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Nov 19

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Solutions to Irodov Problem 1.81 and 1.82

clip_image001The 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 clip_image002 and let the acceleration of body m with respect to an observer on body Mbeclip_image003 along the direction down the inclined plane.
Relation between clip_image002[1] and clip_image003[1] : The relation between the two acceleration can be seen in the figure clip_image004below. 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,
clip_image005
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, clip_image002[2] as it rides along with mass M and ii) its acceleration clip_image003[2] 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.
clip_image006In the direction perpendicular to the incline, there are twp forces acting on the body, i) the component of gravity clip_image007 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, clip_image008 as shown in the figure. Thus we have,
clip_image009
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 clip_image010 pulling it down the incline. The component of net acceleration along this direction is given by clip_image011 as shown in the figure. Thus, we have,
clip_image012
Forces on mass M : For this problem we need consider only the forces acting in the horizontal direction – this is shown in the figure.
clip_image013There 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 clip_image014. The mass M accelerates at a rate w towards the wall. Thus we have,
clip_image015
Now we have all the information needed to solve for w.
From (2) and (4),
clip_image016
From(3) and (5) we have,
clip_image017

Irodov Problem 1.81

clip_image018
In the system, as body 1 moves backwards, body 2 will slide down along the inclined plane. Let the acceleration of body 1 be clip_image002[3] and let the acceleration of body 2 with respect to an observer on body 1 beclip_image003[3] 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.
clip_image019
i) the component of force of gravity clip_image020 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 – clip_image021 as shown in the figure. Thus, we have,
clip_image022
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, clip_image023. 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 – clip_image024 and ii) its acceleration relative to an observer on body 1 along the inclined plane clip_image003[4]. Thus, we have,
clip_image025
clip_image026Forces 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 clip_image027 that is responsible to accelerate the body at a rate clip_image002[4]. Thus, we have,
clip_image028
From (1) and (3)
clip_image029

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