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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

Ladder, Cube & Wall Problem

A ladder of length $l=2\sqrt 6$ unit is resting against a wall just touching the cube of edge 1 unit as shown in the figure. Find h above the cube.


Solution


We have, $\tan \theta  = h$ and $\sin \theta  = \frac{{h + 1}}{l}$

So, $2\sqrt 6 \sin \theta  = 1 + \tan \theta $

$ \Rightarrow \sqrt 6 (2\sin \theta \cos \theta ) = \cos \theta  + \sin \theta $

Squaring, ${(\sqrt 6 \sin 2\theta )^2} = {(\cos \theta  + \sin \theta )^2}$

$ \Rightarrow 6{\sin ^2}2\theta  = 1 + \sin 2\theta $

Let, $\sin 2\theta  = t$

We have, $6{t^2} - t - 1 = 0$

$ \Rightarrow 6{t^2} - 3t + 2t - 1 = 0$

$ \Rightarrow 3t(2t - 1) + (2t - 1) = 0$

$ \Rightarrow (3t + 1)(2t - 1) = 0$

Thus, $t = \frac{1}{2} = \sin 2\theta $ 

[Since, $\sin 2\theta >0$ the other solution is rejected.]

$\therefore 2\theta  = {30^ \circ },{150^ \circ }$

Or, $\theta  = {15^ \circ },{75^ \circ }$

Now, $h = \tan \theta  = \tan {15^ \circ },\tan {75^ \circ } = 2 \pm \sqrt 3 $