# Homework Help: How do you find a = here?

1. Jun 25, 2013

### 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.

Last edited: Jun 25, 2013
2. Jun 25, 2013

### Staff: Mentor

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.

3. Jun 25, 2013

### 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?

4. Jun 25, 2013

### Staff: Mentor

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.

5. Jun 25, 2013

### 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?

#### Attached Files:

• ###### problem1.docx
File size:
438 KB
Views:
75
6. Jun 25, 2013

### Staff: Mentor

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

7. Jun 25, 2013

### 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.

8. Jun 25, 2013

### 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

9. Jun 26, 2013

### Staff: Mentor

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.