G. D. Terskov

Motion of a Body on an Inclined Plane with Longitudinal Oscillations. (General Case).

INTRODUCTION.

The process of motion of a body on an inclined plane with longitudinal oscillations is very often used in harvesting and grain-cleaning machines for transporting and sorting grain and other threshing products. Thus, for example, to transfer a product from one working member to another in threshers, combines, and grain-cleaning installations, so-called chute boards are used—smooth wooden or iron shields placed at some angle to the horizon. In order to reduce the overall height of the machine, the angle of inclination is made small (in any case less than the angle of friction), and in order that the motion of the body downward along the plane should nevertheless be possible, the chute board is set into oscillatory motion. In some machines, for example in the “Advance-Rumely” threshers, stepped chute boards are used, along which the product being processed is transported from one working member to another, passing from one step to the next, with an overall upward rise.

Likewise, for cleaning grain from foreign impurities and for sorting it, so-called screens, sieves, and riddles are used, consisting of metal sheets with holes punched in them and arranged in most cases at a small angle of inclination to the horizon. In this case, in order to obtain relative motion of the product, it is also necessary to impart oscillatory motion to the plane.

In individual cases the problem of transporting the product upward along an incline is combined with the problem of sorting. In such cases (the “Imperator” grain-cleaning installation of the Eger firm) the sieves are made stepped, and the product, rising from one step to another, is at the same time cleaned of finer impurities or grains that fall through the sieve openings.

In designing all the working members listed above, one has to solve the question of how, in general, the motion of the product can be ensured, and in addition to strive for the character of the motion to ensure the best performance of those operations which are carried out on the given working member or on those following it.

It follows from this that, for a complete solution of the problem, it is necessary to know what character the motion can generally have on an inclined plane with longitudinal oscillations and how this character of motion may change depending on various factors; moreover, it is necessary to know which of the possible types of motion on the inclined plane best promotes the accomplishment of the agrotechnical task set before the given working member.

The present article has as its aim the solution of the first part of the question, namely: to investigate, from the standpoint of theoretical and applied mechanics, the character of the motion of a body on an inclined plane, to determine which

types of motion are at all possible and, finally, to reveal the influence of mechanical factors on the character of the motion. In view of the great variety in the structural design of such devices and the difficulty, connected with this, of solving the posed problem in its general form, it is assumed that the oscillatory motion of the plane is effected by means of a crank-and-connecting-rod mechanism with an infinitely long connecting rod, and moreover in such a way that the plane moves parallel to itself, while the trajectory of any point of the plane is a straight-line segment equal to twice the length of the crank. In addition, the coefficient of friction of the body on the plane is regarded as a constant quantity, and air resistance is not taken into account.

1. Possibility of the onset of relative motion.

Let us imagine that a plane inclined to the horizontal at an angle \(\beta\) is set into oscillatory motion by means of a crank-and-connecting-rod or eccentric mechanism in such a way that the plane moves parallel to itself, while the direction of the oscillations forms an angle \(\varepsilon\) with the plane.

If the connecting rod is regarded as infinitely long, and the angles of rotation of the crank \(a\) are reckoned from the left (Fig. 1-a) position of it coinciding with the direction of the oscillations, and the coordinate axes are arranged as shown in the drawing, then the motion of any point of the plane will be determined by the equations

Fig. 1-a              Fig. 1-b.

\[ x = r \cos \varepsilon \cos a;\qquad y = r \sin \varepsilon \cos a. \tag{1} \]

If the crank rotates uniformly with angular velocity \(\omega\), then

\[ a=\omega t \]

The velocity of point \(M\) is determined by the expressions

\[ u_x=-\omega r\cos\varepsilon\sin a;\qquad u_y=-\omega r\sin\varepsilon\sin a, \tag{2} \]

and the acceleration by the components

\[ j_x=-\omega^2 r\cos\varepsilon\cos a;\qquad j_y=-\omega^2 r\sin\varepsilon\cos a \tag{3} \]

If some body lies motionless on the plane at point \(M\), then its motion will also be determined by expressions (1)—(3), and therefore inertia forces will act on the body,

\[ J_x=m\omega^2 r\cos\varepsilon\cos a;\qquad J_y=m\omega^2 r\sin\varepsilon\cos a \tag{4} \]

If the crank is in the first and fourth quadrants, then the inertial force \(J_x\) tends to move the body downward along the plane (Fig. 1-b), while the friction force will oppose this. In order for the body to be able to move downward, the following condition must hold:

\[ m\omega^2 r \cos \sigma \cos \alpha + mg \sin \beta > f\left(mg \cos \beta - m\omega^2 r \sin \sigma \cos \alpha\right). \]

Hence, after the appropriate transformations, we obtain the condition that makes it possible for the body to move downward in the form

\[ \omega^2 r \cos \alpha > g \frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)} . \tag{5} \]

The left-hand side of this inequality will have a maximum at \(\alpha=0\); therefore, it is evident that the body can in general move downward only in the case when

\[ \omega^2 r > g \frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)} , \tag{5a} \]

and, if this condition is satisfied, then the downward displacement of the body can occur only for positions of the crank between

\[ \alpha_3 = 2\pi - \arccos\left[\frac{g}{\omega^2 r}\cdot \frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)}\right] \tag{6} \]

and

\[ \alpha_0 = \arccos\left[\frac{g}{\omega^2 r}\cdot \frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)}\right]. \tag{7} \]

If the crank is in the second and third quadrants, then the inertial force \(J_x\) tends to move the body upward, while the friction force will hinder this. In order for the body to be able to move upward, the condition must hold:

\[ - m\omega^2 r \cos \sigma \cos \alpha - mg \sin \beta > f\left(mg \cos \beta - m\omega^2 r \sin \sigma \cos \alpha\right). \]

Hence, after the appropriate transformations, we obtain the condition that makes it possible for the body to move upward in the form

\[ -\omega^2 r \cos \alpha > g \frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)} . \tag{8} \]

The left-hand side of this inequality will have a maximum at \(\alpha=\pi\); therefore, it is evident that the body can in general move upward only in the case when

\[ \omega^2 r > g \frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)} . \tag{8a} \]

and, if this condition is satisfied, then upward displacement of the body is possible only for positions of the crank between

\[ \alpha_1 = \pi - \arccos\left[\frac{g}{\omega^2 r}\cdot \frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)}\right] \tag{9} \]

and

\[ \alpha_2 = \pi + \arccos\left[\frac{g}{\omega^2 r}\cdot \frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)}\right]. \tag{10} \]

From Fig. 1-b it is seen that the inertia force \(I_y\) tends to tear the body away from the plane. Obviously, separation of the body will be possible only when

\[ m\omega^2 r \sin\sigma \cos\alpha > mg\cos\beta . \]

