Pulley Problem

[anthropos]

Initially these blocks of each mass m are at rest. The tabletop and pulley does not induce friction. So when the system moves, does it accelerate at g? Do both blocks have an acceleration of g? Does the weight of the top block affect the acceleration or exert a force on the bottom block? My book says 5 ms-2 and i am puzzled.

 

[mooeypoo]

mg -- gravity times the mass -- is operating on the Y axis only (hence, top-to-bottom).

That means that the mass that hangs off the edge will be pulled downwards at F=mg

 

Now let's look at the mass on the top of the table -- it's only moving on the X axis, as we can see. So it is in equilibrium on the Y axis.

 

Tabletop Mass M (Y-Axis) - F=mg downwards, and n ("normal")=mg upwards, to account for the equilibrium.

 

Tabletop Mass M (X-Axis) - since it's attached to the rope and to the other mass, the rope exerts tension T on it that is equal to the F=ma of the off-the-table mass.

 

Since there is NO FRICTION from anything, there's no repelling force, so the top-of-the-table-mass will have only the tension T operating on it, and therefore will move at F=T=mg.

 

If there WAS friction it would've been different, since friction is dependant on the mass and operates as a repellent force (opposing the movement).

 

As for the answer, since none of the questions seemed to ask for a time or an acceleration, I am not sure I understand it either. If we have VALUES for the mass (or at least for the distance from the edge of the table, or the length of the rope) we can calculate the amount of time that would pass.

 

Otherwise the acceleration is gravity.

 

~moo

[insane_alien]

the acceleration is half of the gravitational acceleration as gravity only applies an accelerating force to the mass hanging off the edge of the table. and as this has to accelerate the mass on the table in order to move we have a force F acting on 2 masses so the acceleration will be half what it would be if the block was just dropped.

 

assuming the masses are equal of course.

 

in the above scenario i suspect g is 10ms^-2

[mooeypoo]

insane_alien -- the second mass shouldn't interfere or affect much since there's no friction. Since there is no friction, it's as if the on-the-table mass doesn't exist... it's not supposed to be affecting the left-right forces (because no friction exists) and therefore no effect on the T (tension), which is equal to, therefore, the mg of the first mass.

 

if the other mass was suspended in the air it would've been different (because then the F=mg of the second mass was actually affecting the system) but since it's not...

 

Or.. where am I forgetting something here?

[insane_alien]

the mass on the table still has to accelerate doesn't it?

 

say m is 1kg and g is 10ms^-2

 

1kg is over the edge of the table so can be accelerated by gravity. this means we have a force of 10 newtons. BUT this 10 newtons has to accelerate 2 kg of mass. then using F=ma we see an acceleration of 5m/s.

[Mr Skeptic]

Normally, the force of gravity can be given as an acceleration, since a=F/m and gravitational force = gm, then a=mg/m=g. But in the above example, gravity acts on only one mass (the one not on the table), but that force has to accelerate both masses. If both masses are identical, then the acceleration would be half g, or about 5 m/s^2 --- a = mg/2m = g/2

[swansont]

The hanging block has two forces on it: the tension and force due to gravity. Why should it accelerate at g? If T = mg, the block will not accelerate at all! And to accelerate at g, it is necessary to have T=0, which is not the case.

 

Consider the motion if the top block has a mass of 10,000 m. Would you expect it to accelerate very fast?

[mooeypoo]

http://www.sparknotes.com/testprep/books/sat2/physics/chapter8section2.rhtml