mechanicsexerciseclassⅰ

mechanicsexerciseclassⅰ内容摘要：

second block of mass m2 hanging vertically. What (a) the magnitude of the acceleration of each block and (b) The direction of the acceleration of the hanging block? (c) what is the tension of the cord? P96 Solution: Choose m2 to be a system and draw it’s freebody diagram m2 m2g T m1 m2 With Newton’s second law applied to the m2 system, we can obtain 2 2 2m g T m a1m g c os N(1) Choose m1 to be a system and draw it’s freebody diagram and we can write Parallel direction Perpendicular direction 1 1 1T m g si n m a  (2) The acceleration ponents a1 and a2, have the same value since the string does not stretch, thus 1 1 2T m g si n m a  m1g N T 2 2 1 2 2m ( g a ) m g si n m a   22T m ( g a )21212m m s inagmm21 0m m sin21 0m m sinNow we can solve equation (1) and (2) simultaneously for T and a2. First solve equation (1) for T Then substitute for T into equation (2): Solving for a2, we find (3) Discussion : If the direction of the acceleration of the hanging block is vertically downward. On the other hand if the direction is vertically upward. Substituting for a2 for Eq. (3), we find that tension in the rope is of magnitude 12121m m ( s in )TgmmThe force shown in the Fig. has magnitude Fp=20N and makes an angle of 30。