Confused About Weird Acceleration Problem in Physics: mA=14kg, mB=11kg, mC=50kg

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Homework Help Overview

The problem involves a system of three blocks with given masses (mA = 14 kg, mB = 11 kg, mC = 50 kg) and their accelerations. The original poster expresses confusion about the acceleration relationships between the blocks, particularly regarding the acceleration of block B in relation to block A, and questions the validity of their calculations and assumptions about forces acting on the blocks.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationships between the accelerations of the blocks, with some suggesting that the original poster's equations may need adjustments based on the motion of block C. There are inquiries about the normal force acting on block A and the implications of the system's constraints.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting that the original poster needs to reconsider their equations and the relationships between the blocks. There is an ongoing exploration of the correct equations for acceleration and net forces, with no clear consensus reached yet.

Contextual Notes

The original poster has referenced a specific source for the problem and is attempting to reconcile their findings with the provided solutions, indicating a potential misunderstanding of the system's dynamics. There are also hints about the importance of considering the geometry of the setup in relation to the accelerations.

Hafid Iqbal
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http://img849.imageshack.us/img849/9697/physicsy.jpg


The system bodies shown in the figure is released from rest. Assuming all the contacts to be smooth, find the acceleration of each block (mA = 14 kg, mB = 11 kg, mC = 50 kg)



This problem from Problems In Physics by Abhay K. Singh
I really confused about the answer, it said the acceleration in y-axis for B is [tex]a_{By}=-3a_{Ay}[/tex] which acceleration of each block is [tex]a_A = \sqrt{2}, \ a_B = \sqrt{10}, \ a_C = 1[/tex]

I have tried to figure it out and still find [tex]a_{By}=-2a_{Ay}[/tex] and all of my answer didn't match with the answer by the author

I have had :
[tex]a_{Ay}=\frac{(2m_B-m_a)g}{4m_B+m_a}[/tex],

[tex]a_{By}=\frac{(4m_B-2m_a)g}{4m_B+m_a}[/tex],

[tex]a_{C} = \frac{3m_A.m_B.g}{(4m_B+m_A)(m_C+m_B)}[/tex]

One more thing, is it true that normal force for [tex]A[/tex] is zero?
 
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Welcome to PF!

Hi Hafid! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

Your aBy = -2aAy would be correct if block C were not moving.

You'll need to add aCx to your aBy and aAy equation.

Try again (and if you still can't get it, please show your actual calculations :wink:).​
One more thing, is it true that normal force for [tex]A[/tex] is zero?

hmm … if the diagram is correct, then A is accelerating to the right, so the normal force would have to be negative! :rolleyes: Unless A is constrained to move in a slot, A must be hanging at an angle. :redface:
 


tiny-tim said:
Hi Hafid! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

Your aBy = -2aAy would be correct if block C were not moving.

You'll need to add aCx to your aBy and aAy equation.

Try again (and if you still can't get it, please show your actual calculations :wink:).​


hmm … if the diagram is correct, then A is accelerating to the right, so the normal force would have to be negative! :rolleyes: Unless A is constrained to move in a slot, A must be hanging at an angle. :redface:


Hi tiny-tim, thanks for your reply! But I'm still stuck :(
Here's my complete paper work


Phewww... And you know what, it is still wrong :cry:
 
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Wow so neat!:cool: Handwriting that is. And the colours!
 
oooh, I like the colours! :-p

But you're still using aBy = 2aAy. :redface:

You need to mark your diagram with something to show how far across C the middle pulley is. :wink:
 
Iclaudius said:
Wow so neat!:cool: Handwriting that is. And the colours!
Hehe... Thanks Iclaudius... :)
tiny-tim said:
oooh, I like the colours! :-p

But you're still using aBy = 2aAy. :redface:

You need to mark your diagram with something to show how far across C the middle pulley is. :wink:

Hmmm... So, what is the right equation for the acceleration?
And, is my equation for mass C true?
 
Hafid Iqbal said:
Hmmm... So, what is the right equation for the acceleration?

dunno, I haven't worked it out, I was going to let you do it, and then criticise. :wink:

(i'm good at that o:))
And, is my equation for mass C true?

Doing mass C on its own is not a good idea, since you need to know the reaction forces from A and B.

Do the (horizontal) equation for masses A B and C together. :smile:
 
I have tried it all over, and still can't solve it, please if anyone here know the right equation for acceleration and net force on mass C help me...
 
The way to find the equation relating the accelerations is to find the equation relating the positions first, and then differentiate.

So call the distance from the middle pulley to the wall p, and the distance from the middle pulley to the right-hand pulley q …

then what are the x component of C, and the y components of A and B (in terms of p and q)? :smile:
 
  • #10
Weird problem and answer but some crazy fun diagram. Never seen someone solve physics question so colorfully.
For the problems regarding acceleration use this hint: The length of string is constant irrespective of the motion.
 

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