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A cubical block of mass ' m ' and side ' a' is placed on a rough inclined plane of inclination \theta. The direct and shear stresses in the block are _____

Option: 1

f_d=\frac{m g \cos \theta}{a^2}, f_s=\frac{m g \sin \theta}{a^2}


Option: 2

f_d=\frac{m g \sin \theta}{a^2}, f_s=\frac{m g \cos \theta}{a^2}


Option: 3

f_d=\frac{2 m g \cos \theta}{a^2}, f_s=\frac{2 m g \sin \theta}{a^2}


Option: 4

f_d=2 m g \sin \theta, f_s=2 m g \cos \theta


Answers (1)

best_answer

The direct stress \left(f_d\right) is due to m g \cos \theta and the shear stress \left(f_s\right) at the bottom of the cube is due to m g \sin \theta.

Therefore,

\begin{aligned} & f_{\text {direct }}=\frac{m g \cos \theta}{A}=\frac{m g \cos \theta}{a^2} \\ & f_{\text {shear }}=\frac{m g \sin \theta}{A}=\frac{m g \sin \theta}{a^2} \end{aligned}

Posted by

sudhir kumar

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