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satisf
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Homework Statement
Given is the setup in the attachment, I think the values speak for itself. Additionally, there is a friction coefficient of 0.36 that works in on the 2 blocks.
So in short: 2 masses (1 one a slope), connected through a pulley, that does have friction and to complete it, the ground has friction to.
Asked:
a) Accelleration of blocks (0.309m/s2)
b) Tension in both cords (7.67N and 9.22N)
I actually got quite far, I even know what I am missing in my own derivation, a factor of 2 (will explain below). But I can't manage to see why. It's in the derivation for a, a to b Is no problem.
Homework Equations
[tex]F=ma[/tex]
[tex]\tau=RFsin(\theta)[/tex]
[tex]\sum \tau = I\alpha[/tex]
[tex]a_t=r\alpha[/tex]
The Attempt at a Solution
Using free-body calculations I found expressions for T1 (left block) and T2 (right block), as follows:
[tex]T_1=m_1a+0.36m_1g[/tex]
[tex]T_2=-m_2a+m_2g\mathrm{cos}(\frac{\pi}{3})-0.36m_2g\mathrm{sin}(\frac{\pi}{3})[/tex]
I'm quite sure that these are correct, since substituting the solution for a in these gives me the correct tensions.
Now to bring in the pulley I started with this:
[tex]\sum \tau = I\alpha=I\frac{a}{R}=MR^2\frac{A}{R}=MRa[/tex]
Then, knowing that [tex]\tau_1=RT_1\mathrm{sin}(\theta_1)[/tex] and [tex]\tau_2=RT_2\mathrm{sin}(\theta_2)[/tex] and both sinuses will equal one I get to:
[tex]RF_1+RF_2=MRa[/tex]
or the more familiar
[tex]F_1+F_2=Ma[/tex]
But this is where things go wrong, as far as I know it should be [tex]\frac{M}{2}[/tex] on the right hand side. And yes, when I do put in a factor of 2 I get the correct solution.
So now my question is: where the hell in my derivation did I forget the factor?
P.S. while previewing this post the latex got messed up, If it's not solved in final submit, can some moderator adjust it?