Two blocks and pulley systemhelp on small part

[Juntao]

You got to see the picture 1st.

A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.66. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.5 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

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a) With what acceleration does the mass m3 fall?
b) What is the tension in the horizontal string, T1?
c) What is the tension in the vertical string, T3?
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Okay, shouldn't be that HARD of a problem. But this is what I've done so far.

mass 1 (right is positive)
-------
Summation of forces in x direction
T1-u*m1*g=m1a

Summation of forces in y direction
N-m1g=0
N=m1g

mass 3 (up is positive)
------
Summation of forces in y direction
T3-m3*g=m3*a

My biggest question is the following equation correct:

mass 2
------
(T1-T3)R=I*alpha
simplifies to
(T1-T3)R=.5*m2*R^2*(a/R)
or
(T1-T3)=.5*m2*a => is this right for mass 2 for net torque

From here I manipulated the equations until I got a=...
This is what I have so far, but I'm not getting the acceleration value right, unless I made an algebra error.
But the equations look right so far, right?

 

[NateTG]

It looks like you've got your torque acting in the wrong direction. Try
$$\tau_{net}=T_3r-T_1r$$
as your torque formula instead.

 

[Juntao]

Ok, I did what NateTG said, and it gave me a different accerlation of -15.008 m/sec^2 for m3. But its not the right answer. Was there something else I'm supposed to do?

 

[Doc Al]

Originally posted by Juntao
From here I manipulated the equations until I got a=...
This is what I have so far, but I'm not getting the acceleration value right, unless I made an algebra error.

To avoid premature insanity, pick a consistent set of sign conventions. You know that m₁ moves to the right, the pulley moves clockwise, and m₃ goes down. So... use those directions as the positive direction in all your equations.

The way you currently wrote your equations, what you call "a" in one equation equals "-a" in another. Yikes!

 

[Juntao]

Thanks for the tip DocAl! I figured out the answer. Yes!

I learn something new everytime I post here.