Equation Finding Strategies for Solving Physics Problems

[shin777]

so far, I know I have to find 5 equations for each object plus 2 more that I don't know of.
I wrote down everything I know but I just don't understand what to do at this point. please, be more descriptive and help me find equations. So far, I haven't gotten single useful help on the forum.

 

[mfb]

Please keep the units in your equations.

You can related alpha to a via the disk radius.
a and T1 to T4 are 5 unknown constants, (1) to (3) are 3 equations, so you just need two more. You can get those equations from the disk and the ring:
What did you calculate at (4)? T₂ would indicate the acceleration of the disk, but then T₁ is missing in the equation. In addition, where does the formula for I come from? It is a disk, not a ring.

So far, I haven't gotten single useful help on the forum.

Looking at your previous threads, I get a different impression.

 

[shin777]

i got a = 2g(m2 - m1)/(M=2m2+2m1)
a = [2(9.8)x(30-5)]/[5+2(30)+2(5)]
a = 6.53 m/s^s

m1 = 5kg, m2 = 30kg, m3 = 5kg

does it look ok?

 

[mfb]

How did you get that?
It would be a very strange coincidence if that would be right, as I don't see how you used the disk and the ring to get that formula. And even then, I think the prefactors are not right.

 

[shin777]

I followed my school tutor's lead. but more I look at this problem more I wonder it's wrong. left side of circle should calculated as disc and right side should calculated as ring but on here, it seems he count both as disc and end up having 2 boxes with 1 pulley(disc) instead of 3 boxes with 2 pulley(one disc, one ring). Is it just my imagination or is this just done wrong?

 

[mfb]

That would be the solution if the second pulley would not exist at all.

 

[shin777]

that's what i thought. first pulley is disc and second pulley is ring. no matter how many times i look at this, it didn't count second pulley into this calculation.

 

[shin777]

Fd = (1/2) Mp r^2 * a / r^2 = (1/2) Mp
Sum of forces = Mg - mg - mg - Fd - Fh = Mg - 2mg - Mp a - (1/2) Mp a = Mg - 2mg - (3/2) Mp a

Sum of forces also = (M + 2m) a

Mg - 2mg - (3/2) Mp a = (M + 2m) a

Mg - 2 mg = (M + 2m + (3/2) Mp) a

a = [(M - 2m) / (M + 2m + 1.5 Mp)] g

a = 0.41 g

a = 4.1 m/s^2

how about this one? this one seems more likely to me.

 

[mfb]

There is an "a" missing in the first line.
The way you split those forces is .. unconventional (I would sum all parts with *a and all parts with *g instead of a mixture of both), but the result looks right.