Since this is the acceleration in the ξ direction and since F = ma by Newton's second law, it follows that the force in the ξ direction is zero. Hence the force is only along the η direction, which is the criterion for a parallel-force problem. Explicitly, the acceleration in the η direction equals d 2 η η d τ τ 2 = d t d τ τ d d t ( d η η d τ τ ) = − − y 2 y ¨ ¨ = − − y 3 m r F ( r ) because the acceleration in the y -direction equals y ¨ ¨ = 1 m F y = 1 m F ( r ) y r