Hence the condition ensuring the possibility of separation will have the form

\[ \omega^2 r \cos\alpha > g\,\frac{\cos\beta}{\sin\sigma} \tag{11} \]

The left-hand side of the inequality will have a maximum at \(\alpha=0\); therefore, in general, separation of the body from the plane is possible only in the case when

\[ \omega^2 r > g\,\frac{\cos\beta}{\sin\sigma}; \tag{11a} \]

and if this condition is satisfied, then for positions of the crank between

\[ \alpha' = 2\pi-\arccos\left[g\,\frac{\cos\beta}{\sin\sigma}\right] \tag{12} \]

and

\[ \alpha''=\arccos\left[g\,\frac{\cos\beta}{\sin\sigma}\right]. \tag{13} \]

2. Sliding of the body only downward.

Suppose that the body, in one way or another, shared the motion of the plane until the crank occupied the position determined by the angle \(\alpha_3\); at this moment, under the influence of the inertia force \(J_x\), the body will begin to slide downward along the plane with initial absolute velocity

\[ w_3=u_3=-\omega r\cos\sigma\sin\alpha_3 \]

The process of motion of the body along the plane will be associated with the appearance of the friction force

\[ F=f(mg\cos\beta-m\omega^2 r\sin\sigma\cos\alpha) \]

In this connection, when the body slides downward there will be an acceleration directed upward along the plane.

\[ j_n=\frac{F-mg\sin\beta}{m} = g\,\frac{\sin(\varphi-\beta)}{\cos\varphi} -\omega^2 r f\sin\sigma\cos\alpha \tag{14} \]

Therefore, the absolute velocity of the body during downward sliding will change according to the law

\[ w_n=w_3-\int_{t_3}^{t} j_n\,dt = -\omega r\cos\sigma\sin\alpha_3 - g\,\frac{\sin(\varphi-\beta)}{\cos\varphi}(t-t_3) + \]

\[ +\omega r f\sin\sigma(\sin\alpha-\sin\alpha_3); \tag{15} \]

and the relative velocity is determined by the expression

\[ v_n=w_n-u_x = -\frac{\cos(\varphi-\sigma)}{\cos\varphi} \left[ \omega r\sin\alpha_3 + g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)}(t-t_3) -\omega r\sin\alpha \right] \tag{16} \]

The first term in the brackets of expression (16) is a constant quantity, and moreover negative; the second term increases uniformly, and the last changes according to a sine law; therefore, after some time the relative velocity of the body will become equal to zero. The position of the crank \(\alpha_4\) corresponding to this moment is found from the condition \(v_n=0\), or from the expression

\[ -\omega r\sin\alpha_3 - g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\delta)}(t_4-t_3) = -\omega r\sin\alpha_4 \tag{17} \]

This expression is most simply solved graphically in the following way: we construct the curve \(u=-\omega r\sin\alpha\) and find point 3 of this curve corresponding to the position of the crank \(\alpha_3\); from point 3 we draw the inclined line 3–4 at the angle \(-\gamma_k\) to the axis of abscissas, where

\[ \operatorname{tg}\gamma_k = g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\delta)}\,\frac{m_t}{m_v} \tag{18} \]

where \(m_t\) is the scale of time along the axis of abscissas, and \(m_v\) is the scale of velocity along the axis of ordinates; point 4, the intersection of this straight line with the curve \(u=-\omega r\sin\alpha\), determines the required position of the crank \(\alpha_4=\omega t_4\).

In Fig. 2, for \(\omega^2 r=5.13\); with \(f=0.4\); \(\delta=12^\circ\); \(\beta=12^\circ\); \(r=50\ \text{mm}\); the curve \(u=-\omega r\sin\alpha\) is constructed at the scale \(m_v=0.01\ \frac{\text{m/sec}}{\text{mm}}\); \(m_t=0.00667\ \frac{\text{sec}}{\text{mm}}\). \(\operatorname{tg}\gamma_k=1.15\); \(\alpha_3=288^\circ 30'\); \(\alpha_0=71^\circ 30'\); the moment \(\alpha_4=2\pi+162^\circ\) determines the position of the crank at which the relative velocity becomes equal to zero, and the body will be lying motionless on the plane. Subsequently, since \(\alpha_4\) lies outside the zone in which the start of sliding downward is possible, the body will share the motion of the plane until the crank again takes the position determined by the angle \(\alpha_3\), and then the process will repeat again, etc.

Fig. 2.

The motion of the body analyzed here pertains to the case when the body, already located on the plane, has for some reason shared the motion of the plane up to \(\alpha_3\). By analogous reasoning it can also be shown that a body which has fallen onto the plane, for any position of the crank, will eventually begin to move in exactly the same way.

The relative path traversed by the body during each period of sliding (or, what is the same, during each revolution of the crank) will obviously be equal to

\[ S_n = \int_{t_3}^{t_4} v_n\,dt = -\frac{\cos(\varphi-\delta)}{\cos\varphi} \left[ \omega r\sin\alpha_3(t_4-t_3) + g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\delta)} \frac{(t_4-t_3)^2}{2} + r(\cos\alpha_4-\cos\alpha_3) \right] \tag{19} \]

In our example, \(\alpha_3=288^\circ 30'\); \(\alpha_4=522^\circ\); \(t_3=0.481\ \text{sec}\); \(t_4=0.870\ \text{sec}\). By formula (19) we find \(S_n=133\ \text{mm}\).

One can find \(S_n\) graphically; for this it is necessary to planimeter the area \(F_n\) between the straight line 3—4 and the curve \(u=-\omega r\sin\alpha\) (shaded in Fig. 2). In that case

\[ S_n=F_n\cdot m_v\cdot m_t\cdot \frac{\cos(\varphi-\delta)}{\cos\varphi} \tag{19a} \]

In our example

\[ F_n=1920\ \mathrm{mm}^2;\qquad \frac{\cos(22^\circ-12^\circ)}{\cos 22^\circ}=1.06 \]

\[ S_n=1920\cdot 0.00667\cdot 0.01\cdot 1.06=0.135\ \mathrm{m}=135\ \mathrm{mm} \]

i.e., almost the same.

