Q15 (a) (ii)

Criteria	Marks
Provides correct solution	2
Writes the equation from part (i) in terms of $\ell$, $d$ and $h$ OR shows that $\theta>\phi$ OR considers the case where $P$ is $\frac{2h}{3}$ OR equivalent merit	1

Sample answer:
$\tan\theta=\frac{Mg}{T_2\cos\phi}+\tan\phi$
From the diagram,
$\tan\theta=\frac\ell d,\qquad\tan\phi=\frac{h-\ell}{2d}$
Therefore,
$\frac\ell d=\frac{Mg}{T_2\cos\phi}+\frac{h-\ell}{2d}$
$\frac{Mg}{T_2\cos\phi}>0$
Thus,
$\frac\ell d>\frac{h-\ell}{2d}$
$\ell>\frac{h-\ell}{2},\quad\frac{3\ell}{2}>\frac h2,\quad\ell>\frac h3$
$h-\ell<\frac{2h}{3}$
$\therefore P\text{ cannot be lifted to a position }\frac{2h}{3}\text{ metres above the floor.}$