Two fixed frictionless inclined planes making | vedclass

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Question

(B) For a block on a frictionless inclined plane making an angle $	heta$ with the horizontal,the acceleration along the plane is $a = g \sin 	heta$.

Vedclass · Accepted Answer

$4.9 \ m/s^2$ in vertical direction

Vedclass · Answer

(B) For a block on a frictionless inclined plane making an angle $	heta$ with the horizontal,the acceleration along the plane is $a = g \sin 	heta$.The vertical component of this acceleration is $a_v = a \sin 	heta = (g \sin 	heta) \sin 	heta = g \sin^2 	heta$.For block $A$,the angle with the horizontal is $90^\circ - 60^\circ = 30^\circ$. Thus,its vertical acceleration is $a_{vA} = g \sin^2(30^\circ) = g(1/2)^2 = g/4$.For block $B$,the angle with the horizontal is $90^\circ - 30^\circ = 60^\circ$. Thus,its vertical acceleration is $a_{vB} = g \sin^2(60^\circ) = g(\sqrt{3}/2)^2 = 3g/4$.The relative vertical acceleration of $A$ with respect to $B$ is $a_{rel} = a_{vA} - a_{vB} = g/4 - 3g/4 = -g/2 = -4.9 \ m/s^2$.The magnitude is $4.9 \ m/s^2$ in the downward vertical direction.

Two fixed frictionless inclined planes making an angle $30^\circ$ and $60^\circ$ with the vertical are shown in the figure. Two blocks $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$?

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