A bucket with mass m2 and a block with mass m1 are hung on a pulley system.

Forces that affect the bucket $m_{2}$ :

$1.F_{G2}…weight$

$2.T_{1}…tension force$ by which the rope affects the bucket

Forces that affect the block $m_{1}$ :

$1.F_{G1}…weight$

$2.T_{1} …tension force$ by which the rope affects the block

$3.T_{2}…tension force$ by which the rope affects the block

The vector force equation for the bucket:

$F_{G2}+T=m_{2} \times a_{2}(1)$ The vector force equation for the block:

$F_{G1}+T_{1}+T_{2}=m_{1} \times a_{1}(2)$

We choose the y-axis the way it is marked on the picture.

We rewrite equations (1) and (2) to scalar form:

$T-F_{G2}=m_{2}a_{2}(3)$ $F_{G1}-T_{1}-T′_{2}=m_{1}a_{1}(4)$

Because we leave the mass of the pulley system and the rope out of account they have no moment of inertia and don’t affect the tension forces. The following holds for the magnitude of the tension forces:

$|T|=|T_{1}|=|T_{2}|$

We rewrite equations (3) and (4):

$T-F_{G2}=m_{2}a_{2}(5)$ $F_{G1}-2T=m_{1}a_{1}(6)$

Relation between the magnitudes of acceleration $a_{1} and a_{2}$

If the bucket goes up a distance s , the block goes down a distance $\frac{x}{2}$

$s=\frac{1}{2}a_2t^2$

$\frac{s}{2}=\frac{1}{2}a_1t^2$

$2=\frac{a_2}{a_1}$

$a_2=2a_1\tag{7}$

Let's divide the rope in 3 sections - L1, L2 and L3. L1 is the left section pulling M1 up, L2 is the rope between the two pulleys and L3 is section pulling the bucket up. L1 + L2 + L3 make up the length of the rope which is fixed.

Now at time = 1 sec, the lengths of these ropes increase or decrease by the acceleration amounts in these sections. But the overall length of rope is fixed.

So,

(L3- $a_2$ ) + (L2+ $a_1$ ) + (L1+ $a_1$ ) = L1 + L2 + L3

This gives us $a_2$ = 2 $a_1$