1. Jun 19, 2013

jeff.reinecke

1. The problem statement, all variables and given/known data

there are three masses m2 is on a table top with coefficient of friction of 0.350 with two different masses on each side. They are connected by a mass-less string and friction-less pulley
m1=4kg m2=1kg m3=2kg

1.]find the acceleration of all three masses

2.]find tension of the two strings

3.]if the table top were friction-less how would it effect tension (would it increase, stay the same, or decrease )

2. Relevant equations

Ʃf=ma

3. The attempt at a solution
i have attempted to find the acceleration i got 2.28m/s^2 in the negative direction of my CS

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2. Jun 19, 2013

ehild

Hi Jeff, welcome top PF.

Can you show your work in detail?

ehild

3. Jun 19, 2013

jeff.reinecke

with my FBD
i got

Ʃfm1= T1 -m1g=m1a
Ʃfm3=T2 -m3g =m3a
Ʃfm2=T2 +(-T1) +(-fk1) +fk2 + N + m(-g) =m2a

T1 = m1a + m1g + fk2
T2 = m3a + m3g + fk2

fk1 = μm2g
fk2 = μm2g

from there i just substituted in the t1 and t2's equation since everything else cancels out except for ma

4. Jun 19, 2013

voko

Note that the acceleration of the first and the third masses is equal in magnitude, but opposite in direction. That means either the first or the second equation in #3 must have the minus sign on the right hand side. To be consistent with the third equation, it has to be the first one.

Speaking of the third equation, why do you have friction twice in it?

5. Jun 19, 2013

jeff.reinecke

i had no idea which way to put friction since is initially i did not know which direction acceleration would be. Even though obviously the larger mass would force the direction of acceleration to it.

6. Jun 19, 2013

voko

Not knowing which way to put it does not make your trick possible.

Your observation is correct, so go with that.

7. Jun 19, 2013

ehild

So you have got the correct idea. The whole system accelerates along the string, m1 downward, m2 to the left and m3 upward, or it does not accelerate at all if the fiction prevents it.
Friction is against the relative motion between surfaces in contact. It acts between m2 and the tabletop, nowhere else. If m2 moves to the left the friction acts on m2 and it is directed against that motion - to the right. Write up the equations again and solve for a.

ehild

8. Jun 19, 2013

jeff.reinecke

you helped so much

thank you