity and Coulomb's law if charges are involved.

Step-by-step explanation:

Gravitational Force and Third Mass Position

The student's question pertains to the gravitational attraction between masses and how to position a third mass such that the net gravitational force acting upon it from two other fixed masses would be zero. In this physics problem, we find a point where the gravitational forces from two masses (3m and m) balance each other out on a third mass (M). Such a point would lie along the line connecting the two masses.

To solve part (a), we need to apply the principle that the gravitational forces exerted by the two masses (3m and m) on M must be equal in magnitude but opposite in direction for the net force on M to be zero. This would happen if M is placed closer to the smaller mass (m) since force is inversely proportional to the square of the distance according to Newton's law of universal gravitation.

In (b), we analyze the stability of this equilibrium point. (i) Along the line connecting m and 3m, if M is displaced slightly towards one of the masses, the gravitational force from the nearer mass will become stronger than that from the farther mass, leading to an unstable equilibrium. (ii) For points along the line passing through M and perpendicular to the line connecting m and 3m, any small displacement will cause a restoring force to bring M back towards the equilibrium position, indicating a stable equilibrium.

However, the information provided regarding the anchored spheres and electric charges in some parts of the question appears to be a mixed scenario, possibly involving both gravitational and electrostatic forces. In such cases, calculating the total force would require considering both the gravitational force (using Newton's law of gravity) and the electrostatic force (using Coulomb's law).