Question 141: Laws of Motion

Question

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 $$B$$ ?

Accepted Answer

Along inclined plane the equation of motion of the body
$$mg\,\sin \,	heta  = ma$$  $$\,\,\,\,\,\,\,\,$$ $$	herefore$$ $$a = g\,\sin \,	heta $$
where $$a$$ is along the inclined plane.
$$	herefore$$ vertical component of acceleration is $$\left( {g\sin 	heta } ight)\sin 	heta $$ = $$g\,{\sin ^2}	heta $$ (Along vertical)
For block A,
$${a_{A\left( {along\,vertical} ight)}} = g{\sin ^2}60^\circ $$
For block B,
$${a_{B\left( {along\,vertical} ight)}} = g{\sin ^2}30^\circ $$
$$	herefore$$ relative vertical acceleration of $$A$$ with respect to $$B$$ is
$$g{\sin ^2}60^\circ $$ - $$g{\sin ^2}30^\circ $$
=$$g\left( {{{\sin }^2}60 - {{\sin }^2}\left. {30} ight]} ight.$$
$$= 4.9$$ $$\,\,m/{s^2}$$ in vertical direction

Answer

$$4.9m{s^{ - 2}}$$ in horizontal direction

Answer

$$9.8m{s^{ - 2}}$$ in vertical direction

Answer

Zero

Answer

$$4.9m{s^{ - 2}}$$ in vertical direction

Laws of Motion

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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