Great question! In this problem i'm assuming that the pulleys don't have any mass and therefore would have to be pretty small so i didn't include it. The total length of the string should include half the circumference but this won't change the final answer for acceleration or tension because this is extra length is a constant value and when you take the derivatives those constant don't matter. Happy learning.
I tried to get other equations by taking a free body diagram of the whole system. I assumed, that since the machine was not translating (just moving in the inner system), the acceleration of the system would be equal to zero. From that assumption I got that the tension in the pulley of the top (attached to the ceiling) was holding the 3 masses of the blocks. I got my answers wrong. But, I'm still trying to understand how that assumption was incorrect.
@@boraelshani If one side of the pulley is positive when going down, doesn't that make the other side positive when going up? That's the way I was taught.
I first tried the problem considering the mass 1 and 2 like one mass M=m1+m2 and then calculate the acceleration of the block 3 using the same method for an atwood's machine with only two masses. I don´t understand why the acceleration i got is wrong, can you explain me please?
Hi ninja :) , I want to ask a question. Why are the forces acting on pulley A 2T(a) and T(b) and not for example (m1g + m2g) and T(b)? It's something that I don't understand. Thank you
Thanks for your question. I think if you look at the free body diagrams at 3:05 you see that all the tensions are acting up. But keep in mind that the sum of the forces for each block is ma but the mass of each block is different and so can the acceleration. Some of the accelerations are linked together because the length of the rope is fixed but the general expression for the tensions will be more complicated. The simple answer is because of the accelerations.
if you want you can give it a tiny mass (m_pulley) and solve. The acceleration would have to be opposite mass m3 so make it (-a). It just makes the equations a little messier. Sometimes i say 0 but i should really mean small or negligible compares to m2 and m3.
Great question - if you have a look from 6-10mins i explain this point. You're right that the string length is fixed but since the pulley that is connected the two objects is able to accelerate they don't have to be the same. Look at the equation at 10:13 and you'll see the linking between the accelerations. If a_a (acceleration of pulley at is 0) then you would get a1 and a2 to have the same magnitude like in a simple atwood machine. Hope this helps.
why we assume that the masses go downward? I mean the direction of m1a1 and m2a2. We even don't know whether the masses go down or up...I am confusing, help me.
Great question. I started picking a global coordinate system with down as being positive, this choice is arbitrary. As long as you are consisten things will work out. When i find eqns (4) and (5) i also don't assume anything about the direction of motion. At the end, depending on the values of the masses, the accelerations a1, a2, and a3 may end up being positive or negative but this is with respect to my original coordinate system. If they are positive this means that the acceleration of that block is down. Try looking at the just a simple atwood machine (just 2 masses) to clarify this.
thanks for teh video. But why I can not take the system1 (pulley A, mass m1 and mass m2 as one system in the first case) so that the problem reduces to mass m3 and system s1 in the first step with pulley B
Thank you so much. Using length to explain the acceleration relation is the best explanation i have ever seen. Thank you so much!
had an example like this in my book and it had no explanation for how they got accelerations from the strings. thanks for clearing that up
Introduction to Classical Mechanics by David Morin?
your methods are so sophisticated wow
An absolute life saver!
i struggled with this question for 1h the other day and i couldn't do it. Thanks a lot sir
Improve sound quality please in this video..However,it's splendid!!😍😍
Here are the formulas for anyone trying to copy them down...
a1 = g * (4 * m1 * m2 - 3 * m2 * m3 + m1 * m3) / (m1 * m3 + m2 * m3 + 4 * m1 * m2)
a2 = g * (4 * m1 * m2 - 3 * m1 * m3 + m2 * m3) / (m1 * m3 + m2 * m3 + 4 * m1 * m2)
a3 = g * (m1 * m3 + m2 * m3 - 4 * m1 * m2) / (m1 * m3 + m2 * m3 + 4 * m1 * m2)
where a3 is the top mass, a1 & a2 are the bottom masses.
Finally, thankfully i understood this topic :)
maybe, you need to add the (phi*r) in an adding of the length of the wire, because the pulley also has wire in the arc of the circle right?
Great question! In this problem i'm assuming that the pulleys don't have any mass and therefore would have to be pretty small so i didn't include it. The total length of the string should include half the circumference but this won't change the final answer for acceleration or tension because this is extra length is a constant value and when you take the derivatives those constant don't matter. Happy learning.
oh okay because the derivative and the constant will be zero. Great, thanks
@@PhysicsNinja if the mass of the pulley is given,we should have 2Ta-Tb=-mass of pulley (acceleration of Pulley+g)?? right?
