Find the acceleration of B. | vedclass

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Question

(B) Let the acceleration of wedge $A$ be $a$ and the acceleration of sphere $B$ be $b$.From the constraint of contact between the wedge $A$ and sphere

Vedclass · Accepted Answer

$\frac{a \cos \alpha_1}{\cos \alpha_2}$

Vedclass · Answer

(B) Let the acceleration of wedge $A$ be $a$ and the acceleration of sphere $B$ be $b$.From the constraint of contact between the wedge $A$ and sphere $B$,the component of acceleration of both bodies along the normal to the contact surface must be equal.The normal to the vertical surface of wedge $A$ is horizontal. The acceleration of wedge $A$ is $a$ at an angle $\alpha_1$ with the horizontal. The horizontal component of $a$ is $a \cos \alpha_1$.The acceleration of sphere $B$ is $b$ at an angle $\alpha_2$ with the horizontal. The horizontal component of $b$ is $b \cos \alpha_2$.Equating the horizontal components of acceleration along the normal:$b \cos \alpha_2 = a \cos \alpha_1$Solving for $b$:$b = \frac{a \cos \alpha_1}{\cos \alpha_2}$

Find the acceleration of $B$.

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