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**Two boxes connected with a pulley, one hanging *updated***

## Homework Statement

There is a Mass (m

_{1}) on a table. The coefficient of friction between the table and m1 is 0.18. The second mass (m

_{2}) is hanging, suspended by a string around a pulley (m

_{3}) attached to m1. The pulley has a radius (R) and is considered a solid cylinder.

m

_{1}=3.5[kg]

m

_{2}=2.8[kg]

m

_{3}=1.9[kg]

R=0.089[m]

Find:

a) alpha of the pulley

b.)F

_{t1}

c.)F

_{t2}

d.)omega @ (6.9

## Homework Equations

I for solid cylinder = (1/2)M

_{2}R

^{2}

F

_{net}=ma

Torque

_{net}=I*alpha

alpha=a/R

## The Attempt at a Solution

I drew a torque fbd for the pulley, I found I for the pulley, I drew fbd's for the boxes, got my equations like

m

_{1}-->

X) F

_{t1}-F

_{f}=m

_{1}a

Y) F

_{n}-F

_{g}=0

F

_{n}=m

_{1}g

F

_{t1}=m

_{1}a+F

_{f}

F

_{t1}=m

_{1}a+Mew*m

_{1}*g

m

_{2}--->

Y) F

_{t2}-F

_{g}=m

_{2}(-a)

F

_{t2}=-m

_{2}a+m

_{2}g

and finally

Torque

_{net}=I*alpha so

Torque

_{1}=R*sin(90)*F

_{t1}= R*F

_{t1}(k)

Torque

_{2}=R*sin(90)*F

_{t2}= R*F

_{t2}(-k)

==>R*F

_{t1}-R*F

_{t2}=I*alpha

and alpha= a/R

So

R*(m

_{1}a+Mew*m

_{1}*g)-R*(-m

_{2}a+m

_{2}g)=((1/2)M

_{3}R

^{2})*(a/R)

That equation just has one uknown --> a

For a i got 3.97[m/s

^{2}]

For alpha I am getting like 44 [rads/s

^{2}] and others are getting about about 16 [rad/s

^{2}]

Can someone spot what, if anything, I'm doing something wrong?

EDIT: Sorry for the mistakes..

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