Person pulling themselves up a double pulley

[SadDan]

Homework Statement

A man is pulling himself up to a pulley that consists of two disks welded together as shown(same center). The man is currently pulling straight down on the rope in his hands with a force of magnitude 447.2 N (on the bigger disk). The other rope is also vertical and is attached to the man's waist at his center of mass(this is on the smaller disk). The man's mass is 76.3 kg, the pulley's total moment of inertia is 2.74 kg⋅m2, the radius of the small disk is 0.33 m, and the radius of the big disk is 0.62 m.

The man is hanging on a rope attached to the inner disk and is pulling on the rope attached to the outer disk. if you need a better description of the picture let me know.

Homework Equations

a(t) = angular acceleration*R
n2l for rotation and translation

The Attempt at a Solution

R=outer radii r=inner radii
sys: man Tpm+Tpm2-Mg=Ma
a=(Tpm+Tpm2-mg)/M
sys: pulley Tmp*R - Tmp2*r=angular acceleration*I
Tmp*R - Tmp2*r= a*I/r
(r*Tmp*R - Tmp2*r*r)/I= a
so far i have 2 equations and 3 unknowns, how would i solve for a from here?

 

[TSny]

Can you list the three unknowns? I see only 2.

 

[SadDan]

the 2 different tension forces and the acceleration

 

[TSny]

Aren't you essentially given one of the tension forces?

 

[SadDan]

yes, yes i am. Thank you I didn't really make that connection

 

[TSny]

The statement of the problem does not really state what you are supposed to find.
Is the following picture representative of the setup?

 

[SadDan]

sorry i guess i forgot to include that, i am supposed to find the magnitude of acceleration at the mans waist or CM. And yes that picture is the right setup

 

[TSny]

sorry i guess i forgot to include that, i am supposed to find the magnitude of acceleration at the mans waist or CM. And yes that picture is the right setup

OK. I think you have set up the equations correctly.

 

[TSny]

Hold on. Check your signs in your equations to make sure your positive direction for $a$ corresponds to your positive direction for $\alpha$.
[EDIT: Never mind, I think everything is OK!]

 

[SadDan]

So i solved for a=(F*R*r+F-M*g)/(M*r^2+I) and got the wrong answer which is probably a sign problem. If I made a positive clockwise torque and defined positive y to be up would the second tension force be negative in the first equation?

 

[TSny]

So i solved for a=(F*R*r+F-M*g)/(M*r^2+I) and got the wrong answer which is probably a sign problem. If I made a positive clockwise torque and defined positive y to be up would the second tension force be negative in the first equation?

No, both tensions act upward on the person. So, I think you are OK there.

What did you get for the $a$?

 

[SadDan]

-18.99 m/s^2

 

[TSny]

That's now what I get. I get about 5.3 m/s².

 

[TSny]

So i solved for a=(F*R*r+F-M*g)/(M*r^2+I) a

You can see something's wrong with this equation. The terms in the parentheses do not have the same dimensions.

 

[SadDan]

5.3m/s^2 worked but how did you get that? Did you just solve for Tmp in terms of a and plug that into the other equation?

 

[TSny]

I plugged your two equations (from your first post) into Mathematica and let it do the grunt work!

 

[SadDan]

ok i tried again and got the right answer, just an algebra mistake. Thank you so much

 

[TSny]

OK, good work. Did you notice that this person is going to go head-over-heels heels-over-head?

 

[SadDan]

haha yes

 

[TSny]

On second thought, I think it is head-over-heels. (Counterclockwise rotation of the person's body). Oh well.