I tried to get other equations by taking a free body diagram of the whole system. I assumed, that since the machine was not translating (just moving in the inner system), the acceleration of the system would be equal to zero. From that assumption I got that the tension in the pulley of the top (attached to the ceiling) was holding the 3 masses of the blocks. I got my answers wrong. But, I'm still trying to understand how that assumption was incorrect.
You've assumed you knew the tension which is bad. It is obviously not sensible from an energy stability perspective.
Can we arrange this equations using matrix
Because the data is huge (5 equation 😱)
And by using gauss method
Yes, you can solve using matrix algebra. This is a rather lengthy problem
@@PhysicsNinjacan you demonstrate in next video
And what about wedge constrain
beautifully explained!
fun video
Shouldn't either m1 or m2 be negative because one of them has to be going up? How does this affect the calculations?
He picked a coordinate system with down being positive
@@boraelshani If one side of the pulley is positive when going down, doesn't that make the other side positive when going up? That's the way I was taught.
but mass 2 is going to appear to accelerate slower relative to the earth?@@FoxMauser
Thank you kind sir
I first tried the problem considering the mass 1 and 2 like one mass M=m1+m2 and then calculate the acceleration of the block 3 using the same method for an atwood's machine with only two masses. I don´t understand why the acceleration i got is wrong, can you explain me please?
You must got wrong answer because the accelerations of such masses are not equal even though the masses are the same
Thank you
Hi ninja :) , I want to ask a question. Why are the forces acting on pulley A 2T(a) and T(b) and not for example (m1g + m2g) and T(b)? It's something that I don't understand. Thank you
Thanks for your question. I think if you look at the free body diagrams at 3:05 you see that all the tensions are acting up. But keep in mind that the sum of the forces for each block is ma but the mass of each block is different and so can the acceleration. Some of the accelerations are linked together because the length of the rope is fixed but the general expression for the tensions will be more complicated. The simple answer is because of the accelerations.
awesome video, thank you so much!
You said that mass of pully A is zero then how it has a finite acceleration?
if you want you can give it a tiny mass (m_pulley) and solve. The acceleration would have to be opposite mass m3 so make it (-a). It just makes the equations a little messier. Sometimes i say 0 but i should really mean small or negligible compares to m2 and m3.
can you use Hamiltonian function to solve the same question please
Why is a1 and a2 different though? Aren't they the same because the string is inextensible?
Great question - if you have a look from 6-10mins i explain this point. You're right that the string length is fixed but since the pulley that is connected the two objects is able to accelerate they don't have to be the same. Look at the equation at 10:13 and you'll see the linking between the accelerations. If a_a (acceleration of pulley at is 0) then you would get a1 and a2 to have the same magnitude like in a simple atwood machine. Hope this helps.
hi yilun
why we assume that the masses go downward? I mean the direction of m1a1 and m2a2. We even don't know whether the masses go down or up...I am confusing, help me.
Great question. I started picking a global coordinate system with down as being positive, this choice is arbitrary. As long as you are consisten things will work out. When i find eqns (4) and (5) i also don't assume anything about the direction of motion. At the end, depending on the values of the masses, the accelerations a1, a2, and a3 may end up being positive or negative but this is with respect to my original coordinate system. If they are positive this means that the acceleration of that block is down. Try looking at the just a simple atwood machine (just 2 masses) to clarify this.
Shouldnt the acceleration of block 2 and 3 be the same? They are attached to the same string after all.
they have the same acceleration relative to their pulley
u really won my like.
thanks for teh video. But why I can not take the system1 (pulley A, mass m1 and mass m2 as one system in the first case) so that the problem reduces to mass m3 and system s1 in the first step with pulley B
Thanks bro
because you give negative values to tensions?
What if, pullley A had mass?
2Ta-Tb=-Mp(acceleration of Pulley+gravity)
Where Mp=mass of pulley,not forgetting that acceleration of Pulley=-a3
😍🔥
im here from sears and zemanskys!!!
dude why cant u just take the accelerations for m1 and m2 the same??
M1 and M2 are also moving relative to each other if they’re not the same mass.
Crack!
Macher
Thank you