Will the acceleration of two particles be different with a rough surface?

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Homework Help Overview

The discussion revolves around the effects of a rough surface on the acceleration of two particles. Participants are exploring whether the presence of friction alters the acceleration of the particles and how to appropriately model their interactions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning whether to treat the two particles as a single mass or separately when calculating acceleration. There is discussion about the implications of friction and how to account for it in their calculations. Some participants suggest using combined mass and friction to find acceleration, while others explore different equations for acceleration based on individual forces acting on each mass.

Discussion Status

The conversation is active, with participants providing guidance on considering combined forces and masses. There is acknowledgment of different approaches to finding acceleration, and some participants are refining their understanding of how to apply tension and friction in their calculations. Multiple interpretations of the problem are being explored without a clear consensus yet.

Contextual Notes

Participants are navigating the complexities of friction and its impact on acceleration, indicating a need for clarity on when to combine masses versus treating them separately. There is a focus on ensuring correct application of formulas in the context of the problem.

Jadenag
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I need help with part c. Since there is now a rough surface does that mean the acceleration of the two particles will be different? Is it wrong to simply add up the two masses and pretend that its one mass. And then find the net friction force and use f=ma directly?
I don't understand in which cases I have to consider the two particles as one and in which cases separately?
 
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Jadenag said:
27zbsjt.jpg


I need help with part c. Since there is now a rough surface does that mean the acceleration of the two particles will be different? Is it wrong to simply add up the two masses and pretend that its one mass. And then find the net friction force and use f=ma directly?
I don't understand in which cases I have to consider the two particles as one and in which cases separately?

No problem with considering them together when convenient.

Consider the combined mass and the combined friction in connection with the applied force, F, to find the acceleration.
Then use the Tension, the friction force on m2 the mass of m2 and the acceleration to find the value of T.
 
PeterO said:
No problem with considering them together when convenient.

Consider the combined mass and the combined friction in connection with the applied force, F, to find the acceleration.
Then use the Tension, the friction force on m2 the mass of m2 and the acceleration to find the value of T.

I find the acceleration to be
a= [F-μ(m1+m2)g]/m1

Couldnt I also get another equation for a?

/e
Are you suggesting I use this a and put it in f=ma for m2?
Also since I can make FBD for either of the two masses and hence i could get two different equations for a. but both would be correct right?
 
Jadenag said:
I find the acceleration to be
a= [F-μ(m1+m2)g]/m1

Couldnt I also get another equation for a?

You are using a = F/m

I like your force [accounting for the friction on both masses]

I don't like your mass so much, as you have used only m1, rather than the combined mass.
 
PeterO said:
You are using a = F/m

I like your force [accounting for the friction on both masses]

I don't like your mass so much, as you have used only m1, rather than the combined mass.

oh god. lol. sorry yes. the denominator should have the sum of the two masses. now how to approach the T
so this means there's only one correct equation for a. xd
 
Jadenag said:
oh god. lol. sorry yes. the denominator should have the sum of the two masses. now how to approach the T

having found a, now consider only mass m2 as T is the force that accelerates it.
This time you do only consider just one mass.
 
PeterO said:
having found a, now consider only mass m2 as T is the force that accelerates it.
This time you do only consider just one mass.

Considering m2

T-Ff=m2a
T-μm2g=m2a

a= [F-μ(m1+m2)g]/(m1+m2)

Hence:
T= μm2g + m2[F-μ(m1+m2)g]/(m1+m2)

Hope I got it and typed it correct too. Thankyou for the help!
 

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