Let us now see how matters will stand in the same case when

\[ \omega^2 r>g\,\frac{\sin(\varphi+\beta)}{\cos(\varphi+\delta)} \]

In this case, as was proved above, there will be the possibility of the onset of downward sliding from \(\alpha_3\) to \(\alpha_0\) (Fig. 3) and the possibility of the onset of upward sliding from \(\alpha_1\) to \(\alpha_2\). Suppose that at \(\alpha_3\) the body has begun to slide downward. The cessation of downward sliding, generally speaking, may occur at three different positions of the crank, \(\alpha_4\). If \(\alpha_4<\alpha_1\) (\(\alpha'_4\) in Fig. 3), then the body will share the motion of the plane from \(\alpha'_4\) to \(\alpha_1\), and then will begin to slide upward. If \(\alpha_1<\alpha_4<\alpha_2\) (\(\alpha''_4\) in Fig. 3), then at the moment of stopping at \(\alpha''_4\) the body will begin to slide upward. If \(\alpha_4>\alpha_2\) (\(\alpha'''_4\) in Fig. 3), then the body will share the motion of the plane up to \(\alpha_3\), and then will begin to slide downward. In this case, despite the possibility of the onset of upward sliding, this possibility will remain unused, since for crank positions from \(\alpha_1\) to \(\alpha_2\) the body is in the process of sliding downward.

Fig. 3.

Thus it follows that the possibility of upward sliding can be ensured only in the case when \(\alpha_4<\alpha_2\), or, in other words, when the straight line 3—4 passes below point 2. For such a case expression (17) will take the form

\[ -\omega r\sin\alpha_3 - g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\delta)}\,(t_2+T-t_3) < -\omega r\sin(2\pi+\alpha_2) \]

Hence, after the corresponding transformations, we find the condition ensuring the possibility of upward sliding in the form

\[ \omega^2 r< g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\delta)} \cdot \frac{2\pi+\alpha_2-\alpha_3}{\sin\alpha_2-\sin\alpha_3} \tag{20} \]

or, what is the same,

\[ \cos\alpha_3> \frac{\sin\alpha_2-\sin\alpha_3}{2\pi+\alpha_2-\alpha_3} \tag{20a} \]

Expression (20) is solved only by trial and requires the expenditure of a large amount of time; therefore one may bear in mind that if

\(\omega^2 r\) has such a value that it turns expression (8a) into an equality, and the straight line 3–4 at this value intersects the abscissa axis at \(\alpha=\pi\), then condition (8a) will ensure the possibility of upward sliding. In this case we shall have \(\alpha_2=\pi\), and instead of condition (20) we obtain

\[ \omega^2 r < -g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)}\, \frac{3\pi-\alpha_3}{\sin\alpha_3}; \]

hence, taking (8a) into account, after the corresponding transformations we obtain the condition ensuring the possibility of upward sliding in the form

\[ \pi+\arccos\left[ \frac{\sin(\varphi-\beta)\cos(\varphi+\sigma)} {\cos(\varphi-\sigma)\sin(\varphi+\beta)} \right] > \sqrt{ \left[ \frac{\sin(\varphi-\beta)\cos(\varphi+\sigma)} {\cos(\varphi-\sigma)\sin(\varphi+\beta)} \right]^2-1 }. \]

Solving this expression by trial, we finally find

\[ \frac{\sin(\varphi-\beta)\cos(\varphi+\sigma)} {\cos(\varphi-\sigma)\sin(\varphi+\beta)} >0.22 \tag{20b} \]

Thus, if condition (20b) is fulfilled, then the possibility of upward sliding is determined by formula (8a); otherwise upward sliding can occur only in the case when expression (20) or (20a) holds.

For illustration of the foregoing, Fig. 4 shows a graph of the variation of \(\omega^2 r\), ensuring the possibility of upward sliding for the particular case \(f=0.4\) and \(\sigma=0^\circ\).

Fig. 4.

3. Sliding of the body both upward and downward.

If conditions (8a) and (20) are fulfilled, then at the end of downward sliding, at the crank position \(\alpha_4\), the body will begin to slide upward with initial absolute velocity \(w_4=u_4=-\omega r\cos\sigma\sin\alpha_4\). The process of the body’s sliding upward along the plane will be associated with the presence of the friction force \(F\); therefore there will be an acceleration directed downward along the plane,

\[ j_6=\frac{F+mg\sin\beta}{m} = g\,\frac{\sin(\varphi+\beta)}{\cos\varphi} -\omega^2 r f\sin\sigma\cos\alpha. \tag{21} \]

The absolute velocity of the body during upward sliding will vary according to the law

\[ w_6=w_4+\int_{t_4}^{t} j_6\,dt = -\omega r\cos\sigma\sin\alpha_4 + g\,\frac{\sin(\varphi+\beta)}{\cos\varphi}(t-t_4) -\omega r f\sin\sigma(\sin\alpha-\sin\alpha_4), \tag{22} \]

and the relative velocity is determined by the equation

\[ v_6=w_6-u_x = -\frac{\cos(\varphi+\sigma)}{\cos\varphi} \left[ \omega r\sin\alpha_4 - g\,\frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)}(t-t_4) - \omega r\sin\alpha \right]. \tag{23} \]

The instant \(\alpha_5\), when the relative velocity becomes equal to zero and the body comes to rest on the plane, is determined from the condition \(v_{\mathrm{в}}=0\), or

\[ -\omega r \sin \alpha_4 + g\,\frac{\sin(\varphi+\beta)}{\cos(\varphi+\delta)}(t_5-t_4) = -\omega r \sin \alpha_5 . \tag{24} \]

Fig. 5.

Graphically this equation is solved if (Fig. 5), from point 4 on the curve \(u=-\omega r \sin\alpha\) (corresponding to the position of the crank \(\alpha_4\)), one draws the straight line \(4\text{--}5\), inclined to the axis of abscissas at the angle \(\gamma_{\mathrm{в}}\), where

\[ \tg \gamma_{\mathrm{в}} = g\,\frac{\sin(\varphi+\beta)}{\cos(\varphi+\delta)}\,\frac{m_t}{m_v}. \tag{25} \]

The point 5 of intersection of this straight line with the curve \(u=-\omega r\sin\alpha\) will determine the sought position of the crank itself, \(\alpha_5=\omega t_5\).

The relative path traversed by the body during the time of sliding upward will be equal to

\[ S_{\mathrm{в}} = \int_{t_4}^{t_5} v'_{\mathrm{в}}\,dt = -\frac{\cos(\varphi+\delta)}{\cos\varphi} \left[ \omega r \sin\alpha_4 (t_5-t_4) - g\,\frac{\sin(\varphi+\beta)}{\cos(\varphi+\delta)} \frac{(t_5-t_4)^2}{2} + r(\cos\alpha_5-\cos\alpha_4) \right]. \tag{26} \]

or else

\[ S_{\mathrm{в}} = F_{\mathrm{в}}\,m_v\,m_t\, \frac{\cos(\varphi+\delta)}{\cos\varphi}, \tag{26a} \]

where \(F_{\mathrm{в}}\) is the area enclosed between the straight line \(4\text{--}5\) and the curve \(u=-\omega r\sin\alpha\).

