Dude you are the best! I have so much respect for people who dedicate their lives to informing others. Teachers are the best thing that can ever happen to humanity.
I was pleased to find that If you assume m1 slides down the incline (since m1>m2), the signs in the eqns for ma and I x alpha will be reversed and one still finds that a = -1.05 g
I like how Consistent these problems are with each previous problems... In this Example.. if you let THETA = 0 .. then this problem reduces to a previous problem where m1 is on a flat table surface.... the final equations bear that out...... Thank Michel !!..
Why do you consider the acceleration to be positive going to the right (causing a right torque) whenever a clockwise torque is supposed to be negative? You use Torque2 - Torque1, showing that a rotation to the left is considered positive, while an acceleration to the right is negative; so why is the acceleration downwards considered positive, when clearly you set up a framework for it to be negative? In other words, why isn't T1 negative, T2 Positive, mgsin0 positive, and m2g negative?
Whenever you have a system (like this one) where parts of the system move in different directions, it is better to ASSIGN a positive direction for the system (typically the direction the system will accelerate in) as is shown in this video.
Goodday Sir Blezin. Another helpful video. Please could you explain more how you determined the acceleration of the pulley cause won't the 4kg slide down the incline pulling the mass of 3kg down? And more especially how you determined Torque 2 will be greater than Torque 1?
The direction of the acceleration is determined by the net force. Since (3 kg) > (4 kg) (sin (theta)) , the 3 kg mass will accelerate downward and the 4 kg mass will slide up the incline. (See the earlier playlist on Newton's 2nd law and inclined planes.)
Since torque is F x R (R being the radius of the pulley), the magnitude of the torque is proportional to the force on each side, so the torque will be greatest on the side where the force is the greatest (R is the same for both sides)
These videos are great! I've just used this method to solve a problem with 2 pulleys with masses, and an incline with friction. I used 3 different tensions. Hopefully I did it right!
@@MichelvanBiezen I really appreciate you taking the time to reply. Big fan of your videos. I was referring to @5:15 in the video I might be wrong I was just comparing my answer at the end and had that difference
oh i see, so with these types of calculations we decide the direction of acceleration by making the system the whole set up (boxes and pulley) and setting equal the external forces left like m1gsin(theta)=m2g to see which is larger and decide that the larger one is the direction of acceleration?
I guess you confused this with the normal Atwood machine, in which there is no inclined slope, rather 2 masses hanging freely. In that case, whichever weight is greater, would decide the direction of acceleration, but in this case, for the slope, the component of mg (mgsintheta) would cause the acceleration, if it were to slide down the hill but obviously this is smaller than the mg (of the hanging mass).
First determine the direction of the acceleration. Then every force that acts in the opposite direction is then considered a negative force. (Opposing the acceleration)
Tension cannot be negative. But in general tension can be in a negative direction relative to one side or the other side. In this type of example we chose the direction of the acceleration (based on what appears to be obvious). Then all the forces that act in the same direction as the acceleration are considered positive and all the forces that act in the opposite direction as the acceleration are considered negative
@@MichelvanBiezen Thank you! I understand why tension can't be negative, but I still don't understand why mgsintheta has to oppose tension? Why can't it be in the same direction as tension?
Dude you are the best! I have so much respect for people who dedicate their lives to informing others. Teachers are the best thing that can ever happen to humanity.
I was pleased to find that If you assume m1 slides down the incline (since m1>m2), the signs in the eqns for ma and I x alpha will be reversed and one still finds that a = -1.05 g
Woo I love pulley equations. Great job explaining physics concept as always. Keep them informative videos coming!
I like how Consistent these problems are with each previous problems... In this Example.. if you let THETA = 0 .. then this problem reduces to a previous problem where m1 is on a flat table surface.... the final equations bear that out...... Thank Michel !!..
That is incredible, we have someone like u sir💙
Thank you for your encouraging words.
Why do you consider the acceleration to be positive going to the right (causing a right torque) whenever a clockwise torque is supposed to be negative? You use Torque2 - Torque1, showing that a rotation to the left is considered positive, while an acceleration to the right is negative; so why is the acceleration downwards considered positive, when clearly you set up a framework for it to be negative? In other words, why isn't T1 negative, T2 Positive, mgsin0 positive, and m2g negative?
Whenever you have a system (like this one) where parts of the system move in different directions, it is better to ASSIGN a positive direction for the system (typically the direction the system will accelerate in) as is shown in this video.
