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- Thread starter yolo123
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SammyS

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No.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

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.

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Is the answer of the book wrong :O ?

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SammyS

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Is the answer of the book wrong :O ?

Their result looks correct.

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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?

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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SammyS

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They merely used proportions.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?

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

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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?

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SammyS

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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 = mAlso, n = m

Plugging those into -T+Fpush-n=Ma and solving for F

F_{push} = (M + m_{1} + m_{2})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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