The total path traversed by the body downward along the plane will, obviously, be equal to

\[ S = S_{\mathrm{н}} + S_{\mathrm{в}} . \]

In Fig. 5, for:

\[ \omega^2 r = 12.5;\quad f=0.4;\quad \beta=12^\circ;\quad \delta=12^\circ;\quad m_v=0.02;\quad m_t=0.00\,445 \]

thus, it has been found

\[ F_{\mathrm{н}}=2370\ \text{mm}^2;\quad F_{\mathrm{в}}=60\ \text{mm}^2;\quad S_{\mathrm{н}}=225\ \text{mm}.\quad S_{\mathrm{в}}=5\ \text{mm}.\quad S=225-5=220\ \text{mm}. \]

From Fig. 5 it is seen that \(\alpha_2<\alpha_5<\alpha_3\); therefore, after \(\alpha_5\) the body will share the motion of the plane up to \(\alpha_3\), and then the process will repeat again, etc. Thus, in the case under consideration there is sliding both upward and downward, with stops.

If we denote by \(m_0\) the scale of the angles \(\alpha\) along the axis of abscissas, and by \(T\) the period of oscillation, then the time scale along the axis of abscissas is determined by the expression

\[ m_t = m_0\,\frac{T}{360^\circ} = \frac{m_0}{360^\circ}\cdot\frac{2\pi}{\omega} \ldots\ldots\ldots \tag{27} \]

Hence it is clear that, as the number of revolutions of the crank increases (and consequently the angular velocity \(\omega\) increases), \(m_1\) will decrease; and this in turn means that, as \(\omega\) increases (and consequently \(\omega^2 r\)), the angles \(\gamma_e\) and \(\gamma_n\) will decrease. It is also obvious that an increase in \(\omega^2 r\) will bring points 3 and 5 in Fig. 5 closer together, and at some sufficiently large value of \(\omega^2 r\) these points will coincide; and the body, at the moment corresponding to the end of sliding upward, will again begin to slide downward. Thus, by increasing \(\omega^2 r\), one can attain a situation in which the body will slide along the plane both upward and downward without stops.

In fact, if, for example, we take \(\omega^2 r = 21.8\), then for \(r = 50\) mm, \(f = 0.4\), \(\beta = 12^\circ\), \(\delta = 12^\circ\), we shall have: \(\omega = 20.9\); \(\omega r = 1.04\); \(n = 200\) rev/min; \(\alpha_0 = 85^\circ 30'\); \(\alpha_1 = 107^\circ 40'\); \(\alpha_2 = 252^\circ 20'\); \(\alpha_3 = 274^\circ 35'\) (Fig. 6). Suppose, for example, that the body has in one way or another shared the motion of the plane up to \(\alpha_1\); in that case at \(\alpha_1\) the body will begin to slide upward, and the stopping moment will be determined by point 2; then the body will slide downward up to the moment determined by point 3; then there will again be upward sliding up to the moment determined by point 4; subsequently the upward and downward sliding will be repeated continuously, the points at which upward sliding changes to downward sliding, and conversely, approaching certain points \(A\) and \(B\), upon reaching which the motion will become fully established: upward sliding and downward sliding will follow one another without stops.

Fig. 6.

Let us carry out an analytical investigation of this motion. First of all, from expressions (17) and (24) it follows that

\[ -\omega r \sin \alpha_B - g \frac{\sin(\varphi-\beta)}{\cos(\varphi-\delta)} (t_A-t_B) = -\omega r \sin \alpha_A \qquad\qquad (17a) \]

\[ -\omega r \sin \alpha_A + g \frac{\sin(\varphi+\beta)}{\cos(\varphi+\delta)} (t_B-t_A) = -\omega r \sin \alpha_B \qquad\qquad (24a) \]

Let us denote, for the subsequent derivations:

\[ -\omega r \sin \alpha_B = U_B;\qquad -\omega r \sin \alpha_A = U_A;\qquad t_B-t_A=t_v;\qquad t_A-t_B=t_n, \]

where \(t_n\) is the time during which sliding downward occurs, and \(t_v\) is the time during which sliding upward occurs. In that case, from expressions (17a) and (24a), we find

\[ U_B-U_A = g \frac{\sin(\varphi-\beta)}{\cos(\varphi-\delta)}\,t_n = g \frac{\sin(\varphi+\beta)}{\cos(\varphi+\delta)}\,t_v; \qquad\qquad (28) \]

and hence we find the relation

\[ \frac{t_n}{t_v} = \frac{\sin(\varphi+\beta)}{\sin(\varphi-\beta)} \cdot \frac{\cos(\varphi-\delta)}{\cos(\varphi+\delta)} \qquad\qquad (29) \]

Moreover, it is obvious that

\[ t_n+t_v=T. \qquad\qquad (30) \]

Solving (29) and (30) simultaneously, we find

\[ t_v=T\,\frac{\sin(\varphi-\beta)\cos(\varphi+\sigma)} {\sin(\varphi-\beta)\cos(\varphi+\sigma)+\sin(\varphi+\beta)\cos(\varphi-\sigma)} \tag{31} \]

\[ t_n=T\,\frac{\sin(\varphi+\beta)\cos(\varphi-\sigma)} {\sin(\varphi-\beta)\cos(\varphi+\sigma)+\sin(\varphi+\beta)\cos(\varphi-\sigma)}; \tag{32} \]

substituting (31) and (32) into (28), we shall have

\[ U_B-U_A=Tg\,\frac{\sin(\varphi-\beta)\sin(\varphi+\beta)} {\sin(\varphi-\beta)\cos(\varphi+\sigma)+\sin(\varphi+\beta)\cos(\varphi-\sigma)} . \tag{33} \]

Introducing the conventional notations

\[ A=\sin(\varphi-\beta)\cos(\varphi+\sigma);\quad B=\sin(\varphi+\beta)\cos(\varphi-\sigma); \]

\[ C=\sin(\varphi-\beta)\sin(\varphi+\beta), \]

we shall have

\[ t_v=T\frac{A}{A+B};\qquad t_n=T\frac{B}{A+B};\qquad U_B-U_A=Tg\frac{C}{A+B}; \]

therefore,

\[ \alpha_v=\omega t_v=2\pi\frac{A}{A+B}\, . . . . . . . . . . . . . \tag{34} \]

\[ \alpha_n=\omega t_n=2\pi\frac{B}{A+B}\, . . . . . . . . . . . . . \tag{35} \]

