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

AI Thread Summary
The discussion focuses on a physics problem involving three masses connected by a string over a frictionless pulley, with one mass on a table with friction. The user attempts to calculate the acceleration and tension in the system, initially arriving at an acceleration of 2.28 m/s². Participants emphasize the importance of correctly applying friction in the equations and clarifying the direction of forces acting on the masses. They discuss how the absence of friction would affect tension, noting that friction opposes motion and should only be considered where applicable. The conversation highlights the need for clear equations and understanding of dynamics in solving the problem effectively.
jeff.reinecke
Messages
11
Reaction score
0

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
 

Attachments

  • homework help dynamics.jpg
    homework help dynamics.jpg
    61.3 KB · Views: 622
Physics news on Phys.org
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
 

Similar threads

Replies
18
Views
4K
Replies
3
Views
2K
Replies
102
Views
7K
Replies
15
Views
5K
Replies
10
Views
2K
Replies
10
Views
5K
Replies
33
Views
3K
Back
Top