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 $f\ne mg\sin \theta$ 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