Forces with friction problem, two block on each other

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Homework Help Overview

The discussion revolves around a physics problem involving two blocks stacked on each other, with a focus on the forces acting on them, particularly friction and tension. The context includes Newton's Laws and the dynamics of the system, specifically addressing the interactions between the blocks and an applied force.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationships between the applied force, tension, and frictional forces acting on the blocks. There is confusion regarding how to correctly set up the equations of motion, particularly in relation to the tension in the string and the frictional forces between the blocks.

Discussion Status

Multiple interpretations of the forces acting on the blocks are being discussed. Some participants suggest that the tension and friction forces must be balanced, while others emphasize the need to clarify the roles of these forces in the context of the problem. There is an ongoing exploration of how to properly account for the applied force and the interactions between the blocks.

Contextual Notes

Participants note the importance of distinguishing between the forces acting on each block and the potential confusion arising from labeling. The discussion also highlights the need to avoid unnecessary complexity in setting up the equations for the system.

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Homework Statement



The original problem with statement is http://images3a.snapfish.com/232323232%7Ffp733%3B9%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D335%3A728%3A%3B4347nu0mrj".


Homework Equations



Newton's Laws 1-3

The Attempt at a Solution



Note that the table is frictionless, the only friction is between the two boxes

So first:

[tex]\mu = .45[/tex]

and the T on M1 is

[tex]T_{m_{1}}=F_{a}[/tex]

and T on M2 is

[tex]T_{m_{2}}=-F_{a}[/tex]

and

[tex]F_{n_{1}}=m_{1}g[/tex]

[tex]F_{n_{2}}=m_{1}g+m_{2}g[/tex]

So the friction force on M1 from M2 should be:

[tex]F_{ \mu_{1}} = \mu F_{n_{1}}[/tex]

and the friction on M2 from M1 should be:

[tex]F_{ \mu_{2}}=\mu F_{n_{1}}[/tex]

So i am kind of confused as to how I should use the T forces... here is what I did:

[tex]F_{net_{x}} = F_{a}+F_{\mu_{1}-T-F_{\mu_{2}}=0[/tex]

[tex]F_{net_{y}}=F_{n_{1}}+F_{n_{2}}-m_{1}g-m_{2}g=0[/tex]

is that right? I am not sure how to set up T because for m1 t is opposite the friction force and for m2 T is opposite the Fa force... should I set up a relationship that way to do this correctly?
 
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It seems to me that Fa ≠ T. After all, the horizontal forces on block 1 must balance, and according to the FBD that YOU drew for it (which looks correct), there are three such forces. Fa is to the right and BOTH T and the frictional force from block 2 are pointing to the left.
 
Yes,horizontal forces on block 1 should balance.If T is the tension in the string acting to pull the body to left then the frictional force between the bodies will act to the right to counter the tension.So T=f.
Now coming to the second block it has got F acting to the right and tension acting to the left.Since the frictional force is acting to right on block 1,it acts to left on block 2.To sum up the whole thing,your F acts to counter both T(=f) and f.So F equals 2f.Any way i am not sure of this.:biggrin:
 
ajith.mk91 said:
Yes,horizontal forces on block 1 should balance.If T is the tension in the string acting to pull the body to left then the frictional force between the bodies will act to the right to counter the tension.So T=f.

Yeah, but you're forgetting the applied force (also points to the right). EDIT: and you also haven't specified which frictional force you're referring to. The one that acts on block 1 points to the left, in the SAME direction as the tension. I explicitly said that there were THREE horizontal forces acting on block 1 in my previous post. Let's let the OP take stock of things...
 
Block 1 is the top one and block 2 is bottom one right?Force acting on the bottom block cannot directly act on the top one.The only way for it is to use friction.Now if the friction tries to stop the relative motion it should act to right on the top block since it tends to move to left.
 
do not write vertical and horizontal equations for whole system that is unnecessary
just balance vertical and horizontal forces on each block.
Remember the tension in the string is uniform
 
ajith.mk91 said:
Block 1 is the top one and block 2 is bottom one right?Force acting on the bottom block cannot directly act on the top one.The only way for it is to use friction.Now if the friction tries to stop the relative motion it should act to right on the top block since it tends to move to left.

Hey -- I was referring to the bottom block as "block 1" without double-checking the labeling on the picture. Sorry.
 

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