## Wednesday, August 15, 2007

### Irodov Problem 1.81

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)