Three masses, pulleys, dynamics, and tension (Please help)

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Homework Help Overview

The problem involves three masses connected by a mass-less string over a friction-less pulley, with one mass on a tabletop that has a coefficient of friction. The original poster seeks to find the acceleration of the masses, the tension in the strings, and the effect of a friction-less tabletop on tension.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate acceleration and tension using free body diagrams (FBD) and equations of motion. Some participants question the placement of friction in the equations and the direction of acceleration, while others suggest clarifying the role of friction in the system.

Discussion Status

The discussion is ongoing, with participants providing guidance on the setup of equations and the interpretation of forces. There is an acknowledgment of the need to reconsider the direction of friction and acceleration, but no consensus has been reached on the final approach.

Contextual Notes

Participants note the complexity introduced by friction and the uncertainty regarding the direction of acceleration, which affects the setup of the equations. The original poster expresses confusion about the initial conditions and the influence of mass differences on motion.

jeff.reinecke
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Homework Statement



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 )

Homework Equations



Ʃf=ma


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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Hi Jeff, welcome top PF.

Can you show your work in detail?

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

Your observation is correct, so go with that.
 
jeff.reinecke said:
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.

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
 
you helped so much

thank you
 

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