Velocity of mass with pulleys

  • #1

Homework Statement


Kokakola.png


Homework Equations


xcxcxcxcxcxcxcxcxcxcxc[/B]


The Attempt at a Solution


My attempt:
Let displacement of A be x m,
Let displacement of B be y m, and
Let displacement of C be z m.
Step2: [/B]a+b+c+d=(a+x-y)+(b+x)+c+(d+z)
2x-y=-z
Differentiating LHS and RHS components w.r.t dt:
d2x/dt - dy/dt = -dz/dt
2vx- vy= -vz
20-5=-z
15=-z
z= -15m/s (In upwards direction. Is this what the -ve sign means?)

Teacher's attempt:
Let displacement of A be x m,
Let displacement of B be y m, and
Let displacement of C be z m.


Step2: a+b+c+d=(a+x-y)+(b+x-y)+c+(d+z)
0=2x-2y+z
Differentiating by dt on both sides:
2vy-2vx= vz
10-20=vz
-10=vz
v=-10m/s


Please tell why is my attempt wrong (In step two)

Thank you for reading.
 

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Answers and Replies

  • #2
jbriggs444
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As I understand your steps, you have computed the vertical velocity of object C (i.e. vz). Is its resulting motion purely vertical?
 
  • #3
Yes, you are correct.
Also, everything is rigid here: moving bodies, pulleys, strings, everything.
 
  • #4
lewando
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Step2: a+b+c+d=(a+x-y)+(b+x)+c+(d+z) [OP's attempt]
The highlighted term is the only difference in your setup compared with the teacher's setup. This term is also not correct.

Also, jbriggs444 asked an important question, don't overlook.
 
  • #5
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It's easier to do this problem by adopting an inertial frame of reference that is moving to the left with the speed of B (5 m/s), so that B appears stationary. As reckoned from this frame of reference, A would be moving to the left at 5 m/s.
 
  • #6
It's easier to do this problem by adopting an inertial frame of reference that is moving to the left with the speed of B (5 m/s), so that B appears stationary. As reckoned from this frame of reference, A would be moving to the left at 5 m/s.
So if I take B as the reference point, then A would apparently move at 15m/s. Right?
But my real question is in step two of the answer. Why has the teacher subtracted length 'y' from b?
 
  • #7
As I understand your steps, you have computed the vertical velocity of object C (i.e. vz). Is its resulting motion purely vertical?
Yes. you are correct. The motion is purely vertical and the bodies are rigid.
 
  • #8
The highlighted term is the only difference in your setup compared with the teacher's setup. This term is also not correct.
So, if my attempt isn't correct, then what will be the correct term?
And why?
 
  • #9
TSny
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But my real question is in step two of the answer. Why has the teacher subtracted length 'y' from b?
Could it be that the teacher is assuming that the lower, right pulley is somehow attached to B (even though the picture shows no attachment)? If so, then this pulley moves to the left with B.
 
  • #10
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So if I take B as the reference point, then A would apparently move at 15m/s. Right?
Wrong. A would apparently move at 5 m/s.
But my real question is in step two of the answer. Why has the teacher subtracted length 'y' from b?
There is no step two to my answer. The length of strings a and b are both increasing by 5 m/s. The length of string c is constant. So the length of string d must be decreasing by 10 m/s.
 
  • #11
Wrong. A would apparently move at 5 m/s.

There is no step two to my answer. The length of strings a and b are both increasing by 5 m/s. The length of string c is constant. So the length of string d must be decreasing by 10 m/s.
Okay. THank you. Though I still don't get why it's 5m/s.
But thank you anyways.
 
  • #12
Could it be that the teacher is assuming that the lower, right pulley is somehow attached to B (even though the picture shows no attachment)? If so, then this pulley moves to the left with B.
Yes. Perhaps that is the problem. THank you for replying.
 
  • #13
jbriggs444
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Yes. you are correct. The motion is purely vertical and the bodies are rigid.
Think again.
 
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  • #14
lewando
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But my real question is in step two of the answer. Why has the teacher subtracted length 'y' from b?
If you represent the change in length a as a + (x-y), then why would you not think the same for length b? If the pulley between b and c is rigid, as you have stated in post #3 (TSny points out the lack of attachment to B in the graphic), then the right side of b moves with B, same as the right side of a.
 
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  • #15
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Okay. THank you. Though I still don't get why it's 5m/s.
But thank you anyways.
In vector form, the velocity of block A is ##\mathbf{v_A}=-10\mathbf{i_x}##, where ##\mathbf{i_x}## is the unit vector in the positive x direction. Similarly, if vector form, the velocity of block B is ##\mathbf{v_B}=-10\mathbf{i_x}##. Therefore, the relative velocity of block A with respect to block B is ##\mathbf{V_{AB}}=\mathbf{v_A}-\mathbf{v_B}=-5\mathbf{i_x}\ m/s##
 
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  • #16
In vector form, the velocity of block A is ##\mathbf{v_A}=-10\mathbf{i_x}##, where ##\mathbf{i_x}## is the unit vector in the positive x direction. Similarly, if vector form, the velocity of block B is ##\mathbf{v_B}=-10\mathbf{i_x}##. Therefore, the relative velocity of block A with respect to block B is ##\mathbf{V_{AB}}=\mathbf{v_A}-\mathbf{v_B}=-5\mathbf{i_x}\ m/s##
I get it now. Finally. THank you so much for bearing with a 15 year old :D
I was simply adding them earlier. Now I know what's correct.
 
  • #17
Yes. I was wrong to assume that B will only increase in x direction.
If you represent the change in length a as a + (x-y), then why would you not think the same for length b? If the pulley between b and c is rigid, as you have stated in post #3 (TSny points out the lack of attachment to B in the graphic), then the right side of b moves with B, same as the right side of a.
 
  • #18

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