Two blocks on an incline, trying to find acceleration

In summary: Then the normal and tangential forces are just the magnitude of those vectors divided by the cosine of the angle between them.
  • #1
mujadeo
103
0

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!
 
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  • #2
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.
 
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  • #3
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.
 
  • #4
bel said:
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.
 
  • #5
Dick said:
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.
 
  • #6
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).
 
  • #7
Dick said:
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.
 
  • #8
OK please disregard my last post on cancelling g
I'm going to redraw the FBD and work the trig again
thanks
 
  • #9
Yeah, try again. And think carefully about the signs on the forces.
 
  • #10
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.
 
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  • #12
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 [Broken]

See the triangle with the HYPOTENUSE being F_grav?
 
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  • #13
Just because a force lies along the hypotenuse of the inclined plane triangle, doesn't mean it's the hypotenuse of the FORCE triangle.
 
  • #14
Dick said:
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
 

What is the setup for the experiment?

The setup for this experiment involves two blocks of different masses placed on an inclined plane. The angle of incline, mass of the blocks, and any other relevant variables will be determined and recorded.

What is the purpose of the experiment?

The purpose of this experiment is to determine the acceleration of the two blocks as they move down the inclined plane due to the force of gravity. This will allow us to better understand the relationship between mass, angle of incline, and acceleration.

How is the acceleration calculated?

The acceleration of the blocks can be calculated using the formula a = gsinθ, where g is the gravitational acceleration (9.8 m/s^2) and θ is the angle of incline. This formula assumes that there is no friction present on the inclined plane.

What are some potential sources of error in this experiment?

Some potential sources of error in this experiment include measurement errors, air resistance, and friction on the inclined plane. It is important to take multiple trials and average the results to minimize these errors.

How can this experiment be modified for further investigation?

This experiment can be modified by changing the angle of incline, using blocks of different masses, or introducing a frictional force on the inclined plane. These modifications can help us better understand the effects of different variables on the acceleration of the blocks.

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