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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.

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    $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Oct 3 '18 at 8:32
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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 = $180-90-70.57=19.43^\circ$

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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

enter image description here

NOTE

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

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

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  • $\begingroup$ Interesting graph development. I attempted resolving vertical and horizontal force components, then taking moments about B. The answer is 19.4 degree, but I`m unable to calculate that value. $\endgroup$ – tas75 Oct 3 '18 at 14:12
  • $\begingroup$ @tas75 This problem can be solved only with the resultant. No need for moments. Once we know the reaction forces direction, we have $\vec W = \vec F_1 + \vec F_2$ two equations and two unknowns $|\vec F_1|$ and $|\vec F_2|$ $\endgroup$ – Cesareo Oct 3 '18 at 16:40

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