00:01
Okay, so we've got a situation like this.
00:04
We've got a table, pulley, a mass m2 hanging and a mass m1 on the table.
00:12
The pulle is frictionless, the cable's light, but the table does have friction.
00:17
So part a asks for free body diagrams for each of the masses m1 and m2.
00:22
So it asks for axes as well.
00:24
So i think all that means is that tells, i'm going to define right to be positive and down to be positive.
00:32
And so the free body diagram for m1 is going to be the force pulling it forwards is the tension in the cable.
00:44
The force pulling it backwards is the friction from the surface.
00:49
Its weight acts downwards on it and the reaction force from the table acts upwards on it like this.
00:57
For m2, we then have a bit of a simpler diagram.
01:02
There's no horizontal forces acting on m2.
01:04
Its weight, i should put one here.
01:12
Its weight acts down and the tension acts up on it.
01:15
The tension felt by m2 is the same as the tension felt by m1 because the part of the same system.
01:21
Okay, part b then says, assume that mass m2 falls with an acceleration magnitude a, find an expression for the tension in the rope t.
01:29
So we're just going to use newton's second law on m2.
01:32
So this is just going to tell us that the weight of m2 minus the resultant force.
01:38
I've just defined down to be positive.
01:40
So the resultant force on m2 is its weight minus the tension.
01:44
And by newton second law, that equals its mass times its acceleration.
01:49
Its weight is just given by m2g, so we can make that a bit clearer.
01:54
And then we can use that to solve that t is equal to m2 times g minus a.
02:04
Part c then asks us to do the same thing but for m1...