00:01
In this problem we have a smooth inclined plane with inclined angle alpha and blocks a and b have the same mass and they are connected with light rope through a pulley and the pulley has radius r and mass m.
00:22
Using this we have to find out the tension of the rope and the acceleration of b.
00:30
So the forces acting on mass a are weight mg in the downward direction and the tension force ta up the incline and the angle of incline is alpha.
00:52
Now the third force is normal force which is on the mass by the incline and this is normal force fn and the angle of gravitational force with the direction perpendicular to incline is alpha which is the angle of the incline.
01:12
So the component of gravitational force perpendicular to incline is mg cos alpha and the component of gravitational force parallel to incline is mg sin alpha.
01:29
So these are the two components of gravitational force on block a.
01:40
Now for block b the forces are tension tb in the upward direction and weight mg in the downward direction.
01:50
Now if block b is going down with acceleration a then the acceleration of block a will be same up the incline.
02:05
Now we write the equation of motion for block a along the incline taking the direction up the incline as positive x axis.
02:17
So along x axis for block a sum of forces must be equal to mass times acceleration along x axis.
02:26
Forces along x axis are tension ta and mg sin alpha down the incline and this will be equal to mass times acceleration along the x axis which is along the incline.
02:46
So this is equation 1.
02:53
Now we consider the motion of the pulley.
02:56
So the pulley revolves with angular acceleration alpha p.
03:07
This is the angular acceleration of the pulley and the tension forces acting on the pulley are tension ta down the incline and tension tb in the downward direction.
03:27
So the torque due to these two forces about the center of pulley will be torque due to the downward force of tb is force times perpendicular distance from the line of force to the center which is the axis of rotation and this perpendicular distance is radius r.
03:57
So this will be the counter clockwise torque sorry the clockwise torque and we will take the clockwise direction as positive and the other torque is due to tension ta.
04:11
So torque will be force times perpendicular distance.
04:15
So net torque is tb minus ta times r.
04:22
Now torque will be equal to i alpha where alpha p where alpha p is the angular acceleration and i is the moment of inertia of the pulley which is in the shape of a disc.
04:38
So torque is equal to moment of inertia.
04:45
Moment of inertia is mr square upon 2 and angular acceleration is alpha p.
04:59
Now angular acceleration of the pulley will be equal to the linear acceleration which is equal to the linear acceleration of the masses divided by the radius of the pulley.
05:09
So we get tb minus ta equal to times r equal to mr.
05:25
R cancels out.
05:28
Mr a upon 2 then r cancels out and we get tb minus equal to this is equation 2.
05:42
Now we consider the motion of mass b.
05:50
So the forces on b are the downward force of mg.
05:55
So we will take the downward direction as positive to write the equation of mass b.
06:00
So force is mg in the downward direction and tension tb in the upward direction on mass and this will be equal to mass times acceleration.
06:16
This is equation 3.
06:18
Now we have 3 equations...