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Question

(A) The contact force between the block and the plane is the resultant of the normal force $(N)$ and the frictional force $(f)$.For a block of mass $m

Vedclass · Accepted Answer

$m g$

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(A) The contact force between the block and the plane is the resultant of the normal force $(N)$ and the frictional force $(f)$.For a block of mass $m$ on an inclined plane at angle $	heta$ in equilibrium:$1$. The normal force is $N = m g \cos 	heta$.$2$. The frictional force is $f = m g \sin 	heta$.The contact force $(F_c)$ is given by the vector sum of these two perpendicular forces:$F_c = \sqrt{N^2 + f^2}$$F_c = \sqrt{(m g \cos 	heta)^2 + (m g \sin 	heta)^2}$$F_c = \sqrt{m^2 g^2 (\cos^2 	heta + \sin^2 	heta)}$Since $\cos^2 	heta + \sin^2 	heta = 1$,we get:$F_c = \sqrt{m^2 g^2} = m g$.Thus,the correct option is $A$.

$A$ block of mass $m$ is stationary on a rough inclined plane making an angle $\theta$ with the horizontal. Find the contact force between the block and the plane.

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