For the second step I had: -m2a -m1a -1/2Ma = -m2g -m1gsin(theta) -m1gcos(theta)mu Then I divided both sides by -1 since that is what you had done in all of the previous examples, is there a reason why you did things differently in this example because so far I have always gotten the same answer as you until this example. Thanks so much for your help, you have definitely made physics more enjoyable!
Professor, It seems that the u(muoo) in the problem is u kinetic not u static. Should we know the u static to first determine if the system would even move and then if it does, determine acceleration?
That is the standard when considering torque and angular velocity as a vector. Here we are just finding the magnitude of the acceleration (which can never be negative) and consider the direction of acceleration to be positive as chosen, regardless what direction it points.
clockwise or anticlockwise it not relevant with the technique here used to solve the problem. We consider the torque aiding the acceleration positive and the torque opposing the acceleration negative. (Note that when you turn the wedge around you should still get the same answer)
Well I suppose I didn't actually *need* this, but it assured me that I was going the right direction long enough to realize that I screwed up a negative along the way
That is an example we haven't done yet. Probably need to add a few examples like that. The result will depend on how the string is attached to the rolling cylinder. (at the axel or wrapped around the cylinder). If it is at the axel then you have to use conservation of energy, if the string is wrapped, then use the method in this video.
You assume a direction (take you best guess). If you get a negative answer for the acceleration then you picked the wrong direction and work the problem again with the acceleration in the other direction.
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you are the greatest man on the internet
THANKS HOLY MOLY MACARONI YOU ARE THE MAN THE MYTH THE LEGEND THE LEBRON JAMES OF THIS
Thank you. Glad you found our videos.
this playlist is legend
Thank you. Glad you found our videos. 🙂
Sir please check if there's a 'g' missing when the final expression is written in the second last step.
Yes, you are correct. I forgot to write the g there. Thanks.
For the second step I had: -m2a -m1a -1/2Ma = -m2g -m1gsin(theta) -m1gcos(theta)mu
Then I divided both sides by -1 since that is what you had done in all of the previous examples, is there a reason why you did things differently in this example because so far I have always gotten the same answer as you until this example. Thanks so much for your help, you have definitely made physics more enjoyable!
Professor, It seems that the u(muoo) in the problem is u kinetic not u static. Should we know the u static to first determine if the system would even move and then if it does, determine acceleration?
That was not the purpose of this example here, however there are examples in the playlist that take that into account as well.
I thought negative direction is clockwise??
That is the standard when considering torque and angular velocity as a vector. Here we are just finding the magnitude of the acceleration (which can never be negative) and consider the direction of acceleration to be positive as chosen, regardless what direction it points.
I think we should have T1R-T2R instead cz t1 is moving anti clockwise and t2 is moving clockwise
clockwise or anticlockwise it not relevant with the technique here used to solve the problem. We consider the torque aiding the acceleration positive and the torque opposing the acceleration negative. (Note that when you turn the wedge around you should still get the same answer)
Michel van Biezen thanks a lot
Sir,I have a question...
[m1g sin (theta) = μ m1g cos (theta)]
Is this equation always true when a particle slipping down through a incline plan ??
Yes, if the object is sliding at a constant speed.
Well I suppose I didn't actually *need* this, but it assured me that I was going the right direction long enough to realize that I screwed up a negative along the way
Hey, what if the pulley is massless and instead of box on inclined plane we have solid cylinder? Thanks!
That is an example we haven't done yet. Probably need to add a few examples like that. The result will depend on how the string is attached to the rolling cylinder. (at the axel or wrapped around the cylinder). If it is at the axel then you have to use conservation of energy, if the string is wrapped, then use the method in this video.
Great video
how do you know which direction pulley's acceleration will turn?
basically, why would T1R-T2R be in incorrect?
You assume a direction (take you best guess). If you get a negative answer for the acceleration then you picked the wrong direction and work the problem again with the acceleration in the other direction.
sir can you pls tell me how to find the angle at which the smaller mass slides down in this problem
That is a good topic for some future videos.
Very good. Thanks