Goodday Sir Blezin. Another helpful video. Please could you explain more how you determined the acceleration of the pulley cause won't the 4kg slide down the incline pulling the mass of 3kg down?
And more especially how you determined Torque 2 will be greater than Torque 1?
The direction of the acceleration is determined by the net force. Since (3 kg) > (4 kg) (sin (theta)) , the 3 kg mass will accelerate downward and the 4 kg mass will slide up the incline. (See the earlier playlist on Newton's 2nd law and inclined planes.)
@@MichelvanBiezen I understand now sir. Thanks a lot Sir Michel.
And also for determining which Torque is greater, how do i go about that?
Since torque is F x R (R being the radius of the pulley), the magnitude of the torque is proportional to the force on each side, so the torque will be greatest on the side where the force is the greatest (R is the same for both sides)
These videos are great! I've just used this method to solve a problem with 2 pulleys with masses, and an incline with friction. I used 3 different tensions. Hopefully I did it right!
This guy is the greatest, I wish he was my teacher
you make everything easier, thank you !!
It seems that we SHOULD subtract torque1 from torque2, NOT torqe2 from torque1. If we do vice versa, the answer will change.
The method shown in the video will give you the correct magnitude of the acceleration
Am I correct that we are escentially drawing three freebody diagrams, one for each weight plus one for the pulley, without actually drawing them?
No, this is a different technique where we are looking at the whole system at once. We have examples where we use the free body diagram technique.
Is there a slight mistake when subbing in the value for T1 you used (M1 a - Mg Sin theta) .. Should it not be (M1a + MgSin theta)??
That is how it is written in the video.
@@MichelvanBiezen I really appreciate you taking the time to reply. Big fan of your videos. I was referring to @5:15 in the video I might be wrong I was just comparing my answer at the end and had that difference
On that equation, since we are subtracting T1 from T2, we need to place negative signs in front of each term. (thanks for the comment).
thank you, sir! it was very help full. After we got acceleration, can we find torque by substituting it to angular acceleration? thank you.
Sir, shoudn't the the equation of the pulley be T2R-T1cos(30) R = I x alpha ?
the T1 isn't perpendicular with R
At every angle the rope is perpendicular to R. The equations in the video are correct.
@@MichelvanBiezen thank you sir for responding, i got it :)
why isnt the system accelerating down the incline, the mass of m1 is greater than the mass of m2?
The component m1g sin(theta) is smaller than m2g
oh i see, so with these types of calculations we decide the direction of acceleration by making the system the whole set up (boxes and pulley) and setting equal the external forces left like m1gsin(theta)=m2g to see which is larger and decide that the larger one is the direction of acceleration?
That is correct.
I guess you confused this with the normal Atwood machine, in which there is no inclined slope, rather 2 masses hanging freely. In that case, whichever weight is greater, would decide the direction of acceleration, but in this case, for the slope, the component of mg (mgsintheta) would cause the acceleration, if it were to slide down the hill but obviously this is smaller than the mg (of the hanging mass).
You are awesome sir
You are always the best thanks
sir what about the velocities and distances covered by the mass m1 & m2 ,whether velocity of both masses wiil be equal?
their a is constrained to be the same, m1a must = m2a, as such v2 = v1, v is the integral of a
thank you !!!!!!! so clear and amazing.
Your the best!
Thank you. Glad you found our videos. 🙂
great video, thank you !!!
thank you sir. so helpful
Why is m1gsintheta in the negative direction?
First determine the direction of the acceleration. Then every force that acts in the opposite direction is then considered a negative force. (Opposing the acceleration)
@@MichelvanBiezen thank you! However, I still don't understand why mgsintheta would oppose tension
Tension cannot be negative. But in general tension can be in a negative direction relative to one side or the other side. In this type of example we chose the direction of the acceleration (based on what appears to be obvious). Then all the forces that act in the same direction as the acceleration are considered positive and all the forces that act in the opposite direction as the acceleration are considered negative
@@MichelvanBiezen Thank you! I understand why tension can't be negative, but I still don't understand why mgsintheta has to oppose tension? Why can't it be in the same direction as tension?
Thanks 😊
Are you Australian?
Originally I am from Belgium, but now I live in the US.
@@MichelvanBiezen you are one of those physics teachers who give their students the space to think and indulge the subject.
Great teching
i loe ou