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

NTA AbhyasNTA Abhyas 2022

Solution:

Solution
$\overset{ \rightarrow }{a}_{A}=gsin 60^{o}cos ⁡ 60\hat{i}$
$=\frac{g \sqrt{3}}{4}\hat{i}-\frac{3 g}{4}\hat{j}$
$\overset{ \rightarrow }{a}_{B}=gsin 30cos⁡30\hat{i}-gsin^{2}⁡\hat{j}$
$=g\frac{3}{4}\hat{i}-\frac{g \hat{j}}{4}$
$\overset{ \rightarrow }{a}_{\frac{A}{B}}= \, \overset{ \rightarrow }{a}_{\frac{A}{B}}= \, \overset{ \rightarrow }{a}_{A}-\overset{ \rightarrow }{a}_{B}$
$=\frac{- 3 g}{4}\hat{j}+\frac{g}{4}\hat{j}$
$=-29\hat{j}$
$\overset{ \rightarrow }{a}_{\frac{A}{B}}=-4.9\hat{j}$