- #1
How Do You Calculate Cursor Velocity in Physics Problems?
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Homework Help Overview
The discussion revolves around calculating cursor velocities in a physics problem involving pulleys and cable systems. The original poster presents a scenario where a cursor B moves at a specified speed, and participants are asked to determine the velocities of points A and C, as well as the relationship between these velocities.
Discussion Character
- Exploratory, Assumption checking, Conceptual clarification
Approaches and Questions Raised
- Participants express the need for the original poster to show their work before assistance can be provided. Some suggest using relationships involving the velocities of A and B based on pulley mechanics. Others question the assumptions made about the number of strings affecting the velocities.
Discussion Status
There is an ongoing exploration of the relationships between the velocities of points A, B, and C. Some participants have provided partial reasoning and attempted to clarify the mechanics involved, while others seek further explanation of specific terms and relationships. The discussion is productive, with multiple interpretations being explored.
Contextual Notes
Participants note the importance of understanding the mechanics of the pulley system and the implications of string lengths on velocity calculations. There is also mention of the need for clarity in the original poster's phrasing of the problem, particularly regarding the meaning of "the velocity of C on B."
- #2
Pengwuino
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Please show some work first. We are not here to do your homework for you.
- #3
Apprentice123
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Pengwuino said:
Because of Pulleys, I think
2Vb + 4Va = 0
But not find the answer
- #4
kNYsJakE
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Apprentice123 said:
People can't help you unless you post your attempted solutions. It's a place where people help you on homework, not do your homework for you.
- #5
Apprentice123
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kNYsJakE said:
But I do not know how to solve this problem. As I will show something is not understood the problem ?
- #6
tiny-tim
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Hi Apprentice123! 
Tell us your reason (in words) for thinking it's 2Vb + 4Va = 0, and then we can see what went wrong, and we'll know how to help.
Apprentice123 said:
Tell us your reason (in words) for thinking it's 2Vb + 4Va = 0, and then we can see what went wrong, and we'll know how to help.
- #7
Apprentice123
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tiny-tim said:Hi Apprentice123!
Tell us your reason (in words) for thinking it's 2Vb + 4Va = 0, and then we can see what went wrong, and we'll know how to help.
I thought the speed at B is 1 pulley (2 * speed) and speed in A has 2 pulleys (4 * speed)
- #8
tiny-tim
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Apprentice123 said:
?? I've no idea what you're talking about
it's the number of strings that matters
Hint: if A moves distance x to the left, how much longer do the two strings on the right of A get?
- #9
Apprentice123
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tiny-tim said:?? I've no idea what you're talking about
it's the number of strings that matters
Hint: if A moves distance x to the left, how much longer do the two strings on the right of A get?
Yes. I think
B (have 2 strings)
A (have 4 strings)
2x = 150
4x = 300 Ok.
But the C portion (have 1 string) x = 75 not the answer
- #10
tiny-tim
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Apprentice123 said:
No, A has 3 strings (1 on the left, 2 on the right), and only 2 of them are on a pulley
Anyway, answer my question …if A moves distance x to the left, how much longer do the two strings on the right of A get?
- #11
Apprentice123
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tiny-tim said:No, A has 3 strings (1 on the left, 2 on the right), and only 2 of them are on a pulley
Anyway, answer my question …if A moves distance x to the left, how much longer do the two strings on the right of A get?
If A have strings (1 on the left and 2 on the right) move x to left and 2x to right. It is ?
- #12
tiny-tim
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Apprentice123 said:
It is very difficult to understand your English
"If A moves a distance x to the left, then each string on the right must get x longer, so the two strings together get 2x longer"
ok, call the length of the string on the left D, and the length of each string on the right E
then C + D + 2E = constant,
and D + E = … ?
- #13
Apprentice123
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tiny-tim said:It is very difficult to understand your English
"If A moves a distance x to the left, then each string on the right must get x longer, so the two strings together get 2x longer"
ok, call the length of the string on the left D, and the length of each string on the right E
then C + D + 2E = constant,
and D + E = … ?
Thank you.
A)
1 a left
two for right
Va = 2 x 150 = 300 mm/s
B)
1 a left
2x2 for right
Vc = 4 x 150 = 600 mm/s
C)
(2 + CB cable) = 3
Vc/b = 3 x 150 = 450 mm/s
- #14
tiny-tim
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Yes, your answers to (a) and b) are correct, but I have no idea how you got them.
In the exam, to get full marks, you will need to explain clearly how you get your answers.
Unfortunately, I do not understand what question (c) means, so I can't say whether your answer is correct … what is "the velocity of C on B"? … can you please explain it?
In the exam, to get full marks, you will need to explain clearly how you get your answers.
Apprentice123 said:
Unfortunately, I do not understand what question (c) means, so I can't say whether your answer is correct … what is "the velocity of C on B"? … can you please explain it?
- #15
Apprentice123
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tiny-tim said:Yes, your answers to (a) and b) are correct, but I have no idea how you got them.
In the exam, to get full marks, you will need to explain clearly how you get your answers.
Unfortunately, I do not understand what question (c) means, so I can't say whether your answer is correct … what is "the velocity of C on B"? … can you please explain it?
The speed of the string C on B
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