How Do You Calculate Cursor Velocity in Physics Problems?

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The discussion revolves around calculating the velocities of different points in a pulley system, specifically for points A, B, and C. The velocities are determined to be 300 mm/s for A, 600 mm/s for C, and 450 mm/s for C relative to B. Participants emphasize the importance of showing work to receive assistance, as the forum is intended for collaborative problem-solving rather than direct answers. Clarifications about the mechanics of the pulley system and the relationships between the velocities are also discussed, highlighting the need for clear explanations in problem-solving. Understanding the mechanics of the strings and their lengths is crucial for accurate calculations.
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In the position shown in Figure, the cursor B moves to the left with a speed of 150 mm/s. Find (a) the velocity of the A (b) the velocity C portion of the cable (c) the velocity of C on B

Answer:
(a) 300 mm/s
(b) 600 mm/s
(c) 450 mm/s
 

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Please show some work first. We are not here to do your homework for you.
 


Pengwuino said:
Please show some work first. We are not here to do your homework for you.

Because of Pulleys, I think

2Vb + 4Va = 0

But not find the answer
 


Apprentice123 said:
In the position shown in Figure, the cursor B moves to the left with a speed of 150 mm/s. Find (a) the velocity of the A (b) the velocity C portion of the cable (c) the velocity of C on B

Answer:
(a) 300 mm/s
(b) 600 mm/s
(c) 450 mm/s

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.
 


kNYsJakE 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.

But I do not know how to solve this problem. As I will show something is not understood the problem ?
 
Hi Apprentice123! :wink:
Apprentice123 said:
Because of Pulleys, I think

2Vb + 4Va = 0

But not find the answer

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. :smile:
 


tiny-tim said:
Hi Apprentice123! :wink:


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. :smile:

I thought the speed at B is 1 pulley (2 * speed) and speed in A has 2 pulleys (4 * speed)
 
Apprentice123 said:
I thought the speed at B is 1 pulley (2 * speed) and speed in A has 2 pulleys (4 * speed)

?? I've no idea what you're talking about :confused:

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? :smile:
 


tiny-tim said:
?? I've no idea what you're talking about :confused:

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? :smile:

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
Apprentice123 said:
B (have 2 strings)
A (have 4 strings)

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? :smile:
 
  • #11


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? :smile:

If A have strings (1 on the left and 2 on the right) move x to left and 2x to right. It is ?
 
  • #12
Apprentice123 said:
If A have strings (1 on the left and 2 on the right) move x to left and 2x to right. It is ?

It is very difficult to understand your English :redface:

"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 = … ? :smile:
 
  • #13


tiny-tim said:
It is very difficult to understand your English :redface:

"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 = … ? :smile:

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
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.
Apprentice123 said:
Find … (c) the velocity of C on B

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? :smile:
 
  • #15


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? :smile:

The speed of the string C on B
 
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