00:01
Hi guys, now in this film, this is a diagram of the problem i have, where the theta is equal to what, this is 30 degree, this angle is 30 degree, this is m of a is equal to what, this is equal to what, this is equal to what, this is equal to what 4 kc, or m of b is equal to what, 1 .2 kj.
00:25
So i'll say this one is equal to what m.
00:29
So this one is equal to what 4 of m.
00:34
Now this pulley has a radius of what r which is about 5 cm and mass of this one is equal to this is in 0 .9 kg let's draw the fbd for this one there is a mg acting downward this is mg acting downward components of gmg one will be like this second one will be like this this is normal i have and when you this moves in downward direction with the acceleration of a so friction will work like this is friction i have and this is tension in the string similarly mfbg is acting downward this moves up with the acceleration of a this is tension of what i'll say that this is tension of t1 there is also tension working here now this tension i'll say this is t1 i'm gonna say t1 and this is what t2 friction here is not negligible i'll say this is t2 again this is t1 i have this moves with the acceleration of a this is m of b g this is acceleration of a this one we have m of a times of a this is m of a g sign of 30 degree this one i'll say that this is m of b g g cause of 30 degree this is normal i have f of n that is a friction force f of k let's write down newton's second law for each object so i'll say this one i'll say a that t1 minus m of bj or i'll say only m g is equal to m times of a for this one i'll say for this object it moves downward m of ag so this is what for m g sign 30 degree so this one comes out of 2mg so here forces are what this is 2 of mg minus this is what we have this is a friction force which but mu of k this is what 4 of m g i'll say that this is is 4 of mj, a square root of 3 divided by 2 minus tension t 2 is equal to 4m times of a.
03:12
Now it is obvious that when it moves downward this pulley will be what rotated in this direction so we have angular of acceleration alpha and it is moving without slipping which means i'll say that acceleration is just equal to this alpha is equal to a divided by r.
03:30
So let me write down torque is equal to i alpha so i'll say that t2 times of r minus t1 times of r and torque i alpha which is what m r square this is the mass of the pulley and say this is m of p r square divide by alf divide by two alpha we have what this is a divide by r from each expression this r and r gets cancelled out we have t2 minus t1 is equal to this is m of p is at 0 .9 which is at 0 .45 times of a this is equation first this is equation second this is equation third these are the all three newton second law equations we have now once i know this value i can easily find out this is a value of the kinetic friction if we're willing to find out what you need to do just you need to add them all this is first second third add them all t1 and t2 gets cancelled allowed.
04:49
This is 2 of m g minus mj, which is only mg...