Question

Abody is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^{\circ}$ to $90^{\circ}$ the contact force between the block and the plane.

• A. remains constant
• B. first remains constant then decreases
• C. first decreases then increases
• D. first increases then decreases

Answer 1

Answer: (B)

before the block starts slipping, friction is static and just sufficient to balance the force $mg \sin \theta$ along the plane.
$\therefore $ contact force $F =\sqrt{ N ^{2}+ f ^{2}} $
$ =\sqrt{( mg\,\sin\, \theta)^{2}+( mg\, \cos\, \theta)^{2}} $
$ F = mg$
but as the angle is increased block will start slipping at some angle $\theta$.
Non $f \neq \,mg\, \sin \theta$ but $f =\mu \,mg \,\cos \theta$
$\therefore F =\sqrt{( mg \,\cos \theta)^{2}+(\mu\, mg \cos \,\theta)^{2}} $
$F =\sqrt{1+\mu^{2}} \times \,mg \,\cos \,\theta$
as $\theta$ increases $\cos \theta$ decreases, contact force, $F$ decreases