Question

A homogenous block having its cross-section to be a parallelogram of sides 'a' and 'b' (as shown) is lying at rest and is in equilibrium on a smooth horizontal surface. Find the angle $\theta$, where $\theta$ is acute angle.

Answer 1

If R be the resultant of the parallelogram.

$R^2=a^2+b^2+2ab\cos \theta R^2-a^2-b^2=2ab\cos \theta \cos \theta =\frac{R^2-a^2-b^2}{2ab}\theta ={\cos }^{-1}\left(\frac{R^2-a^2-b^2}{2ab}\right)$