ME Dynamics - Multiple ideal pulleys + an inclined plane

[Feodalherren]

Homework Statement

Find the acceleration of both blocks

Homework Equations

Dynamics

The Attempt at a Solution

Everything in black is given in the problem. The red/blue is my work.

First I want to relate the acceleration of block A to block B by finding the constraint equations:

Lrope = 2L1 + L2 + 2L3 + constants(lengths around pulleys)

taking the derivative twice to find the acceleration:

0 = 2L1 + L2 + 2L3

I feel like I have too many variable here. I should be able to knock it down to two so that I can relate the acceleration of A to B, correct? How would I do that?

The forces on B, all in Y using the normal xy-coordinate system.

MbAb = 3T - Mb(g)

The forces on A, using a coordinate system along the incline where n1 is "up" and n2 is positive along the slope DOWN:

Sum of the forces in n1 = ma(n1) = Na - mgCos30
Sum of the foces in n2 = ma(n2) = -2T - Fa + mgSin30

Relating the two coordinate systems:

n1 = Cos 30 j - Sin30 i
n2 = -sin30 j - cos30 i

Have I done everything correctly so far?

 

[Feodalherren]

There's a small error in the drawing. The small pulley right above L3 is fixed to the large pulley above it.

 

[gneill]

To investigate your constraint equation, consider the situation if block A moved upslope by some distance x. Two strings are "giving up" a length x, so that makes a length of 2x available to pass over the top pulley. That "new" length has to be distributed over how many lengths as the bottom pulley moves down? By how much must each extend in order to accommodate the "new" 2x?

 

[Feodalherren]

They must extend by 2x/3 ?

 

[Feodalherren]

They must extend by 2x/3 ?

 

[gneill]

Looks good.

 

[Feodalherren]

So then

a1 = 2/3 a2?

 

[gneill]

So then

a1 = 2/3 a2?

Well, depending upon which block is a1 and which is a2, that seems to be a valid conclusion.

 

[Feodalherren]

Ops :). a1 would be the acceleration of block A and a2 wold be the acceleration of block B.