Newton's Laws; ramps and pulleys

AI Thread Summary
The discussion focuses on solving a physics problem involving Newton's laws, specifically regarding a system with ramps and pulleys. The user initially struggles with understanding how to apply forces and tension in the context of a frictionless ramp system. After creating a free-body diagram and setting up the equations of motion, they derive the equations for the forces acting on both masses. Ultimately, they solve for the mass m2, arriving at a value of 11.0 kg, which they confirm is likely correct based on the method used. The conversation highlights the importance of free-body diagrams and proper application of Newton's laws in solving physics problems.
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Homework Statement


http://img201.imageshack.us/img201/6315/diagramyk1.gif
-Assuming the system is frictionless, find m2 when it is accelerating down the ramp at 3.4 m/s2.

Homework Equations


-... F = ma?

The Attempt at a Solution


-In this problem, I tried tilting the axis so the sliding-down part aligned with the y-axis and the m1 part on the x-axis. Then, I thought of using the components to the Fw, to figure out the Tension on the rope for the m2 side (which, I guessed, to be m2a.).
I couldn't figure out what to make the Tension equal to, because m2gcos(\theta) wasn't coming out the same as 3.4*m2.

...I then realized I was quite retarded at physics (I'm being perfectly honest, now.) and needed help.
 
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did you make the free body diagram?
 
oh, sorry! I forgot to attach it... >_<;;;

http://img297.imageshack.us/img297/3359/diagram2im0.gif

I suppose what confuses me is the fact that they're on ramps; I don't know how/where to draw the gravity part.

(apologies if this is frustrating for you. I'm just really bad at physics.)
 
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nooooooooooo
this is what a free body diagram for such a situation looks like.
now make the other one. write the appropriate equation ie. Fnet = m*a.
eliminate T from both the equations. you have "a", find "m" and show us what you got. ok??
 

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The weight acts at the centre of gravity (mass) of the block.
 
i guess OP is asking the direction of gravitational force andrevdh
 
(well, my retardation may have reached a new low.)

I figured out how to do the free-body diagram right this time, and got:

http://img254.imageshack.us/img254/5139/diagramca9.gif

\SigmaFx = m2gsin\alpha - T = m2a.
and
\SigmaFx = T- m1gsin\beta = m1a.

so if I solve for T it would be m1a + m1gsin\beta, and substitute that into the first equation to figure out m2a.

with grouping like terms, plugging in numbers, solving, &c, I got 11.0 kg.

...would that be right?

(and a big thank you for the help!)
 
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i didnt calculate the values, but the method you described, answer should be alright.
 

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