I like so much this exercise about the massive rope. Very well explained. Problems like this must be explained by professors like you ! Thank you MIT !!
Why the Tension on the direction of Mg is T(x+deltax)?? I tried other way, but I am no sure (I am a lego) T(x)=Mg((L-x)/L). 1º eq T(x+deltax)=Mg((L-x-deltax)/L) 2º eq if I subtract 2º eq-1º eq I get T(x+deltax)-T(x)=Mg((L-x-deltax-L+x)/L) so i get T(x+deltax)-T(x)=-(Mg*deltax)/L T(x+deltax)-T(x)/deltax= -Mg/L this is the same eq. that he got. please help me to know if this is correct
Ji hoo Lee you're totally right. That's because what he said doesn't make sense at all. What we want is to find the tension for a small piece of the rope, hence T(x+delta(x))-T(x). Then we add'em up in order to find the function T(x)
Tension at a point is the magnitude of an action-reaction pair. The upward tension at position x+deltax -- which is T(x+deltax) -- is equal in magnitude to the downward tension.
@Ji Hoo Lee The function T(x) is formulated as the tension in the rope based on the positive magnitude “x” below the rope support. I think you’re confused with how the function is written as x is inversely proportional with the value of T(x), which you are bringing up assuming T(x) is directly proportional with x. Not sure what Zonnymaka’s comment is talking about as the free body diagram is drawn and explained correctly in the lecture. The small section of rope (delta x) will encounter the tensile forces of the rope acting in opposite directions, with the difference made up of that segment’s delta m times g. To Zonnymaka: If you’re going to claim someone is giving wrong information, you should be much more clear in your explanation, because it seems like you’re the one who doesn’t make sense at all.
I had trouble understanding what’s happening here but I think I finally figured it out. The tension T(x) is caused by all the rope above the point x and the tension T(x + delta x) is caused by all the rope below x + delta x. We also have to consider the weight of the element itself delta mg. Since the element is static, the total force on the element sums to 0.
Here's the link to 13.1 from the playlist: th-cam.com/video/9NS0JcjNdp4/w-d-xo.html. For more info, see the course on MIT OpenCourseWare: ocw.mit.edu/8-01F16. Best wishes on your studies!
unbelievably clear for a relatively hard concept. well done
I like so much this exercise about the massive rope. Very well explained. Problems like this must be explained by professors like you ! Thank you MIT !!
Why the Tension on the direction of Mg is T(x+deltax)?? I tried other way, but I am no sure (I am a lego)
T(x)=Mg((L-x)/L). 1º eq
T(x+deltax)=Mg((L-x-deltax)/L) 2º eq
if I subtract 2º eq-1º eq I get
T(x+deltax)-T(x)=Mg((L-x-deltax-L+x)/L) so i get
T(x+deltax)-T(x)=-(Mg*deltax)/L
T(x+deltax)-T(x)/deltax= -Mg/L
this is the same eq. that he got.
please help me to know if this is correct
Damn these lectures are saviour
U r a Great teacher👏👏Kudos to u
Why is the direction of the tension force upward for T(x) and downward for T(x+delta x) ?
you saved me, thank you Professor
Amazing explanation. Thanks!
5:49 Why is the downward force T(x+delta x), shouldn't it be T(L-(x+delta x))?
Ji hoo Lee you're totally right. That's because what he said doesn't make sense at all.
What we want is to find the tension for a small piece of the rope, hence T(x+delta(x))-T(x).
Then we add'em up in order to find the function T(x)
Tension at a point is the magnitude of an action-reaction pair. The upward tension at position x+deltax -- which is T(x+deltax) -- is equal in magnitude to the downward tension.
JEE students be like-noobs 😂
@Ji Hoo Lee The function T(x) is formulated as the tension in the rope based on the positive magnitude “x” below the rope support. I think you’re confused with how the function is written as x is inversely proportional with the value of T(x), which you are bringing up assuming T(x) is directly proportional with x. Not sure what Zonnymaka’s comment is talking about as the free body diagram is drawn and explained correctly in the lecture. The small section of rope (delta x) will encounter the tensile forces of the rope acting in opposite directions, with the difference made up of that segment’s delta m times g. To Zonnymaka: If you’re going to claim someone is giving wrong information, you should be much more clear in your explanation, because it seems like you’re the one who doesn’t make sense at all.
I had trouble understanding what’s happening here but I think I finally figured it out. The tension T(x) is caused by all the rope above the point x and the tension T(x + delta x) is caused by all the rope below x + delta x. We also have to consider the weight of the element itself delta mg. Since the element is static, the total force on the element sums to 0.
Can someone please give me the link to the video lecture before this one?
Here's the link to 13.1 from the playlist: th-cam.com/video/9NS0JcjNdp4/w-d-xo.html. For more info, see the course on MIT OpenCourseWare: ocw.mit.edu/8-01F16. Best wishes on your studies!
@@mitocw That's very kind! I somehow managed to utterly confuse myself and not find it.
how can something be that simple and yet so beautiful
Indians rock 🤘🤘
MMMMMM , lovely
Proud to be a bengali
❤️