First slide

Application of Newtons laws

Question

In figure, the acceleration of A is ${\vec{a}}_{A} = 15 \hat{i} + 15 \hat{j}$ . Then the acceleration of B is (A remains in contact with B)

Moderate

A

$6 \hat{i}$
B

$- 15 \hat{i}$
C

$- 10 \hat{i}$
D

$- 5 \hat{i}$

Solution

From constraint, the acceleration of both block and wedge should be same in a direction perpendicular to the inclined plane as shown in figure.
${(a_{A})}_{⊥} = {(a_{B})}_{⊥}$ ,
$a_{AX} = 15, a_{AY} = 15$
$a_{AX} \cos 53^{\circ} - a_{AY} \cos 37^{\circ} = a_{B} \cos 53^{\circ}$
or $a_{B} = - 5 {ms}^{- 1} or {\vec{a}}_{B} = - 5 \hat{i}$

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