# Horzontal Massive Pulley Problem with Friction

1. Nov 1, 2012

### Kristenx2

1. The problem statement, all variables and given/known data
Consider 2 masses (m1=13kg, m2=26kg) connected by a string. The pulley is a uniform disc of radius R=0.05m and mass m=1kg. μ between the horizontal surface and the larger mass is 0.2.
Find Frictional force acting on larger mass.
Find the linear acceleration of the blocks.
Find tension of the string acting on both weights.
Calculate the angular acceleration of the pulley in rad/s2.
Calculate the time it will take block M to move 1m when released from rest.
Calculate the number of revolutions made by the pulley during the time period of the precious problem.

2. Relevant equations
F=μ*m*g
m2g-T2=m2a
α=τ/I
I=1/2mr2
T2-T1=1/2mpa
That's all I could really figure out to use. Would I also use m1g*μ-T1=m1a?

3. The attempt at a solution

I got the friction force fine, 51.012N.

I messed up the next part thinking the pulleys were massless. I did a=(m2g-μm1g)/(m1+m2 and got 1.962 m/s2. Naturally, for the tension part, I ended up with the same tensions of 102N. I know that's not right.

So I did T2-T1=1/2mpa and ended up with 1.94m/s2. That seems wrong as well, because my T1 ended up being 0.646N (mgμ-T=ma). It just seems really off and I am not sure what I am doing wrong. I am almost positive it has something to do with the way I am incorporating the friction, though.

This is what the diagram looks like:

O\_______|m1(26kg)|____
|//////////////////////////////////
|
|
|M2(13kg)|

2. Nov 1, 2012

### grzz

Of course you HAVE also to use T$_{1}$ - Frictional force = 26a.

Otherwise you would not have enough equations to solve for the all the unknown physical quantities that you have.

3. Nov 1, 2012

### Kristenx2

Hi, grzz. Sounds good. Does that mean when I go to find the linear acceleration I would use m2g-m2a-m1gμ-m1a = 1/2mpa?

4. Nov 1, 2012

Yes.

5. Nov 1, 2012

### Kristenx2

So my linear acceleration really is 1.92m/s2?
It just seems strange because when I plug that into (26)(9.81)(0.2)-T=(26)(1.92) I get an obviously ridiculous number, and for the other tension, (13)(9.81)-T=(13)(1.92) I get 102N.

6. Nov 1, 2012

### Kristenx2

Oh! I am doing it backwards. You said T-Fμ=26a, right? So I was actually wrong about (26)(9.81)(0.2)-T=(26)(1.92) and I will not get a low or negative number.

7. Nov 1, 2012

### Kristenx2

For the angular acceleration of the pulley {alpha} I did R2*T2-R1*T1=I{alpha} and got 105.2 where R=0.05m and I used 102 and 49.4 for my Ts, and 1.94m/s2 for my acceleration. However, when I use {alpha}=a/R, using the same values, I end up with 38.8rad/s2... Which one? ;c

8. Nov 1, 2012

### grzz

My values are
T1 = 101.4N
T2 = 102.3N
a = 1.94m/s2