Two blocks on an incline, trying to find acceleration

[mujadeo]

Homework Statement

2 identical blocks tied together with a string that passes over a pulley at creast of inclined planes
One makes an angle of (theta1) 28deg to horizontal
other makes angle of (theta2) 62deg to horizontal

Homework Equations

if no friction, then with what acceleration do the blocks move?

The Attempt at a Solution

OK here's what I did (but its wrong apparantely).

I orientated axix on both blocks so that Normal force points in pos y direction
tension points along x axis

SO, according to how i orientated axises, I can diregard Fsuby entirely, because I am only concerned with motion along the x-axis.

so forces in x direction are:
Fsubxsuba= Tsuba - -mg/sin28 = ma
Fsubxsubb = -Tsubb + mg/cos28 = ma

I know Tsuba = Tsubb, so I'll just call it "T" now
So after I canceled out mg on both sides i was left with 2 unknons and 2 equations.
I set them equal to each other to find a:

After doing the algebra I ended up with:

2a = 1/cos28 + 1/sin28

thus a = 1.63 ms/s

this is wrong though.
Can sum1 help thanks alot!

 

[Dick]

1) why did you DIVIDE by the trig functions? Isn't that a little eccentric? 2) How can you cancel g?? It doesn't occur on both sides of the equation. And 3) shouldn't the mg forces be pointed in opposite directions? It looks like you added the equations together to cancel T. Well done. But then shouldn't the mg forces appear as a difference rather than a sum? Finally, think about simple tests to make on your answer, e.g. if the angles are equal the acceleration should be zero.

 

[bel]

The way you set up your axes sounds good, but try not to worry about the tension, and focus on the acceleration in the x direction (given by the x-axis, i.e., along the planes). Having found those, add the acceleration vectors together to find net acceleration.

 

[Dick]

The way you set up your axes sounds good, but try not to worry about the tension, and focus on the acceleration in the x direction (given by the x-axis, i.e., along the planes). Having found those, add the acceleration vectors together to find net acceleration.

How can the OP "not worry about the tension"? The problem is being done basically correctly - but has some odd technical errors.

 

[mujadeo]

1) why did you DIVIDE by the trig functions? Isn't that a little eccentric? 2) How can you cancel g?? It doesn't occur on both sides of the equation.

OK this had me worried too, but when i worked the trig, it always ended up that way.
So, if angle from horizontal is 28deg, and I know adjacent side (mg).. then wouldn't Hypotenuse be mg / sin 28deg?

Thats how I was doing it, i don't know another way.

Also, even if its an odd way of doing it, shouldn't it still give correct acceleration?

Thanks a lot for the help.

 

[Dick]

No, it's odd enough to give you a completely wrong answer. Draw a force diagram. There is an mg vertical force (hypotenuse!), the normal and tangential parts are mg*cos(theta) and mg*sin(theta).

 

[mujadeo]

2) How can you cancel g?? It doesn't occur on both sides of the equation. .

Tsuba = Tsubb so I called it T.

So T = ma - mg / sin28
and T = mg/cos28 - ma

Then I equated the right sides of the equations:

ma - mg/sin28 = mg/cos28 - ma

So mg is on both sides of equations, that's how i canceled it.

 

[mujadeo]

OK please disregard my last post on cancelling g
I'm going to redraw the FBD and work the trig again
thanks

 

[Dick]

Yeah, try again. And think carefully about the signs on the forces.

 

[mujadeo]

I am not understanding why the trig should not be mg/cos62.
Heres my diagram, along with my free body diagram.

http://www.imagination3.com/LaunchP..._172908170_1134883715_usa&transcript=&_lscid=

Heres is a drawing that shows how I arrive at mg/cos 62
(I am using costheta = adj/hyp to get the hyp angle, which runs along the neg x-axis)

http://www.imagination3.com/LaunchPage?aFileType=&_nolivecache&sessionID=&message=trig 1&room_email=&from_email=mujadeo@yahoo.com&from_name=muji&to_email=mujadeo@yahoo.com&to_name=&aDrawingID=20070711_173539785_855161508_usa&transcript=&_lscid=

I don't understand why it should be mg*cos 62??

I think this is source of what's messing up whole problem for me.

Thanks in advance for any more help.

 

[mujadeo]

http://www.imagination3.com/LaunchP...1_180442270_230643192_usa&transcript=&_lscid=

(last link to trig drawing didnt work for sum reason)

 

[Dick]

You didn't draw your forces in such a way as to form a triangle at all.

Try this site:

http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l3e.html

See the triangle with the HYPOTENUSE being F_grav?

 

[Dick]

Just because a force lies along the hypotenuse of the inclined plane triangle, doesn't mean it's the hypotenuse of the FORCE triangle.

 

[mujadeo]

Just because a force lies along the hypotenuse of the inclined plane triangle, doesn't mean it's the hypotenuse of the FORCE triangle.

aha.

i'm going to rework problem again