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(B) Let the brick be divided into two halves,upper and lower. Let $M$ be the mass of the brick. Each half has mass $M/2$.Let $N_1$ and $N_2$ be the no

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Lower

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(B) Let the brick be divided into two halves,upper and lower. Let $M$ be the mass of the brick. Each half has mass $M/2$.Let $N_1$ and $N_2$ be the normal forces on the upper and lower halves,and $f_1$ and $f_2$ be the friction forces on them.For the upper half: $N_1 = (M/2)g \cos \alpha$ and $f_1 + R = (M/2)g \sin \alpha$,where $R$ is the internal interaction force between the two halves.For the lower half: $N_2 = (M/2)g \cos \alpha$ and $f_2 = R + (M/2)g \sin \alpha$.Since the brick is in equilibrium,the friction forces must balance the component of gravity along the plane. The internal force $R$ acts to push the lower half down the plane.Comparing the friction forces: $f_2 = R + (M/2)g \sin \alpha$ and $f_1 = (M/2)g \sin \alpha - R$. Clearly,$f_2 > f_1$.Since $N_1 = N_2 = (M/2)g \cos \alpha$,the contact force $C = \sqrt{f^2 + N^2}$ will be greater for the part with greater friction.Thus,the lower half exerts a greater contact force.

$A$ homogeneous rectangular brick lies on a rough inclined plane of very small inclination. Which half part of the brick exerts a greater contact force on the plane?

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