Question

A cylinder and a wedge with a vertical face, touching each other, move along two smooth inclined planes forming the same angle a with the horizontal (Fig. 19). The masses of the cylinder and the wedge are m₁ and m₂ respectively.

Determine the force of normal pressure N exerted by the wedge on the cylinder, neglecting the friction between them.

Answer 1

The cylinder is acted upon by the force of gravity m₁g, the normal reaction N₁ of the left

inclined plane, and the normal reaction N₃ of the wedge (force N₃ has the horizontal direction). We shall write the equation of motion of the cylinder in terms of projections on the x₁-axis directed along the left inclined plane (Fig. 153):

where a₁ is the projection of the acceleration of the cylinder on the x₁ - axis.

The wedge is acted upon by the force of gravity m₂g, the normal reaction N₂ of the right inclined plane, and the normal reaction of the cylinder, which, according to Newton's third law, is equal to - N₃. We shall write the equation of motion of the wedge in terms of projections on the x₁-axis directed along the right inclined plane:

During its motion, the wedge is in contact with the cylinder. Therefore, if the displacement of the wedge along the x₁-axis is Δx, the centre of the cylinder (together with the vertical face of the wedge) will be displaced along the horizontal by Δx cos a. The centre of the cylinder will be thereby displaced along the left inclined plane (x₁-axis) by Δx. This means that in the process of motion of the wedge and the cylinder, the relation

is satisfied.

Solving Eqs. (1) - (3) simultaneously, we determine the force of normal pressure N = N₃ exerted by the wedge on the cylinder: