# Homework Help: Newton's Laws; ramps and pulleys

1. Jul 22, 2007

### glacialis

1. The problem statement, all variables and given/known data
http://img201.imageshack.us/img201/6315/diagramyk1.gif [Broken]
-Assuming the system is frictionless, find m2 when it is accelerating down the ramp at 3.4 m/s2.

2. Relevant equations
-... F = ma?

3. 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.

Last edited by a moderator: May 3, 2017
2. Jul 23, 2007

### ank_gl

did you make the free body diagram???

3. Jul 23, 2007

### glacialis

oh, sorry! I forgot to attach it... >_<;;;

http://img297.imageshack.us/img297/3359/diagram2im0.gif [Broken]

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.)

Last edited by a moderator: May 3, 2017
4. Jul 23, 2007

### ank_gl

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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5. Jul 23, 2007

### andrevdh

The weight acts at the centre of gravity (mass) of the block.

6. Jul 23, 2007

### ank_gl

i guess OP is asking the direction of gravitational force andrevdh

7. Jul 23, 2007

### glacialis

(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 [Broken]

$$\Sigma$$Fx = m2gsin$$\alpha$$ - T = m2a.
and
$$\Sigma$$Fx = 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!)

Last edited by a moderator: May 3, 2017
8. Jul 24, 2007

### ank_gl

i didnt calculate the values, but the method you described, answer should be alright.