# Homework Help: Two blocks on an incline, trying to find acceleration

1. Jul 10, 2007

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. The attempt at a solution

OK heres 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 cancelled 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 10, 2007

### 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.

Last edited: Jul 10, 2007
3. Jul 10, 2007

### 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.

4. Jul 10, 2007

### Dick

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

5. Jul 10, 2007

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 dont know another way.

Also, even if its an odd way of doing it, shouldnt it still give correct acceleration?

Thanks alot for the help.

6. Jul 10, 2007

### 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).

7. Jul 10, 2007

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, thats how i cancelled it.

8. Jul 10, 2007

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

9. Jul 10, 2007

### Dick

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

10. Jul 11, 2007

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)

I dont understand why it should be mg*cos 62??

I think this is source of whats messing up whole problem for me.

Thanks in advance for any more help.

Last edited by a moderator: Apr 22, 2017
11. Jul 11, 2007

Last edited by a moderator: Apr 22, 2017
12. Jul 11, 2007

### 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 [Broken]

See the triangle with the HYPOTENUSE being F_grav?

Last edited by a moderator: May 3, 2017
13. Jul 11, 2007

### 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.

14. Jul 11, 2007