**m**pushes the body **M**backwards. As the body**M** 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 **m** with respect to an observer on body **M**be**Relation between**** **** ****and**** **** ****:**** **The relation between the two acceleration can be 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 string**AB + 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 mass**m**** **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, **M** and ii) its acceleration **M.**** **The net acceleration is the summation of these two accelerations. From **(1)**** **however, the magnitude of both these accelerations is the same. **N** from the surface of mass **M.**** **The component of the net acceleration of mass **m** along the perpendicular direction is given by,

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 **Forces on mass M :**** **For this problem we need consider only the forces acting in the horizontal direction – this is shown in the figure. **M** that effect its motion in the horizontal direction, i) the normal reaction from mass **m**and 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 **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 **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 **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 –

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, **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

From **(1)** and **(3) **