Step 2 : Components.
Again, our vectors line up nicely, so on to Step 3 .
Step 3 : Equations.
Before we write any equations, we must be careful about signs: we shall call counterclockwise rotation
of the pulley “positive.”
For the more massive block, we know that, because it is not flying off the table or tunneling into it, it
is in equilibrium in the up–down direction. But it is not in equilibrium in the right–left direction.
For the less massive block, we only have one direction to concern ourselves with: the up–down direction.
F (^) net = T − mg = ma
We can solve for T from the “F (^) net, (^) x ” equation for the more massive block and plug that value into the “F
net ” equation for the less massive block, giving us
(−Ma ) − mg = ma
We rearrange some terms to get
Now we plug in the known values for M and m to find that
To finish the problem, we plug in this value for a into the “F (^) net, (^) x ” equation for the more massive block.
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