Tension in pulley systems Concept help

  • #1
I need help understanding why in problems tension is modeled as T-mg = ma.

I understand there are different variations, but the main concept is what I need help with.

For example:

If i have 2 masses hanging off a pulley (m1=5kg & m2=10kg) then m1 has a downward force of 49N down. The formula given is T-m1g=ma. and i understand that t-m1a=g is the force, which is why it is set = to ma, BUT i cannot grasp the concept of how tension is found that way and why for m2 it is m2g-T2=m2a..

I need a good concrete explanation. Baby steps and spell it out for me to understand please!!
 

Answers and Replies

  • #2
arildno
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Okay:
I think,perhaps, your lecturer is a bit too swift on this, and has made some simplifications that can seem rather confusing.

We'll start with the following observation:
The rope is taut, and of the same length all the time (agreed to that one?).
Now, try to formulate, in your own words:
If we call object 1's acceleration a_1, object 2's acceleration a_2, what must be the relation between these two accelerations?
 
  • #3
Okay:
I think,perhaps, your lecturer is a bit too swift on this, and has made some simplifications that can seem rather confusing.

We'll start with the following observation:
The rope is taut, and of the same length all the time (agreed to that one?).
Now, try to formulate, in your own words:
If we call object 1's acceleration a_1, object 2's acceleration a_2, what must be the relation between these two accelerations?
considering its in a system where a is constant wouldn't the acceleration be the same?
 
  • #4
jhae2.718
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The blocks will have different accelerations. You apply Newton's 2d Law for each block. It's possible they will be the same, but not necessary.

I would suggest you start attacking the problem by drawing free-body diagrams of each blocks. Then define your coordinate system/reference frame(s), get the kinematics of the block (this is very simple for rectilinear motion!), and apply Newton's 2d Law.

You will get your "formula" from the free-body diagrams.
 
  • #5
arildno
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"wouldn't the acceleration be the same?"

They are equally "large", right, but while one block accelerates upwards, the other accelerates downwards. Agreed?

This gives us the equation I'll call (1):

a_1=-a_2 (1) (Here, the minus sign is included to indicate acceleration in opposit directions!)

"considering its in a system where a is constant "
Do you understand that this is irrelevant?

The crucial factor is that IF block 1 didn't accelerate upwards as much as block 2 accelerates downwards (or vice versa), THEN the distance between them, measured as the length of the rope, would change. It does not matter whether the acceleration is constant or not. Do you understand that?

-----------------------------------------
" It's possible they will be the same, but not necessary."
Completely incorrect. That the rope is taut and remains of the same length throughout the period, is a kinematic constraint on the problem at hand; as necessary to handle as everything else. You have three basic unknowns in your problem, hence you need three equations, not just two, in order to solve it.
Think about it..:smile:
 
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  • #6
jhae2.718
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" It's possible they will be the same, but not necessary."
Completely incorrect. That the rope is taut and remains of the same length throughout the period, is a kinematic constraint on the problem at hand; as necessary to handle as everything else. You have three basic unknowns in your problem, hence you need three equations, not just two, in order to solve it.
Think about it..:smile:
I know; I'm not sure why I wrote what I did last night, but under the assumptions that were made that statement is indeed a falsehood.

Thanks for pointing that out.
 

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