Irodov Problem 1.74

Let the force of friction between the bead and the rope be f, the tension in the string be T and let the accelerations of the masses M and m be and respectively directed towards the floor (downwards as shown in the picture).

Forces acting on M: There are two forces acting M, i) the tension in the string T and the force of gravity pulling it downwards Mg. The mass accelerates at a rate downwards, hence its dynamics can be written as,



Force acting on the bead : There are two forces acting on the bead, i) the force of friction f opposing the downward motion of the bead as it slides along the rope and ii) the force of gravity pulling it downwards mg. Thus, it dynamics are given by,




Relating the force of friction to the tension in the string : The missing piece is the relation between the force of friction and the tension in the string. For this let us consider the small section of the string that holds the bead. as shown in the figure. There are two forces acting on this section of the string, i) the tension T from the rest of the string pulling the section up and ii) the force of friction f from the bead pulling it down (as it opposes motion of the string relative to the bead and the string is moving up relative to the bead). The mass of the section of string is 0 and its acceleration is upwards. Thus, we have,




From (3) and (1) we can solve as follows,








Now the relative acceleration with which the bead moves up with respect to the rod is thus we have,





From (5) and (6) we have,

Irodov Problem 1.73

Let be the tension in string connected to mass and let T be the tension in the string connected to masses and . Further let be the acceleration of the mass and let w be the acceleration of the mass relative to the moving pulley B (holding masses and ) directed towards pulley B. The net acceleration of the masses and with respect to a stationary frame are and respectively.

Forces on mass : Theres only one force on this mass the tension in the string over pulley A. Thus, the dynamics of the mass are given as,





Forces on mass : There are two forces acting on this mass, i) the tension in the string T pulling it up and the force of gravity pulling it downwards. The net acceleration of this mass is towards the floor. Thus,



Forces on mass : There are two forces acting on this mass, i) the tension in the string T pulling it up and the force of gravity pulling it downwards. The net acceleration of this mass is is towards the floor. Thus,



Forces on the Pulley B: There are three forces acting on pulley B, i) tension pulling it towards pulley A and ii) tensions T on each of the two parts of the string. The mass of the pulley is 0 and its acceleration is acting downwards. Thus, we have,



Now we have all the information to solve for all the unknowns as follows,



From (1), (4) we have,




From (2) and (5) we have,








Similarly, from (3) and (5) we have,








From (6) and (7) we can now find w as,





Now we can use (6) and (8) to find T as,




From (9) and (5) we have,




Finally we can determine the accelerations of both the masses and respectively as,