Old AP Physics Problem

<p>I can't figure out two questions on an old AP Physics test essay section.</p>

<p>First of all, here are the diagram and question:</p>

<p><a href="http://img.villagephotos.com/p/2004-11/878653/physics-problem.jpg%5B/url%5D"&gt;http://img.villagephotos.com/p/2004-11/878653/physics-problem.jpg&lt;/a&gt;&lt;/p>

<p>"Blocks 1 and 2 of masses m1 and m2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks 1 and 2 move with a constant velocity v down the inclined plane, which makes an angle θ with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2*f*."</p>

<p>The two questions I'm stuck on are:</p>

<p>(1) Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane.
(2) The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.</p>

<p>Using an acceleration of 0 is getting me nowehre. Please help. :)</p>


<p>for part 1, you need to componentize your forces for all 3 masses. it's the same rope connecting all of them so the T will cancel out.</p>

<p>you're on the right track, acceleration = 0 because the masses are moving at a constant velocity, which means all the forces are balanced.</p>

<p>so you should get something that looks like:</p>

<p>Mg + f + 2f = m1gsinθ + m2gsinθ
Mg = gsinθ(m1+m2) - 3f
M = sinθ(m1+m2) - 3f/g</p>

<p>part 2 is pretty simple, you just treat m1 individually. If you componentize your forces, f is pulling it up the ramp while m1gsinθ is pulling it down, so do a net force equation</p>

<p>Fnet = ma
m1gsinθ - f = m1a
a = gsinθ - f/m1</p>

<p>I think that's right, my mechanics is a bit rusty so someone might need to doublecheck me on this, lol......... ;)</p>

<p>Wow, you just demolished that problem. ;) Now that you've laid it all out for me, it makes perfect sense. Thanks, bud.</p>