In addition,

\[ U_B-U_A=-\omega r\sin(\alpha_A+\alpha_v)+\omega r\sin\alpha_A =-2\omega r\cos\left(\alpha_A+\frac{\alpha_v}{2}\right)\sin\frac{\alpha_v}{2}, \]

whence

\[ \alpha_A=\pi+\arccos\left[\frac{U_B-U_A}{2\omega r\sin\frac{\alpha_v}{2}}\right]-\frac{\alpha_v}{2} \tag{36} \]

and consequently

\[ \alpha_B=\pi+\arccos\left[\frac{U_B-U_A}{2\omega r\sin\frac{\alpha_v}{2}}\right]+\frac{\alpha_v}{2}. \tag{37} \]

Transforming formulas (19) and (26) as applied to the case under consideration, we find that the total relative path of the body for each revolution of the crank will be equal to

\[ \begin{aligned} S=S_n+S_v ={}&-\frac{\cos(\varphi-\sigma)}{\cos\varphi} \left[ -U_Bt_n +g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)}\frac{t_n^2}{2} +r(\cos\alpha_A-\cos\alpha_B) \right] \\ &-\frac{\cos(\varphi+\sigma)}{\cos\varphi} \left[ -U_At_v -g\,\frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)}\frac{t_v^2}{2} +r(\cos\alpha_B-\cos\alpha_A) \right] \\ ={}&\cos\sigma \left\{ \frac{U_A+U_B}{2}\left[T+f\tan\sigma\,(t_n-t_v)\right] +2rf\tan\sigma(\cos\alpha_B-\cos\alpha_A) \right\}. \end{aligned} \]

Substituting here the corresponding values, we find

\[ \begin{aligned} S=2\pi r\cos\sigma \Bigg\{& \cos \frac{\pi A}{A+B} \left(1+f\tg\sigma\,\frac{B-A}{A+B}\right) \\ &+\frac{2}{\pi} f\tg\sigma \sin \frac{\pi A}{A+B} \Bigg\} \sqrt{ 1-\left[ \frac{\pi g}{\omega^{2}r}\cdot \frac{C}{(A+B)\sin \dfrac{\pi A}{A+B}} \right]^2 } \end{aligned} \tag{38} \]

If \(f,\ \beta,\ \sigma\) and \(\omega^2 r\) are known, then from formula (38) it is not difficult to calculate the path traversed by the body; for example, for \(f=0.4;\ \beta=12^\circ;\ \sigma=12^\circ\),

\[ \frac{S}{2\pi r}=0.978\{0.8(1+0.05)+0.032\} \sqrt{1-\frac{52}{(\omega^2 r)^2}}; \]

if \(\omega^2=21.8\), then

\[ \frac{S}{2\pi r}=0.852\sqrt{1-0.109}=0.8, \]

and for \(r=50\ \text{mm}\)

\[ S=0.8\cdot 2\pi\cdot 50=251\ \text{mm}. \]

Thus, in the established motion of the body along the plane without stops, the path traversed by the body will be the greater, the larger \(r\) and \(\omega^2 r\) are; moreover, as \(\omega^2 r\) increases, the ratio \(\dfrac{S}{2\pi r}\) rapidly approaches the limit

\[ S_{\max}=2\pi r\cos\sigma \left\{ \cos \frac{\pi A}{A+B} \left(1+f\tg\sigma\,\frac{B-A}{A+B}\right) +\frac{2}{\pi} f\tg\sigma \sin \frac{\pi A}{A+B} \right\}. \tag{39} \]

Let us now see what conditions ensure the possibility of the body sliding up and down the plane without stops. From all that has been set forth above it is clear that such a possibility will occur only in the case when \(a_B>a_3\), or, taking into account expressions (6) and (37), when

\[ 2\pi-\arccos\left[ \frac{g}{\omega^2 r}\, \frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)} \right] \le \pi+\arccos\left[ \frac{U_B-U_A}{2\omega r\sin \dfrac{\alpha_B}{2}} \right] +\frac{\alpha_B}{2}. \]

Hence, after the corresponding transformations, one can obtain the condition ensuring the possibility of sliding both upward and downward without stops in the form

\[ \omega^2 r \ge -\frac{g}{\sin \dfrac{\pi A}{A+B}} \times \]

\[ \times \sqrt{ \left[ \frac{\pi C}{(A+B)\sin \dfrac{\pi A}{A+B}} \right]^2 + \frac{\sin^2(\varphi-\beta)}{\cos^2(\varphi-\sigma)} + \frac{2\pi C}{(A+B)} \cdot \frac{\sin(\varphi-\beta)} {\cos(\varphi-\sigma)\tg \dfrac{\pi A}{A+B}} } \tag{40} \]

For example, for \(f=0.4,\ \beta=12^\circ,\ \sigma=12^\circ\), \(\omega^2 r>14.3\ \text{m/sec}^2\) ensures the possibility of the body sliding along the plane both upward and downward without stops.

4. Motion of the body in the presence of separation.

a. In the absence of sliding.

In order to clarify the essence of the process of motion of the body in the presence of separation, let us first analyze a simpler case, when at the instant of separation the body is at rest on the plane (and in general sliding on the given plane is impossible).

It was proved above that the instant of separation is characterized by the position of the crank \(a'\), for which from formula (11) we have

\[ \cos \alpha'=\frac{g}{\omega^2 r}\frac{\cos\beta}{\sin\sigma}; \qquad \sin \alpha'=\sqrt{1-\left(\frac{g}{\omega^2 r}\frac{\cos\beta}{\sin\sigma}\right)^2} \tag{41} \]

Up to the instant of separation the body shared the motion of the plane, and therefore at the instant of separation the absolute velocity of the body will be equal to the velocity of the plane:

\[ w_x'=-\omega r\cos\sigma \sqrt{1-\left(\frac{g}{\omega^2 r}\frac{\cos\beta}{\sin\sigma}\right)^2}; \qquad w_y'=-\omega r\sin\sigma \sqrt{1-\left(\frac{g}{\omega^2 r}\frac{\cos\beta}{\sin\sigma}\right)^2} \tag{42} \]

The position of the body at the instant of separation is evidently determined by the coordinates

\[ x'=\frac{g}{\omega^2}\frac{\cos\beta}{\operatorname{tg}\sigma}; \qquad y'=\frac{g}{\omega^2}\cos\beta \tag{43} \]

From the instant of separation only one force, gravity, will act on the body; therefore the velocity of the body in flight will be determined by the expressions

\[ w_x=w_x'+g\sin\beta\,(t-t'); \qquad w_y=w_y'-g\cos\beta\,(t-t'), \tag{44} \]

and the trajectory described by the body—by the equations of motion

\[ x=x'+w_x'(t-t')+g\sin\beta\,\frac{(t-t')^2}{2}; \qquad y=y'+w_y'(t-t')-g\cos\beta\,\frac{(t-t')^2}{2} \tag{45} \]

Since the velocity of the body along the \(Y\)-axis gradually decreases, the instant of greatest rise \(t_1'\) will be determined from the condition

\[ w_y=w_y'-g\cos\beta\,(t_1'-t')=0, \]

whence the time spent in rising is determined by the expression

\[ t_{\text{rise}}=t_1'-t'=\frac{w_y'}{g\cos\beta} = -\frac{\omega r\sin\sigma}{g\cos\beta} \sqrt{1-\left(\frac{g}{\omega^2 r}\frac{\cos\beta}{\sin\sigma}\right)^2} \tag{46} \]

Substituting (46) into (45), one can find the coordinates of the body at the instant of greatest rise.

