# Acceleration of two masses on pulley.

1. Mar 16, 2015

1. The problem statement, all variables and given/known data
Two masses are connected by massless and frictionless pulleys and string. The coefficients of friction are µs = 0.35 and µk = 0.25 between mass m1 and the table. The masses are m1 = 8.0 kg and m2 = 8.0 kg.

Find the acceleration of each mass and the tension in the string. (Note: Because of the pulley, if m1 moves a distance L, then m2 moves half that amount, a distance L/2. Think about how this affects the relative acceleration of the two blocks.)

2. Relevant equations
a1=-1/2a2
fk=ukn=-ukm1g
T-ukm1g=m1a1
T-m2g=m2a2

3. The attempt at a solution

I think the acceleration of block 2 will be negative since it is falling and block 1 will have a positive acceleration since it is moving towards the right.
Ok, I hope I am right in assuming that the accelerations are related as I've written above.
I substituted the a2 into into the first T equation to get T-ukg=-(1/2)m1a2.To cancel the T out, I subtracted the equations to get: (m2-ukm1)g=-1/2 (m2+m1)a2

Then punching in my variables I got a2=-7.35 m/s^2 which I was pleased to see is negative. But then if I use my relationship as a1=-1/2 a2 then I get a1=3.7m/s^2, which is at least positive. But I don't think it makes sense that the acceleration of block 2 would be faster than block 1?

I can't decide where I went wrong? I think it might be my relationship between the two accelerations, but today I asked my professor and she said it looked OK. So now I think I messed something up with the friction... I'm using kinetic but the problem also gave static... do I need to use both? Or does fk not equal ukmg?

I'd appreciate any suggestions. Thanks.

Last edited: Mar 16, 2015
2. Mar 16, 2015

### andrewkirk

The first three equations look OK. It is not obvious to me how you got your fourth equation, as the tension T on the string is not directly pulling the pulley upwards but rather is going around the pulley. I would expect the upward force on $m_2$ to be either half or double of T, but it could be something else. I doubt that it's T though.

When you subtract the two T equations you get
$(m_2-\mu_km_1)g=-\frac{1}{2}(m_2+m_1)a_2$
but the correct subtraction gives
$(m_2-\mu_km_1)g=-\frac{1}{2}(2m_2+m_1)a_2$

I find problems like this easier to do by using conservation of energy, so that we don't need to try to work out difficult things like what is the upward force on $m_2$.

Approaching it that way, we can write an equation:

KE of m1 + KE of m2 = loss of PE of m2 + distance moved by m1 x sliding frictional force on m1

This gives an equation in tems of $s_1,s_2,\dot{s}_1,\dot{s}_2$ where $s_i$ is the displacement of mass $i$ from its starting position (so that $a_i=\ddot{s}_i$). Using your first equation, we can reduce that to an equation in $s_1$ and $\dot{s}_1$, which is easily solved by integrating and using the initial conditions that, presumably $\dot{s}_1(0)=0$ and defining $s_1(0)$ to be 0.

You can then get the accelerations of the two masses by differentiating, and get the tension from your third equation.