i was failing this course, there is a lot of videos on youtube that confuses us a lot however...now that I found you ... my hope increased exponentially. best wishes.
See the above description of the video for the answer to your question. I just added more text to it so that it addressed your question more directly. Thanks.
Thank you very much. My text's explanation of the Atwood machine was incomprehensible and then I found you. You integrate the algebraic approach to solving the problem with the use of numbers and make the problem easy to solve. Thanks a lot. I'll check out your other videos now.
thank you very much ! this video helped me understanding this test and the tension , but i'm still a little bit confused by the tension force. Due to Newton's third law , the forces are paired and equal, so when both of m1 and m2 are trying to get each other down, what should tension force be depended on? For instance,when m1 is trying to get down, it also get the reaction force from m2. When m2 is trying to get m1 down , it would probably get the reaction force from m1, lol. I'm just confused , hope to hear from you.
Tension is a force and forces are vectors, so if something is accelerating in the direction that a string is pulling, acceleration and tension will be the same.
It became clearer for me, I checked rotational dynamics part III, and noticed that atwood's is a bit simpler than of the problem that the pulley is massive.
At 0:58 it's easy to assume that the acceleration would be greater for the system on the right because m1 is on a level surface instead of hanging down. The system on the left has m1 hanging down., which makes it easy to assume that the acceleration would be slower. Would such assumptions be right?
okay, can you explain how you got had m2 in the denominiator to be positive? I had a negative m2. I used the technique of summation of forces along the y-axis.
Nice vid but i got a question i cant seem to figure out... so for my questions the givens are m1=5kg Ft= 78.4 and a=5.9m/s the unknown is m2.. it seems so simple but i cant get it together, think u could give me a formula??
Why can't I use the equatn Ft-mg=Fnet? Because Ft is pointing upward(Assumed +ve) and mg is pointing downward(Assumed -ve) if we are to take the vector direction in mind.
I tried to apply the Atwood Machine in the field at my job. I have to carry my 50 pound backpack up and down a 175' tower. So I thought to create an Atwood Machine using a 200' rope, two pulleys, my backpack(M1) and some weights(M2). The objective was to establish equilibrium so that I could easily take my backpack up and down the tower with little to no effort. My atwood machine failed miserably. I was able to raise my backpack (lowering the weights) with very little effort. But lowering my backpack (raising the weights) took a tremendous amount of effort - there had to have been at least 25lbs of resistance when lowering my backpack. I'm baffled in how it worked one way but not the other.
How do you come to the conclusion that 3kg box has a 30N force and the 2kg box has a 20N force? Did you just round the acceleration of gravity from 9.81 to 10 and multiply that by the mass? Or should the force down on each box be given? Thanks.
thank you so, so, so much! our provided (ripoff) textbook is of no use in working through problems like this. thank you for your time and contribution to public education!
Both masses affect the tension in the string. Once you know the acceleration of the system (by analyzing both masses), then you can use either mass and Newton's 2nd law to determine the tension in the string.
yeah i agree, i'm in highschool physics and we were taught ma= fa- fw -ft+ ft -fw which is basically the same but way more confusing this video helped a lot
For different scenarios the tension will vary but for a given scenario the tension will be the same throughout the rope. This is only true if the rope's mass is negligible and the pulley's mass is negligible or the rope slides across the pulley with no friction.
@@notloc8993 I'm on a trip but when I get home in a few days, I'll make a video to answer your question. For now, imagine breaking the rope into 100 little segments. Segment 2 pulls on segment 1 (s1) with the same force that s1 pulls on s2 because of Newton's 3rd law. Now if s2 has an almost zero mass then the net force on it would be m*a which would be almost 0 because m is 0. That means that the force of s1 on s2 has to be almost equal to the force of s3 on s2. Next we say that the force of s2 on s3 is equal to the force of s3 on s2 because of the 3rd law. And so on down the rope. You can see why a video would be easier than typing this in the comments.
If the elevator were accelerating upward at 2 m/s/s then it is equivalent to being on a planet with a g of (9.8+2)m/s/s. So you do the same problem but use 11.8 m/s/s. If it were accelerating downward at 2 m/s/s then you would use (9.8 - 2)m/s/s. In other words if the elevator were in free fall, then the system wouldn't move relative to the elevator (9.8 - 9.8)m/s/s or 0 m/s/s. Of course I haven't told you why this is the case. I would need to make a video to do so.
