<p>Can somebody please help me on #2 on the following webpage:</p>
<p>Thank you in advance!</p>
<p>For block A,
2Mg - T(v) = 2Ma
So T(v) = 2Mg - 2Ma (answer to question a) </p>
<p>For the pulley, both tensions on vertial and horizontal directions work against each other. Using Torque = (Force)(Radius) = Iα, with a = Rα, (non-slipping)</p>
<p>(T(v) - T(h))R = (2Mg - 2Ma - T(h))R = (3MR^2)(a/R)
So 2Mg - 2Ma - T(h) = 3Ma
So T(h) = 2Mg - 5Ma (answer to question b)</p>
<p>On block, there are two horizontal forces acting: T(h) to the right and the frictional force from block C to the LEFT. (action-reaction pair: block C doesn't want to slip to the left relative to block B, so fricitional force acts to the right; so in return, block B experiences a frictional force to the left) </p>
<p>T(h) = 2Mg - 5Ma = 2Mg - 5M(2) = 2Mg - 10M
Force of friction = uN = 4Mg(u) </p>
<p>So T(h) - (force of friction) = (2Mg - 10M) - 4Mg(u) = 3M(2)
so setting g=10, then u = 0.1 (answer to c)</p>
<p>The only force block C experiences horizontally, is the above mentioned frinctional force to the right. So, </p>
<p>4Mg(u) = 4Mg(0.1) = (4M)a
a = 1 (answer to d)</p>
<p>I would answer but he said it perfectly ^</p>
<p>Thank you so much vju1988! I get it now!</p>
<p>yay. i'm happy to hear that:)</p>