- #1
Still stuck on two blocks and pulleys
- Thread starter tigerseye
- Start date
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- Tags
- Blocks Pulleys Stuck Two blocks
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Homework Help Overview
The discussion revolves around a problem involving two blocks and pulleys, specifically focusing on determining the vertical acceleration of mass 1 in relation to mass 1, gravity, and tension in the system.
Discussion Character
- Exploratory, Mathematical reasoning, Assumption checking
Approaches and Questions Raised
- Participants are attempting to derive expressions for the acceleration of mass 1, with various formulations presented. Questions are raised regarding the correctness of these expressions and the underlying reasoning behind them.
Discussion Status
There is an ongoing exploration of different equations for acceleration, with some participants providing algebraic manipulations and others questioning the derivation process. Guidance has been offered regarding the relationship between the accelerations of the two masses and the implications of string length constraints.
Contextual Notes
Participants are working under the assumption that the system involves two forces acting on mass 1 and that the relationship between the accelerations of the two masses is critical to solving the problem. There is a noted lack of consensus on the correct expressions for acceleration and tension.
- #2
Yegor
- 147
- 2
a_1= ((m_1*g)-T)/m_1
a_1= (4/5)*(m_1*g)/(m_1+m_2)
Is it correct?
a_1= (4/5)*(m_1*g)/(m_1+m_2)
Is it correct?
- #3
Yegor
- 147
- 2
Sorry. I'd like to say:
a_1= 4*(m_1*g)/(4*m_1+m_2)
If it is correct, i'll try to explain
a_1= 4*(m_1*g)/(4*m_1+m_2)
If it is correct, i'll try to explain
Last edited:
- #4
tigerseye
- 16
- 0
Can you explain how you got that? 
- #5
Yegor
- 147
- 2
there is important, that a_2=(a_1)/2 (look carefull at string).
On first mass acts 2 forse: gravity and tension. So:
m_1*a_1= (m_1*g)-T
But on the 2nd mass acts F=2*T
m_2*a_2=2*T
m_2*(a_1)/2=2*T
We have a system:
m_2*(a_1)/2=2*T
m_1*a_1= (m_1*g)-T
We don't know a_1 and T.
If we solve:
T=m_2*(a_1)/4
And a_1= 4*(m_1*g)/(4*m_1+m_2)
Am I right?
On first mass acts 2 forse: gravity and tension. So:
m_1*a_1= (m_1*g)-T
But on the 2nd mass acts F=2*T
m_2*a_2=2*T
m_2*(a_1)/2=2*T
We have a system:
m_2*(a_1)/2=2*T
m_1*a_1= (m_1*g)-T
We don't know a_1 and T.
If we solve:
T=m_2*(a_1)/4
And a_1= 4*(m_1*g)/(4*m_1+m_2)
Am I right?
- #6
tiger_striped_cat
- 49
- 1
I didn't check your algebra on your last step but it looks right.
to derive your accelearation relation, you don't have to "look at the strings" and think it through. in these more difficult problems you should derive a constant from the problem. In this case, the length of the string is constant. If I call the length from the wall to the stationary pully L1 the length from the stationary pully to the moving pully x and the length from the stationary pully to m2 y then I can derive the formula:
L1+x+x+y=LT
Where LT is the total length of the string. Takeing time derivatives of both sides twice:
[tex]2\ddot{x}+\ddot{y}=0[/tex]
then negnative sign you get should be argued away. So then you get the relations for acceleration that you derived.
to derive your accelearation relation, you don't have to "look at the strings" and think it through. in these more difficult problems you should derive a constant from the problem. In this case, the length of the string is constant. If I call the length from the wall to the stationary pully L1 the length from the stationary pully to the moving pully x and the length from the stationary pully to m2 y then I can derive the formula:
L1+x+x+y=LT
Where LT is the total length of the string. Takeing time derivatives of both sides twice:
[tex]2\ddot{x}+\ddot{y}=0[/tex]
then negnative sign you get should be argued away. So then you get the relations for acceleration that you derived.