Inclined planes stuck on someting

  • Thread starter NUBIG10
  • Start date
  • #1
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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 heres 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
 

Answers and Replies

  • #2
4
0
one mistake the friction coefficient is multiplied to F parallel
 
  • #3
4
0
for my final equation I get a= -m2g-Ffriction+Fparallel/Mass1+Mass2

Help
 
  • #4
4
0
figured it out. Complete waste of time on here... Get an infraction quickly but no response.
 

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