Ladder on smooth inclined plane

Question

A uniform ladder AB of weight W and length 2d rests in equilibrium with its upper end A against a smooth vertical wall and its lower end B on a smooth inclined plane. The inclined plane is 10 degrees to the the horizontal. Find what angle the ladder makes with the wall.

Because the wall and inclined plane are smooth surface`s, only normal reactions will exist at the wall and inclined plane.

Answer 1

In the attached figure, the inclined floor is represented by line $L_1$ Line $L_3$ is perpendicular to the supporting vertical wall. Line $L_2$ is perpendicular to $L_1$ and intersects $L_3$ The intersection point ($p_i$) has the same $x$ coordinate as the ladder mass center ($G$).

$$ A = \lambda_4(0,1)\\ L_1\to B=\lambda_1(v_x,v_y) \\ L_2\to p_i = B+\lambda_2(-v_y,v_x)\\ L_3\to p_i = A+\lambda_3(1,0)\\ G = \frac{1}{2}(A+B)\\ G_x = (p_i)_x\\ \lVert A-B\rVert = L $$

Solving for $\lambda_1,\lambda_2,\lambda_3,\lambda_4$ we have

$$ \lambda_2 = \frac{L}{\sqrt{v_x^2+4v_y^2}}\\ \lambda_1 = \frac{2v_yL}{v_x\sqrt{v_x^2+4v_y^2}}\\ \lambda_3 = \frac{v_yL}{\sqrt{v_x^2+4v_y^2}}\\ \lambda_4 = \frac{(v_x^2+2v_y^2)L}{v_x\sqrt{v_x^2+4v_y^2}} $$

After that we can build the plot

NOTE

$L$ represents the ladder lenght. $v = (v_x,v_y) = (1,\tan(\frac{10\pi}{180}))$

$$ \angle BAO=\alpha = \arccos\left(\frac{(0,1)\cdot(B-A)}{\lVert B-A\rVert}\right) = 19.4254^{\circ} $$

Answer 2

vertical resolution: $wg=R \cos10\cdots(1)$

horizontal resolution: $S=R \sin10\cdots(2)$

moments about B: $wg d \cos\alpha = S \times 2d \sin\alpha$

$\therefore \quad wg \cos a=2S \sin\alpha$

$\therefore \quad \frac{wg}{2S}=\tan\alpha\cdots(3)$

(1) and (2) $\Rightarrow$ (3) $\frac{R \cos 10}{2R \sin 10}=\frac{1}{2} \cot 10=\ \tan\alpha$

$\therefore \quad \tan\alpha=\frac{5.67128}{2}=2.83564$ so $\alpha=70.57^\circ$

Thus by considering right-angle triangle, Angle A = $180-90-70.57=19.43^\circ$