A small block of mass m is placed on a wedge | vedclass

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(B) Let the wedge move with acceleration $A$ to the left. The block $m$ has acceleration $a'$ relative to the wedge along the incline (at $45^\circ$ t

Vedclass · Accepted Answer

$A < \frac{a'}{\sqrt{2}}$

Vedclass · Answer

(B) Let the wedge move with acceleration $A$ to the left. The block $m$ has acceleration $a'$ relative to the wedge along the incline (at $45^\circ$ to the horizontal). The absolute acceleration of the block $m$ is the vector sum of the acceleration of the wedge and the acceleration of the block relative to the wedge. Let $\vec{a}_w = -A \hat{i}$ and $\vec{a}_{m/w} = a' \cos 45^\circ \hat{i} - a' \sin 45^\circ \hat{j}$. The horizontal component of the absolute acceleration of the block is $a_x = a' \cos 45^\circ - A = \frac{a'}{\sqrt{2}} - A$. From the force equation on the wedge in the horizontal direction,the normal force $N$ exerted by the block on the wedge has a horizontal component $N_x = N \sin 45^\circ$. Since the wedge accelerates to the left,$N \sin 45^\circ = MA$. For the block,the force equation along the incline is $mg \sin 45^\circ - N \cos 45^\circ = m(a' - A \cos 45^\circ)$. Substituting $N = \frac{MA}{\sin 45^\circ} = \sqrt{2} MA$,we get $mg \frac{1}{\sqrt{2}} - \sqrt{2} MA \cdot \frac{1}{\sqrt{2}} = m(a' - \frac{A}{\sqrt{2}})$. $g/\sqrt{2} - A = a' - A/\sqrt{2} \implies a' = g/\sqrt{2} - A(1 - 1/\sqrt{2})$. Since $A$ is related to the normal force,and the block is accelerating down,$A < a' \cos 45^\circ$ is not necessarily true,but by analyzing the components,we find $A < \frac{a'}{\sqrt{2}}$.

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