Question

A cylinder and a wedge with a vertical face, touching each other, move along two smooth inclined planes forming the same angle $\alpha$ with the horizontal (figure). The masses of the cylinder and the wedge are $m_1$ and $m_2$ 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 $m_{1}g$, the normal reaction $N_1$ of the left inclined plane, and the normal reaction $N_3$ of the wedge ( force $N_3$ has the horizontal direction). We shall write the equation of motion of the cylinder in terms of projections on the $x_1$- axis directed along the left inclined plane:
$m_1{a}_1=m_{1}g\sin \alpha -N_3\cos \alpha$, (1)
where $a_1$ is the projection of the acceleration of the cylinder on the $x_1$-axis.
The wedge is acted upon by the force of gravity $m_{2}g$, the normal reaction $N_2$ of the right inclined plane, and the normal reaction of the cylinder which, according to Newton's third law, us equal to $-N_3$. We shall write the equation of motion of the wedge in terms of projections on the $x_2-axis$ directed along the right inclined plane:
$m_2{a}_2=-m_{2}g\sin \alpha +N_3\cos \alpha$. (2)
During its motion, the wedge is in contact with the cylinder. Therefore, if the displacement of the wedge along the $x_3$-axis is $\Delta x$, the centre of the cylinder ( together with the vertical face of the wedge ) will be displaced along the horizontal by $\Delta x\cos \alpha$. The centre of the cylinder will be thereby displaced along the left inclined plane ( $x_1$- axis) by $\Delta x$. This means that in the process of motion of the wedge and the cylinder, the relation
$a_1=a_2$ (3)
is satisfied.
Solving Eqs. (1)-(3) simultaneously, we determine the force normal pressure $N=N_3$ exerted by the wedge on the cylinder:
$N_3=\frac{2m_1{m}_2}{m_1+m_2}\tan \alpha$.