Help me in understanding use of vector in this problem 



In the question here, velocity of block A is $5 \;\text{m/s}$. To find the velocity $v$ of block B, why do we use $v \cos 53° = 5$ instead of $5 \cos 53° = v$. Isn't the $5 \;\text{m/s}$ velocity acting on the block B the resultant velocity? So why can't we take components of $5 \;\text{m/s}$ here?
 A: If mass $B$ was pulled directly towards the pully $P$ at 5m/s, then yes, its component to the left would be $5\cos{53}^{\circ}$.
But it's different if the mass stays on the table.
Imagine $B$ slides a small distance $dx$ to the left, to point $C$, in a time $dt$.
From triangle BCD can see that $$dx \cos{53}^{\circ} = BD\tag1$$.
Since triangle $DPC$ is isosceles, if the angle $DPC$ is small, then $BD$ is the distance moved by $A$, so dividing both sides of $(1)$ by $dt$ $$V_B \cos{53}^{\circ} = 5$$

A: This problem will be easier to consider the instantaneous displacement: if the displacement of A is $\Delta s$ (in the interval $\Delta t$), then what is the displacement required for the block B (constrained to the horizontal line)?
\begin{align}
\Delta s &= \frac{h}{\sin(53^o-\Delta \theta)} - \frac{h}{\sin 53^o} \\
&= h \csc(53^o-\Delta\theta) - h \csc 53^o\\
&= h \left\{\csc53^o +\csc53^o\cot53^o \Delta\theta \right\} - h \csc 53^o\\
&= h \csc53^o\cot53^o \Delta\theta \tag{1}
\end{align}
The displacement along the horizontal for block B:
\begin{align}
\Delta s_B &= \frac{h}{\tan(53^o-\Delta \theta)} - \frac{h}{\tan 53^o} \\
&= h \cot(53^o-\Delta \theta) - h \cot 53^o\\
&= h \left\{ \cot53^o + \csc^2 53^o\Delta\theta \right\} - h \cot53^o\\
&= h\csc^253^o \Delta\theta \,\,\,\,\,\,\,\text{ using Eq.(1) for} \Delta\theta\\
&= h\csc^253^o \frac{\Delta s}{ h \csc53^o\cot53^o} \\
&= \frac{\Delta s}{ \cos53^o}
\end{align}
Divide $\Delta t$
$$
  v_B = \frac{\Delta s_b}{\Delta t} = \frac{\Delta s}{ \Delta t\cos53^o}=\frac{v_A}{\cos53^o}
$$