After the instant of greatest rise, the body, descending, will begin to approach the plane and at some instant \(t_2'\) will fall onto it; at this instant the ordinates of the body and of the plane will be equal, and therefore the position of the crank \(a_2'=\omega t_2'\), corresponding to the instant of fall of the body, is found from the condition

\[ y'+w_y'(t_2'-t')-g\cos\beta\,\frac{(t_2'-t')^2}{2} = r\sin\sigma\cos\omega t_2' \tag{47} \]

The simplest way to solve this equation is to construct, point by point, the curves corresponding to the right- and left-hand sides, and to find the point of intersection of these curves. The point thus obtained will determine the required position of the crank (Fig. 7).

Fig. 7.

Knowing \(\alpha_2'\) and \(t_2'\), it is not difficult to determine the range of the body's flight by the formula

\[ S=x_{\text{body}}-x_{\text{plane}} =x' + w_x'(t_2'-t') + g\sin\beta\,\frac{(t_2'-t')^2}{2} -r\cos\sigma\cos\alpha_2'. \tag{48} \]

The trajectory of the body's flight can be constructed (Fig. 8) point by point with the aid of equation (45); thus, all the factors characterizing the body's flight in the air can be determined.

In the figure: trajectory of the body's flight in the absence of sliding; \(\omega^2 r=40\), \(\beta=12^\circ\), \(\sigma=20^\circ\).
Other labels: body trajectory; horizon; direction of vibrations.

Fig. 8.

Of particular interest is the case when, at the moment of falling, the body will again be thrown up. Such a situation will occur only in the case when

\[ \alpha_2'=\alpha' + 2\pi \quad \text{or} \quad t_2'-t'=\frac{2\pi}{\omega}=T. \tag{49} \]

Substituting (49) and (43) into (47), we find the condition ensuring continuous tossing, in the form

\[ \omega^2 r = g\,\frac{\cos\beta}{\sin\sigma}\sqrt{\pi^2+1} = 32.3\,\frac{\cos\beta}{\sin\sigma}. \tag{50} \]

Thus, if

\[ \omega^2 r > g\,\frac{\cos\beta}{\sin\sigma}, \]

then, in the absence of sliding, the body will separate from the plane at \(\alpha'\); the flight of the body will continue until the crank takes the position determined by the angle \(\alpha_2'\). If \(\alpha_2'\) lies outside the zone \(\alpha'—\alpha''\), then after falling the body will share the motion of the plane up to \(\alpha'\), and then the process will repeat again; if, however, \(\alpha_2'=2\pi+\alpha'\), then the body will immediately be thrown up again and will fly along the same trajectory as the first time; if \(\alpha_2'\) lies within the zone \(\alpha'—\alpha''\), then the body will be thrown upward, and the trajectory will be determined by the position of the crank at the moment of separation.

6. In the presence of sliding.

Let us suppose that the body has in one way or another shared the motion of the plane up to \(\alpha_3\) (the beginning of the possibility of sliding downward); at \(\alpha_3\) the body will begin to slid—

downward, and the motion along its \(Y\)-axis will be the same as the motion of the plane, while along the \(X\)-axis the absolute velocity is determined by expression (15).

If there were no separation, then, by the method set forth above, one could find that the body would cease sliding at \(\alpha_4=2\pi+230^\circ\) (Fig. 9) for \(\omega^2 r=40\), \(f=0.4\), \(\beta=12^\circ\), \(\delta=20^\circ\), and the relative path traversed by the body would be \(s=302\) mm.

However, owing to the possibility of separation, the body will slide only up to \(\alpha'_2=314^\circ 90'\), and then will separate and will execute free flight. In the course of sliding, the absolute velocity \(w_x\), in accordance with formula (15), is represented in Fig. 9 by the segment of the curve \(AB\), while during the flight, in accordance with formula (44), \(w_x\) is represented by the straight line \(BC\), where \(w'_x\) is found from (15) for \(t=t'\). The fall of the body onto the plane, just as in the absence of sliding, will occur at the moment determined by the position of the crank \(\alpha'_2=2\pi+139^\circ\) [determined from eq. (47)]. At the moment of falling, in the direction of the \(X\)-axis, the body will obviously have velocity \(w'_2x\) [found from (44) for \(t=t'_2\)], and therefore subsequently sliding of the body downward along the plane will again take place, the velocity of the body being determined by the expression

\[ w_x=w'_{2x}-g\frac{\sin(\varphi-\beta)}{\cos\varphi}(t-t'_2)+ \]
\[ +\omega r f\sin\delta(\sin\alpha-\sin\alpha'_2) \tag{50} \]

Fig. 9.

Constructing the curve \(CD\) in Fig. 9 from this formula, we find the point \(4'\) of its intersection with the curve \(\dot u_x=-\omega r\cos\delta\sin\alpha\). The position of the crank \(\alpha'_4=\omega t'_4=2\pi+302^\circ\), corresponding to this point, determines the moment when the absolute velocity of the body becomes equal to the velocity of the plane and sliding downward ceases. If \(\alpha'_4>\alpha_3\), then subsequently the body will share the motion of the plane up to \(\alpha_3\), and then the process will repeat again. If \(\alpha'_4<\alpha_3\), then the body will begin to slide upward; the position of the crank corresponding to the moment when upward sliding ceases is found by the method analyzed above. In this case two possibilities can occur: either the upward sliding ceases before the crank reaches position \(\alpha_3\)—in this case the body will share the motion of the plane up to \(\alpha_3\), and then the process will repeat again; or the upward sliding ceases after \(\alpha_3\) (as in Fig. 9)—in this case, at the moment when upward sliding ceases, downward sliding will begin again, with an initial velocity somewhat smaller than the first time; therefore the line \(ABCD\) will pass somewhat lower than the first time. After several revolutions the motion will become steady.

