How Does the Pulley Rotate in a Modified Atwood's Machine with Friction?

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Homework Help Overview

The problem involves a modified Atwood's machine with a mass sliding on an inclined surface, where friction plays a role in the dynamics of the system. The setup includes two masses, M1 and M2, and the question focuses on the direction of pulley rotation and the time it takes for one mass to reach the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze forces acting on both masses and set up equations of motion. They express uncertainty about the signs of forces, particularly regarding friction, and seek clarification on their approach to solving for acceleration and tension.

Discussion Status

Some participants have provided feedback on the original poster's calculations, including a note on the application of the friction coefficient. However, the original poster expresses frustration with the lack of responses and indicates they have resolved their issue independently.

Contextual Notes

The original poster mentions a known answer for acceleration and is focused on improving their problem-solving skills for future tests. There is an indication of imposed homework rules, as the discussion is framed within a homework help context.

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Homework Statement



2) In the variant of Atwood’s machine shown below, a mass slides on an inclined surface 40θ above the horizontal with a coefficient of kinetic friction μk = 0.15. The masses M1 and M2 are 1 kg and 2 kg, respectively. Which way does the pulley rotate? If the lowest point of each mass is initially 2m above the ground (vertically), and the system has a very small but non‐zero initial speed, how long does it take the first mass to hit the ground?

Homework Equations


F=MA



The Attempt at a Solution



So here's how I went about the problem. I first got a force diagram going and found all the forces

F-parallel = 15.01N
F-Perpend=12.59N
F-gravity=19.6N
F-Tens=?
This is for Mass 2 on the incline

For mass 1 I got

F-gravity 9.8
F-Tens=?

So I have been trying to use the Ʃ all forces=ma for both equations, this gives me

M2= ƩF=ma=Ftens-Ffrict+Fparallel
ƩF=2kg*a=Ftens+(-.15*12.59N)+15.01N

M1=ƩF=ma=Ftens-Fgrav
ƩF=1kg*a=Ftens-1N

Then I put the above M1 equation into M2 and solve for acceleration then plugging in a into either equation to get Ftens. I already know the answer a=-.183 I just want to beat on my craft so when the test rolls around I can set up quick variable templates then plug and chug. Areas of doubt are,

-signs
-what to do about friction in the sum I already know its going anticlockwise so Friction is negative correct?
-Plugged it in many times and cannot get the known answer.

Thanks for any help and I will keep this post updated if I come across the answer before I receive help
 
Physics news on Phys.org
one mistake the friction coefficient is multiplied to F parallel
 
for my final equation I get a= -m2g-Ffriction+Fparallel/Mass1+Mass2

Help
 
figured it out. Complete waste of time on here... Get an infraction quickly but no response.
 

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