Two masses connected by string on a cliff

[issisoccer10]

[SOLVED] Two masses connected by string on a cliff...

This is a general question about physics...and I'm a little confused about it..

Two equal masses (A and B) are attaced by a massless string of length L. Mass A is on a cliff and is L/2 away from the edge. Mass B is being held even horizonatally with the cliff but a length L/2 away from the edge. If you drop mass B from this position, would mass B swing down and hit the edge of the cliff before mass A reaches the edge of the cliff?

Mass A would reach the edge of the cliff because as mass B falls and pulls on A, the radius of the arc that B would need to make to reach the edge of the cliff would continually increase, thus not allowing B to hit the edge of the cliff.

Does this make sense or am I just making stuff up? Is there a better "logical" explanation that would seem to fit better with physics?

 

[pixel01]

If I understand the stuff right, mass B can not reach the edge of the cliff again because it transfers some energy to mass A (so that mass A can move closer to the edge).

 

[Mephisto]

hmmm this is a pretty interesting question. I don't think there is an easy answer without actually working out the problem. I kinda did it on my table right here in my room and mass B hit the table before A slipped away, but I suspect Friction as the cause. I'm going to guess it hits just as A slips off, but maybe I'll try to work this out later :)

 

[pmbasa]

Hmm. Very interesting. Try out the experiment. That's the only way you can get an answer which you can trust, since you did the experiment.

 

[tomprice]

The forces acting on mass B are the force of gravity and the force of tension from the rope. The forces acting on mass A are the tension, gravity, and normal force, the last two of which cancel out. The horizontal component of the force on mass a is the same as the tension in the rope, since the rope is pulling horizontally.
The horizontal component of the force on mass b is the tension in the rope times the sin of the angle that the rope makes with the vertical face of the cliff (call this angle x).
Before mass B hits the edge of the cliff:
0 <= x < 90 degrees.
Therefore, sin x < 1.
Therefore, the magnitude of the horizontal component of the force on mass B will be less than on mass A.
Therefore, the horizontal acceleration on mass B will be less than that on mass A.
Therefore, the magnitude of the velocity of mass B in the horizontal direction will be less than that of mass A (since the start out with the same velocity).
Therefore, the horizontal displacement of mass B from the moment it was dropped will be less than the horizontal displacement of mass A from that same moment.
Therefore, by the time mass B travels a distance of L/2 horizontally, mass A will have traveled a greater distance horizontally.
Therefore, it will not hit the edge of the cliff before mass A falls.

 

[issisoccer10]

thanks a lot tomprice...that really makes sense to me

 

[tomprice]

No, thanks to you, it was an interesting problem.