Solve an Unusual Pulley Problem: Finding the Timing of Two Weights in Motion

[z_particle]

I came across this problem recently. It's a pulley problem, albeit a slightly unusual one.

The situation is sort of hard to explain without a diagram, but I'll do my best.

There's a pulley on the edge of a table, and there's a string that's stretched across it. The string has two weights attached to its two ends - of equal mass M. A length L of the string stretches on either side of the pulley, and initially the masses are held so that the string is horizontal (i. e. one mass is kept at rest on the table, and the other which is outside the table is supported so that the string remains horizontal, and just touching the pulley). From this position, the system is released. Two things happen - the mass outside the table swings down to hit the wall, and the mass on the table slides forward to hit the pulley. There is no friction involved.

The question is, which happens earlier? And how do we find that out rigorously?

I can't get the hang of it because:

1. I can't figure out if the tension remains constant throughout the duration of the fall of the outside block.

2. The length of the string outside the table also varies during the fall, so I can't apply anything related to circular motion on this block.

Can someone please help me out?

Many thanks in advance,

Z

 

[Doc Al]

Hint: Consider and compare the vertical and horizontal forces acting on each mass.

 

[astrokreng]

oe

I would approach this problem by first identifying the key variables involved: mass (M), length of string (L), and time (t). From the given information, we can also assume that the system is in a state of equilibrium before it is released, meaning that the tension in the string is equal on both sides.

To solve for the timing of the two weights in motion, we can use the equations of motion for both blocks. For the block on the table, we can use the equation s = ut + 1/2at^2, where s is the distance traveled, u is the initial velocity (which is 0), a is the acceleration (due to gravity), and t is the time. We can solve for t by setting s = L (the length of string on the table) and a = g (the acceleration due to gravity). This gives us t = √(2L/g).

For the block outside the table, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0), a is the acceleration (due to gravity), and s is the distance traveled. We can solve for v by setting s = L (the length of string outside the table) and a = g (the acceleration due to gravity). This gives us v = √(2gL).

Since the block outside the table will hit the wall when it reaches the ground, we can equate the final velocities of both blocks (v = √(2gL)) and solve for t. This gives us t = √(2L/g).

Therefore, we can conclude that the block on the table will reach the pulley first, and the block outside the table will reach the wall after a time of 2√(2L/g).

In summary, by using the equations of motion and considering the system's initial state of equilibrium, we can solve for the timing of the two weights in motion in this unusual pulley problem. It is important to carefully consider all the variables and use appropriate equations to accurately solve the problem.