What is the equation for M and why is there a T?

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Homework Help Overview

The discussion revolves around the dynamics of a system involving masses, tension, and forces, specifically focusing on the role of tension (T) in the equations of motion for a mass (M) connected via a pulley system. Participants are examining the relationships between forces acting on the masses and questioning the assumptions made in their equations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to derive equations for the forces acting on mass M, questioning the presence and role of tension in their equations. Some are verifying their methods and results against those in a textbook, while others are exploring the implications of a frictionless pulley.

Discussion Status

The discussion is active, with participants sharing their methods and questioning the correctness of their results compared to a textbook. There is a mix of agreement on certain equations, but also ongoing exploration of the implications of tension and the setup of the problem.

Contextual Notes

Some participants express confusion regarding the assumptions about the pulley being frictionless and the resulting equations for forces. There is mention of gravitational force relationships and proportions that may influence their understanding of the problem.

yolo123
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Please see picture.
Here is my method. Is it good?
T=m2g
m2g=m1a
a=m2g/m1
-n+Fpush=Ma. (n=normal force of m2 on M).
n=m2a=(m2)^2g/(m1a)

-n+Fpush=Ma.
Fpush=Ma+(m2)^2g/m
 

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yolo123 said:
Please see picture.
Here is my method. Is it good?
T=m2g
m2g=m1a
a=m2g/m1
-n+Fpush=Ma. (n=normal force of m2 on M).
n=m2a=(m2)^2g/(m1a)

-n+Fpush=Ma.
Fpush=Ma+(m2)^2g/m
No.

There is a horizontal force of T exerted to the left on M via the pulley.

There is also a downward force of T exerted on M, but that doesn't come into play here. It would if there was friction opposing the movement of M.

attachment.php?attachmentid=68689&d=1397609466.png
 
Is the answer of the book wrong :O ?
 
yolo123 said:
Is the answer of the book wrong :O ?

Their result looks correct.
 
Perfect, I will come back to it!
What about this problem? I solved the force values:
I get -(5/8)M^2G/r^2 for the first attraction of the big mass and M^2G/(2r^2) for the attraction of the small mass.

Where did they get 5/16 and 1/4?
 

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yolo123 said:
Perfect, I will come back to it!
What about this problem? I solved the force values:
I get -(5/8)M^2G/r^2 for the first attraction of the big mass and M^2G/(2r^2) for the attraction of the small mass.

Where did they get 5/16 and 1/4?
They merely used proportions.

Gravitational force is directly proportional to mass & inversely proportional to the square of distance.

attachment.php?attachmentid=68693&d=1397616243.png
 
Ok perfect!

Coming back to the previous question:
Would the equation for M be:

-T+Fpush-n=Ma?

Why is there a T? :O Is the pulley not frictionless?
 
yolo123 said:
Ok perfect!

Coming back to the previous question:
Would the equation for M be:

-T+Fpush-n=Ma?

Why is there a T? :O Is the pulley not frictionless?

Yes, the equation is
-T+Fpush-n=Ma .​
Furthermore, T = m1a as well as m2 g .

Also, n = m2a .

Plugging those into -T+Fpush-n=Ma and solving for Fpush gives
Fpush = (M + m1 + m2)a​
in agreement with the book.

Yes, the pulley is frictionless, but there is thension T to the left on it an tension T downward on it. Both of these are transmitted to block M.
 

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