85.5 applications of Newton's law Solution Draw the forces diagram Choose triangle block as the reference frame W=mg F=mosin= ma F=N-mgcoso=0 The results: N=mgcos a=gsin 6 Are these results right? 85.5 applications of Newtons law Choose the ground as the reference frame For the triangle block 6 =N sin 6= Mam =Q-Mg-Nc0s6=0(2) M*O, M is not inertial reference frame.15 N = mgcosθ a′ = gsinθ cos 0 sin = − = = = ′ θ θ F N mg F mg ma y x x y m θ a′ r N r W mg r r = Solution: Choose triangle block as the reference frame Draw the forces diagram The results: Are these results right? §5.5 applications of Newton’s law M θ x y Choose the ground as the reference frame: cos 0 (2) sin (1) = − − = = = θ θ F Q Mg N F N Ma y x M aM r For the triangle block θ N N r r ′ = − W r Q r aM ≠ 0 , M is not inertial reference frame. §5.5 applications of Newton’s law