## Friday, November 19, 2010

### Solutions to Irodov Problem 1.81 and 1.82 The body mpushes the body Mbackwards. As the bodyM is pushed backwards,m is forced to slide down the incline on M. Let the tension in the string be T and let the normal reaction between the surfaces be N. Further let the acceleration of body M be and let the acceleration of body m with respect to an observer on body Mbe along the direction down the inclined plane.
Relation between and : The relation between the two acceleration can be seen in the figure below. As seen in the figure,AB and CD are two section of the string before and after the pulley. After the mass M and hence the pulley moves back by x units the length CD shortens to C'D = CD-x. Since, the total length of the stringAB + CD is to remain constant AB must extend to A'B' = AB + x. In other words if the mass M moves x units towards the wall, the massm slides the same x units on the inclined plane. Thus, we have, Forces on mass m : We will resolve the forces acting on m in the parallel and perpendicular direction to the incline. This mass experiences two kinds of i) accelerations, as it rides along with mass M and ii) its acceleration as it slides on the incline relative to M. The net acceleration is the summation of these two accelerations. From (1) however, the magnitude of both these accelerations is the same. In the direction perpendicular to the incline, there are twp forces acting on the body, i) the component of gravity and ii) the normal reaction N from the surface of mass M. The component of the net acceleration of mass m along the perpendicular direction is given by, as shown in the figure. Thus we have, In the direction parallel to the incline, there are two forces acting on the mass m, i) the tension in the string T and ii) the component of force of gravity pulling it down the incline. The component of net acceleration along this direction is given by as shown in the figure. Thus, we have, Forces on mass M : For this problem we need consider only the forces acting in the horizontal direction - this is shown in the figure. There are three forces acting on the mass M that effect its motion in the horizontal direction, i) the normal reaction from mass mand ii) the tensions of magnitude T in the parts of the string after and before the pulley directed along the direction of the string. The component of tension in the part string connecting the mass m and the pulley is given by . The mass M accelerates at a rate w towards the wall. Thus we have, Now we have all the information needed to solve for w.
From (2) and (4), From(3) and (5) we have,  In the system, as body 1 moves backwards, body 2 will slide down along the inclined plane. Let the acceleration of body 1 be and let the acceleration of body 2 with respect to an observer on body 1 be along the direction down the inclined plane. Further let the normal reaction between the two bodies be N.
Forces acting on body 2 : We shall resolve all forces in directions parallel and perpendicular to the incline of body 1. There are two forces acting on this body in the perpendicular direction. i) the component of force of gravity and ii) the normal reaction N between the surfaces. The only acceleration experienced by the body in this direction is the component of its acceleration as it rides on body 1 - as shown in the figure. Thus, we have, In the direction parallel to the inclined plane, there is only one force acting on the body - the component of force of gravity that's pulling it down the inclined plane, . The net acceleration of the body along this direction is sum of two accelerations it is being subject to, i) its acceleration as it rides on body 1 - and ii) its acceleration relative to an observer on body 1 along the inclined plane . Thus, we have,  Forces acting on body 1: We shall consider only the horizontal direction for this part. There is only one force acting on the body - the component of normal reaction that is responsible to accelerate the body at a rate . Thus, we have, From (1) and (3) 