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

AI Thread Summary
In a modified Atwood's machine with a mass sliding on an incline, the pulley rotates counterclockwise due to the differing weights of the masses (1 kg and 2 kg) and the influence of kinetic friction (μk = 0.15). The user calculated forces acting on both masses, including gravitational, tension, and frictional forces. They struggled with the signs in their equations and the correct application of friction in their calculations. Ultimately, they found the acceleration to be -0.183 m/s² but expressed frustration over the lack of responses and support in the forum. The discussion highlights the complexities of analyzing forces in a system with friction and inclined planes.
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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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