Block on inclined plane with pulleys and friction

[phlstr]

Homework Statement

coefficient of kinetic friction = 0.15

Find the acceleration of block a.

Homework Equations

see below

The Attempt at a Solution

http://www.wolframalpha.com/input/?i=-200sin30%2B0.15*200cos30%2B2x%3D-200%2F32.2*a%3Bx-40%3D40%2F32.2*b%3Ba%3D0.5*b

I used x for tension and a and b for the acceleration of the blocks.

I assumed that the acceleration of block a is down the plane and so block b would go up, but the answer I got was negative! I tried assuming that block a goes up and block b goes down, but I got a negative answer again! Are the equations that I used wrong? or the blocks don't move at all?

 

[tiny-tim]

hi phlstr!

your geometric constraint b = 0.5a is wrong, it should be b = -0.5a

(look at the diagram! )

 

[phlstr]

Hi :). But I already made the net mass*acceleration negative for A and the net force for B positive. I used the geometric constraint just for the magnitude..

and I think you mean a=0.5b. :shy:

 

[tiny-tim]

Hi :). But I already made the net mass*acceleration negative for A and the net force for B positive.

no, both equations are measuring up as positive

(we can see that eg from the fact that your x (tension) is positive both cases)

so if a increases, b decreases ​

 

[phlstr]

What do you mean by measuring up as positive?

 

[tiny-tim]

What do you mean by measuring up as positive?

i mean uphill for A, and directly up for B

 

[phlstr]

So these are my equations:

$-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+2T=-\frac{200}{32.2}a$
I use negative values for forces pointing downhill, and positive for pointing up. Since I assumed that the net force for block A is going down, I made the right side negative..

$T-40=\frac{40}{32.2}b$
Negative is down and positive is up. Since block A is going downhill, then block B is accelerating upwards and the net force is upwards.

$a=0.5b$
The magnitude of the acceleration of block A is half of the acceleration of block B.

This is how I understand the equations that I'm using.. Can you point where it's wrong?

 

[tiny-tim]

ah, now that i see it on the same page, it does looks right

ok, i get positive times a on the LHS = a positive number on the RHS,

what did you get?

 

[phlstr]

I plugged the equations in wolfram alpha and got
a=-0.534946 b=-1.06989, T=38.6709

I also tried using substitution and got the same for A...
and it's negative??

 

[tiny-tim]

I also tried using substitution

let's see

 

[phlstr]

$-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+2T=-\frac{200}{32.2}a$
$T-40=\frac{40}{32.2}b$
$a=0.5b$

$T=\frac{40}{32.2}b+40$
$2a=b$
$T=\frac{40}{32.2}2a+40$

$-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+2\left(\frac{40}{32.2}2a+40\right)=-\frac{200}{32.2}a$

$-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+\frac{160}{32.2}a+80=-\frac{200}{32.2}a$

$-200sin\left(30\right)+(0.15)(200)cos\left(30\right)+80=-\frac{200}{32.2}a-\frac{160}{32.2}a$

$5.980762114=-11.18012422a$

$a=-0.5349459$

 

[tiny-tim]

oh … very small and negative!

ie uphill

so if the kinetic friction is uphill, the motion would have to be uphill also, which is impossible

and if we make the kinetic friction downhill, then that +25 becomes -25, and the motion is also downhill, again impossible

so the blocks don't move …

you could find the friction force F by solving the same pair of equations, with 0 on the RHS and F in the middle of the LHS ​