Thus, depending on the value of \(\alpha'_4\) (in other words, depending on \(\omega^2 r\)), various motions may occur:

a) sliding downward—flight—sliding downward—stop;

b) sliding downward—flight—sliding downward—sliding upward—stop;

c) sliding downward—flight—sliding downward—sliding upward.

The path traversed by the body during the period of sliding downward is determined by the area enclosed between the line \(ABCD\) and the curve \(u_x=-\omega r\cos\sigma\sin\alpha\). In the case considered in Fig. 9, this area is equal to \(5620\ \mathrm{mm}\), and therefore the path will be

\[ S=5620\cdot 0.02\cdot 0.00248=0.280\ \mathrm{m}=280\ \mathrm{mm}. \]

From the foregoing it follows that, when \(\omega^2 r\) is increased, the angle \(\alpha'\) decreases, while the angle \(\alpha'_2\) increases; therefore, as \(\omega^2 r\) increases, the points \(D\) and \(A\) in Fig. 9 will approach each other and, consequently, for a sufficiently large value of \(\omega^2 r\), sliding downward and flight will follow one another, while stops and sliding upward will not occur at all. Such a situation will evidently occur only in the case when the curve \(CD\) passes through the point \(A\), or above it; and for this, in turn, it is necessary that the conditions

\[ w'_{2x} - g\frac{\sin(\varphi-\beta)}{\cos\varphi}(t_3-t'_2) + \omega r f\sin\sigma(\sin\alpha_3-\sin\alpha'_2) \ge -\omega r\cos\sigma\sin\alpha_3 \]

be satisfied.

Substituting here \(w'_{2x}\), and then \(w'_x\), after the corresponding transformations we obtain the condition necessary for the motion of the body without stops and without sliding upward in the form

\[ \omega^2 r < g \frac{(\alpha'_2-\alpha')\sin\varphi\cos\beta-2\pi\sin(\varphi-\beta)} {\sin\varphi\sin\sigma(\sin\alpha'_2-\sin\alpha')} \tag{51} \]

Thus, for example, when \(\omega^2 r=40\), the right-hand side is \(5.05\), and therefore stops and sliding upward will occur. Conversely, when \(\omega^2 r=55\), the right-hand side will be equal to \(89\). Consequently, the motion will proceed according to the scheme “sliding downward—flight.”

Let us consider this case in more detail: Fig. 10 shows the graph of the variation of the velocity of the body in the direction parallel to the \(X\)-axis for

\[ \omega^2 r=55;\quad f=0.4;\quad \beta=12^\circ;\quad \sigma=20^\circ;\quad r=50\ \mathrm{mm}. \]

As was to be expected, in this case there is neither sliding upward nor stopping of the body on the plane; first the body slides downward (curve \(AB\)), then it undergoes flight (straight line \(BC\)), then it again slides downward (curve \(CB_1\)) until \(\alpha_1\), again undergoes flight (straight line \(B_1C_1\)), again slides downward (curve \(C_1B_2\)), and so on. At the same time, with each revolution the absolute velocity of the body increases by a certain amount \(\Delta w_x\). The quantity \(\Delta w_x\) can be determined on the basis of the following considerations: during flight \(w_x\) increases by the amount \(\Delta w'_x=g\sin\beta(t'_2-t')\) (see formula 44), while during sliding it decreases by the amount

\[ \Delta w_{x\,\mathrm{sl}} = g\frac{\sin(\varphi-\beta)}{\cos\varphi}\,[T+t'-t'_2] - \omega r f\sin\sigma(\sin\alpha'-\sin\alpha'_2); \]

therefore

\[ \Delta w_x = \Delta w'_x-\Delta w_{x\,\mathrm{sl}} = g\left[ f\cos\beta(t'_2-t') - T\frac{\sin(\varphi-\beta)}{\cos\varphi} \right] + \omega r f\sin\sigma(\sin\alpha'-\sin\alpha'_2). \tag{52} \]

Fig. 10.

Thus the motion of the body along the plane is accelerated; the magnitude of the acceleration (conditionally) will evidently be equal to

\[ j_{\text{cond}}=\frac{\Delta v_x}{T} = g\left[ f\cos\beta\,\frac{\alpha_2'-\alpha'}{2\pi} - \frac{\sin(\varphi-\beta)}{\cos\varphi} \right] + \frac{\omega^2 r}{2\pi} f\sin\sigma\,(\sin\alpha'-\sin\alpha_2') \tag{53} \]

With an increase of \(\omega^2 r\), a case may occur in which

\[ \alpha_2'=2\pi+\alpha'; \]

in this case the sliding of the body downward disappears altogether (the section of the curve \(CB_1, C_1B_2\), etc., will be absent from the graph), and continuous tossing of the body will take place. From formula (53), for such a case we find, as was to be expected,

\[ j_{\text{cond}}=g\sin\beta \tag{53a} \]

5. Motion of a body along a stepped plane.

It was proved above that a body sliding both upward and downward along an inclined plane nevertheless, as a result, creeps downward; meanwhile, in designing grain-cleaning machines one has to deal with cases in which it is necessary to transport the processed product along the machine while simultaneously raising it upward.

The accomplishment of such a task is possible only in the case when the possibility of sliding downward is eliminated in one way or another. For this purpose, for example, the inclined plane may be made stepped (Fig. 11); thereby the body is protected from the possibility of sliding downward and, consequently, will share the motion of the plane until the crank assumes the position determined by the angle \(\alpha_1\). At \(\alpha_1\) the body will begin to slide upward; the end of the upward sliding will be determined by the point \(\alpha\) of intersection of the straight line \(1-\alpha\), drawn from point 1 at the angle \(\gamma_8\) to the axis of abscissae (Fig. 6), with the curve \(u=-\omega r\sin\alpha\). The path traversed by the body is determined by formula (26), which for the given case takes the form

\[ S_8 = - \frac{\cos(\varphi+\sigma)}{\cos\varphi} \left[ \omega r\sin\alpha_1\,(t_a-t_1) - g\frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)} \frac{(t_a-t_1)^2}{2} + r(\cos\alpha_a-\cos\alpha_1) \right] \tag{54} \]

After \(\alpha_a\) the body will share the motion of the plane until the crank again assumes the position \(\alpha_1\), and then the process will be repeated again. The length of the step must obviously be equal to, or somewhat less than, \(S_8\), so that during each revolution of the crank the body moves along the plane by one step. It is self-evident that the use of a stepped plane for transporting or sorting a product with upward feed will be possible only in the case when condition (8a) is fulfilled.

In some cases of design practice there may arise a need to use a stepped plane for feeding the product downward. This may be either because, in order to increase strength, the inclined plane is made stepped (the screen of large threshing machines)

or, because it is undesirable for the transported product to be rubbed away, while the required speed of displacement is so great that on a smooth plane the body could slide both upward and downward (Fig. 12).

