A question about the perpendicular vector

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The discussion revolves around the challenge of drawing a perpendicular vector correctly in relation to a reverse vector. Participants express uncertainty about the original drawing due to a lack of visibility in the attachment. There is a suggestion to start with a clear reference point to illustrate the desired perpendicular vector. Additionally, the importance of understanding the cross product is highlighted, with an invitation to share examples for better clarity. Overall, the conversation emphasizes collaborative learning in vector mathematics.
student120
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Homework Statement
I tried to draw it perpendicular to the reverse, but I'm not sure if it's right.
Relevant Equations
I tried to draw it perpendicular to the reverse, but I'm not sure if it's right.
I tried to draw it perpendicular to the reverse, but I'm not sure if it's right.

1729584309179.png

I would be very grateful if you could help me. I am in the learning process
 
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It's not clear from the attacement what is the full question?
 
:welcome:

student120 said:
I tried to draw it perpendicular to the reverse, but I'm not sure if it's right.
Neither am I, because I cannot see what you did!
Start with number (1) and show us where you want to draw ##\vec B##:

1729592936130.png

student120 said:
I am in the learning process
We all are :smile: !
If it is the cross product itself that causes you problems, post a typical example.

##\ ##
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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