85.5 applications of Newton's law (b) because a m2G4+, or a B If the direction of a, is downward then a2=a (1) T because a3=43+a1 Assume a2 is upward then-a3=-3-a1(2) From equations(1)and (2), we can get a2 an a 85.5 applications of Newtons law I Exercise 5: A small metal tube is sliding down along the string with the acceleration a with respect to the string as shown in tigure The mass of the block is Fi ei, and the mass of the tube is the string is massless, the kinetic friction force between the pulley and the string is ignored. Find the accelerations of the The string is block and the tube with respect not a inertial to the ground, the tension of the reference string and the friction force frame! between the string and the tube20 §5.5 applications of Newton’s law B T1 r T2 r a1 r T3 r 3 3 1 a a a r r r = ′ + (2) 3 3 1 −a = −a′ −a a3 r because Assume is upward then From equations (1) and (2) ,we can get a2 an a3. 2G 2p pG a a a m m r r r because = ′ + 2 2 1 a a a r r r = ′ + (1) 2 2 1 a = a′ − a a2 r If the direction of is downward then or (b) m1 m2 o A The string is not a inertial reference frame! §5.5 applications of Newton’s law Exercise 5:A small metal tube is sliding down along the string with the acceleration a with respect to the string as shown in Figure. The mass of the block is m1, and the mass of the tube is m2, the string is massless, the kinetic friction force between the pulley and the string is ignored. Find the accelerations of the block and the tube with respect to the ground, the tension of the string and the friction force between the string and the tube