Fig. 11.

Fig. 12.

In such a case all the arguments set forth above are applicable, with the only difference that sliding will begin at \(\alpha_3\) and end at \(\alpha_4\) (see Fig. 5), and the distance traveled will be determined by formula (19).

Figure 13 shows a graph indicating the variation of \(S_{\mathrm{в}}\) and \(S_{\mathrm{н}}\) for a stepped plane, in the particular case—for \(\omega^2 r=35;\ r=50;\ f=0.4\).

Fig. 13.

From this graph it follows that

a) \(S_{\mathrm{в}}\) will be the greater, the smaller \(\beta\) is and the closer \(\sigma\) is to \(-\psi\);

b) \(S_{\mathrm{н}}\) will be the greater, the larger \(\beta\) is and the closer \(\sigma\) is to \(+\psi\).

6. Conclusions.

The principal conclusion from all that has been set forth above is that the character of the motion of a body on a plane with longitudinal oscillations may be very varied and depends on: the centrifugal acceleration of the crank pin \(\omega^2 r\), the coefficient of friction \(f\) of the body on the plane, the angle of inclination \(\beta\) of the plane to the horizon, and the angle \(\sigma\) between the direction of the oscillations and the plane.

The influence of \(\omega^2 r\) consists in the following:

1. The body will not move along the plane at all if

\[ \omega^2 r < g \frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)} . \]

1. The body will move along the plane only downward, if

\[ g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)} < \omega^2 r < g\,\frac{\sin(\varphi+\beta)}{\cos(\varphi+\sigma)} \quad \text{when} \quad \frac{\sin(\varphi-\beta)\cos(\varphi+\sigma)} {\cos(\varphi-\sigma)\sin(\varphi+\beta)} \ge 0.22 \]

or

\[ g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)} < \omega^2 r < g\,\frac{\sin(\varphi-\beta)}{\cos(\varphi-\sigma)} \cdot \frac{2\pi+\alpha_2-\alpha_3}{\sin\alpha_2-\sin\alpha_3} \]

when

\[ \frac{\sin(\varphi-\beta)\cos(\varphi+\sigma)} {\cos(\varphi-\sigma)\sin(\varphi+\beta)} <0.22 . \]

1. The body will slide along the plane both upward and downward with stops, if \(\omega^2 r\) is greater than in the preceding case, and

\[ \omega^2 r < \frac{g}{\sin\dfrac{\pi A}{A+B}} \times \]

\[ \times \sqrt{ \left[ \frac{\pi C}{(A+B)\sin\dfrac{\pi A}{A+B}} \right]^2 + \frac{\sin^2(\varphi-\beta)}{\cos^2(\varphi-\sigma)} + \frac{2\pi C}{A+B}\cdot \frac{\sin(\varphi-\beta)} {\cos(\varphi-\sigma)\tg\dfrac{\pi A}{A+B}} }\, . \]

1. The body will slide along the plane both upward and downward without stops, if

\[ \frac{g}{\sin\dfrac{\pi A}{A+B}} \times \]

\[ \times \sqrt{ \left[ \frac{\pi C}{(A+B)\sin\dfrac{\pi A}{A+B}} \right]^2 + \frac{\sin^2(\varphi-\beta)}{\cos^2(\varphi-\sigma)} + \frac{2\pi C}{A+B}\cdot \frac{\sin(\varphi-\beta)} {\cos(\varphi-\sigma)\tg\dfrac{\pi A}{A+B}} } < \omega^2 r < g\,\frac{\cos\beta}{\sin\sigma}. \]

1. In the process of sliding downward the body will separate from the plane and make a flight; in this case stops and upward sliding may occur, if

\[ g\,\frac{\cos\beta}{\sin\sigma} < \omega^2 r > g\, \frac{(\alpha'_2-\alpha')\sin\varphi\cos\beta-2\pi\sin(\varphi-\beta)} {\sin\varphi\sin\sigma\,(\sin\alpha'_2-\sin\alpha')}. \]

1. The body will slide downward and make a flight without stops on the plane, if

\[ g\, \frac{(\alpha'_2-\alpha')\sin\varphi\cos\beta-2\pi\sin(\varphi-\beta)} {\sin\varphi\sin\sigma\,(\sin\alpha'_2-\sin\alpha')} \ge \omega^2 r < g\,\frac{\cos\beta}{\sin\sigma}\sqrt{\pi^2+1}. \]

1. The body will continuously bounce\(^{1)}\) along the plane without stops and downward slidings, if only

\[ \omega^2 r = \frac{g\cos\beta}{\sin\sigma}\sqrt{\pi^2+1}. \]

\(^{1)}\) The general motion of the body will be directed downward and will be uniform for cases 1–5, and accelerated for cases 6–7.

The influence of the angle of inclination \(\beta\) of the plane to the horizontal is visible from Fig. 14 and consists in the following:

1. Increasing the angle \(\beta\) decreases the value of \(\omega^{2}r\) required for the possibility of sliding downward, and increases the value of \(\omega^{2}r\) required for the possibility of sliding upward, especially when \(\beta\) is close to \(\varphi\).

Fig. 14.

1. Increasing the angle \(\beta\) decreases the value of \(\omega^{2}r\) required for the possibility of separation, and this decrease is not especially large.

2. Increasing the angle \(\beta\) decreases the value of \(\omega^{2}r\) required for accelerated motion; for \(\beta > \varphi\), accelerated motion will occur in the absence of oscillations, while for \(\beta = 0\) accelerated motion is altogether impossible.

The influence of the angle \(\sigma\) between the direction of the oscillations and the plane is also visible from Fig. 14 and consists in the following:

1. A change in \(\sigma\) has only a slight effect on the value of \(\omega^{2}r\) that ensures the possibility of sliding downward; the minimum value of \(\omega^{2}r\) is obtained at \(\sigma = +\varphi\).

2. Increasing \(\sigma\) increases the value of \(\omega^{2}r\) required for the possibility of sliding upward, and decreases the value of \(\omega^{2}r\) required for the possibility of separation; the minimum value ensuring the possibility of sliding upward is obtained at \(\sigma = -\varphi\); the minimum value of \(\omega^{2}r\) ensuring the possibility of separation occurs at \(\sigma = \dfrac{\pi}{2}\), while at \(\sigma = 0\) separation is altogether impossible. Increasing the angle of friction \(\varphi\) causes an increase in the values of \(\omega^{2}r\) ensuring the possibility of sliding downward and upward. The angle \(\varphi\) does not affect the possibility of separation.