00:01
In this question, we are given the following figure with masses ma and mb which are joined to a pulley.
00:08
So first let us take a look at all the forces acting on mass mb.
00:12
So first we have gravitational force mbg acting downwards and the normal force acting upwards and here we have frictional force fb acting.
00:24
Next tension t2 from the pulley and this gives an opposite tension t2 as well as well as so here we have t1 and t1 acting on each side.
00:36
For mass ma we have mag acting here, mag.
00:44
Here we are taking an angle of q and a normal force acting upwards.
00:53
Also on the pulley we have mass mpg that is mass of pulley acting on the pulley.
01:00
So now using this we can write the force equations.
01:04
For b we have the force equations tension t2 minus frictional force denoted by f is equal to mb a also we have in the vertical direction normal force nb is equal to mass of mb into g so for the pulley we have the torque equation for pulley we have talk equation r times t1 minus r t2 is equal to i alpha.
01:39
Force equation for a will be m .ag sine q, m .a .g, sine q.
01:46
Since q is the angle here, so we are going to take vertical and horizontal components since a is inclined.
01:52
So, minus t1 will be equal to n .a of a.
01:56
And here we have normal n .a is equal to m .a .g.
02:01
Cos g or cos q...