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- Thread starter aliciaw0
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marlon

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So i am assuming that the pulley has no friction and its rotation does not need to be taken into account, also the pulley does has no mass and can be treated as a point particle.

Just draw a FBD for each mass seperately.

The first mass has friction f with the table (i assume this, if it is not the case than f =0 ofcourse). T is the tension in he rope

[tex]m_1a_1 = T-f[/tex]

For mass 2 you get

[tex]m_2a_2=2T-m_2g[/tex]

hence [tex]m_1 = \frac{m_2(a_2+g)-2f}{2a_1}[/tex]

This is the most general form. I do not know whether extra info was given on the masses standing still or not ? If so, some variables in the above formula will vanish

marlon

Just draw a FBD for each mass seperately.

The first mass has friction f with the table (i assume this, if it is not the case than f =0 ofcourse). T is the tension in he rope

[tex]m_1a_1 = T-f[/tex]

For mass 2 you get

[tex]m_2a_2=2T-m_2g[/tex]

hence [tex]m_1 = \frac{m_2(a_2+g)-2f}{2a_1}[/tex]

This is the most general form. I do not know whether extra info was given on the masses standing still or not ? If so, some variables in the above formula will vanish

marlon

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- #3

aliciaw0

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and there was no other information about the masses standing still

- #4

Doc Al

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Show how you arrived at this answer.aliciaw0 said:i don't understand why

(-m_2*g)/(((-m_2)/2)-m_1) isn't right

That's correct. (Be careful with signs.)because the a of mass#2 is (-a_1)/2 right?

- #5

aliciaw0

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x m_1: T=m_1*a_1

y m_2: T-m_2*g=m_2*a_2

2(m_1*a_1)*g=m_2*a_2

a_2= a_1/2

2 (m_1*a_1)*g=m_2*(a_1/2)

y m_2: T-m_2*g=m_2*a_2

2(m_1*a_1)*g=m_2*a_2

a_2= a_1/2

2 (m_1*a_1)*g=m_2*(a_1/2)

- #6

Doc Al

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OK.aliciaw0 said:x m_1: T=m_1*a_1

Two problems here:y m_2: T-m_2*g=m_2*a_2

(1) The rope pulls

(2) Careful with your sign convention. If you want to use up for positive, then be consistent. (You chose a positive a_1 going to the right.)

If m_1 moves to the right, then m_2 moves down: this means that if you call a_1 positive, then by your sign convention for m_2, its acceleration should bea_2= a_1/2

Regarding sign conventions, here's what I do: I always assume a_1 and a_2 to be positive numbers. I also visualize how the accelerations relate to each other. If a_1 is to the right, then a_2 is down. Then I write my equations accordingly.

Try it again.

- #7

aliciaw0

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m2 y: 2T-m_2*g=m_2*a_2

since T= m_1*a_1 i subsituted that it and solved for a_1 and got

a_1= (2*m_2*g)/(4m_1+m_2) is this right?

- #8

Doc Al

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Looks good to me.

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