Problem 531 | Friction | Engineering Mechanics Review at MATHalino

Problem 531
A uniform plank of weight W and total length 2L is placed as shown in Fig. P-531 with its ends in contact with the inclined planes. The angle of friction is 15°. Determine the maximum value of the angle α at which slipping impends.

Plank resting at some angle on two inclined planes

Solution 531

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The force polygon below is isosceles, thus, $R_B = W$

$\Sigma M_A = 0$

$(R_B \cos 30^\circ)(2L \cos \alpha) = W(L \cos \alpha) + (R_B \sin 30^\circ)(2L \sin \alpha)$

$(W \cos 30^\circ)(2L \cos \alpha) = W(L \cos \alpha) + (W \sin 30^\circ)(2L \sin \alpha)$

$2WL \cos 30^\circ \cos \alpha = WL \cos \alpha + 2WL \sin 30^\circ \sin \alpha$

$2 \cos 30^\circ \cos \alpha = \cos \alpha + 2 \sin 30^\circ \sin \alpha$

$2 \cos 30^\circ \cos \alpha - \cos \alpha = 2 \sin 30^\circ \sin \alpha$

$(2 \cos 30^\circ - 1) \cos \alpha = 2 \sin 30^\circ \sin \alpha$

$\dfrac{2 \cos 30^\circ - 1}{2 \sin 30^\circ} = \dfrac{\sin \alpha}{\cos \alpha}$

$\tan \alpha = \sqrt{3} - 1$

$\alpha = \arctan (\sqrt{3} - 1)$

$\alpha = 36.21^\circ$ answer

Problem 531 | Friction

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