LE and q LG, thenB4. Motion on precribed trajectory
Static moment. Equilibrium of two forces acting on a rotatable body
Why does the action of P at D (Fig. 50) generate motion ? P can be decomposed into the two components radially, DR and DT. DR acts and does not cause motion; it could displace the entire body, but that is not possible due to the body being fixed at A and B, or it deform the body perpendicular to, but the body is rigid. DT acts at D the radius along which D CD and is tangent at D to the circular path travels if the body rotates, whence it D and thus causes motion of rotation of the entire body.
You do have to decompose P in order to understand this. The direction of P is tangent to the circle which you can describe about C with the distance CE as radius. If you shift the point of attack of P from D to E, then P acts at E in the direction of the path along which E travels during the rotation of the body about its axis; hence P must cause motion of E and rotation of the body.
The action of the force DT(=T), which acts tangentially at D, is equal to that of P which acts tangentially at E. This equality is shown by the triangles CED and DTP. They are similar, because their angles are equal, whence: T/P = p/t or Tt = Pp, that is, the products of the forces and their perpendicular distances from the axis are equal. Thus you can replace a turning force P in its action by a smaller force T; you only need to increase the distance of the smaller force from the axis in a certain proportion to that of the larger force. The equation above yields the proportion.
The products Tt and Pp are called the static moments or turning moments of T and P with respect to the axis AB. A force has a static moment only with respect to an axis of rotation. With respect to the same axis, the forces may replace each other, provided their static moments with respect to this axis are equal and equally directed, so that each tends to rotate the body in the same direction. If the force P acts on the rigid body, rotatable about the axis AB, at the point D and the body is to remain at rest, then there must act in addition to P yet another force, which by its action cancels that of P - naturally again only its tangential component (Fig. 51).
What must be the magnitude of this force and its direction? The tangential component of the force, which is to balance the action of P, say Q, must obviously tend to rotate the body in the direction of the arrow II. Let it act at F. During the rotation, F would draw with FC a circle about C. The tangent to this circle at F is the straight line which is perpendicular to CF at F. The tangential component of Q must act along the line in the direction from F to U, in order to act in the direction of the arrow II.
If the body is to remain at rest in spite of the action of the two forces, the resultant of the two tangential components must intersect the axis, that is, it must pass through C. But this resultant must also pass through the point of intersection N (to which the points of attack of the two forces can be shifted by shifting them along their lines of action, as is shown above), because they share with their components the point of application. The resultant of the tangential components thus is along the line XY which passes through N and C. Let their point of attack be at L, the intersection with the direction of P, and shift also that of the force P to L, so that P is represented by LE. But also the force Q, which must have its point of attack at F, must pass through L, because it yields with the force at L a resultant, which lies along LN. Its direction is thus determined by the line through L and F.
As regards the three forces at L, you know the size and direction of the component P (= LE), the direction of Q, because it must lie on LF, and that of the resultant R, because it must lie along LF. This knowledge is sufficient for the construction of the forces parallelogram LEOG and the magnitudes of Q = LG and the resultant R = LO. The force Q is shifted in its direction to L. You only need to shift it along its line of action to F, in order to recognize what force must act at F to balance the force at D.
The resultant R is a pressure on the axis the action of which is removed by the body's rigidity.
In order to find the relationship of P and Q, draw in Fig. 52 ( with only the necessary lines of Fig. 51) the lines CE and CG.
In this way, we arrive at the
triangles CLE and CLG with sides LE = P
and LG = Q.. Compute now their areas. If you draw
from C the line p LE and q LG, then
DLEC = ½LE · p, DLGC = ½LG · q.
If you now draw a LC and b LC, then
DLEC = ½LC · a, DLGC = ½LC· b.
The equality of the areas of the (as halves of parallelograms) congruent triangles LOE and LOG shows that DLEC = DLGC , whence ½LE · p = ½LG · q, that is, if you replace LE and LQ by P and Q, then P·p = Q·q. The heights p and q measure the distance of the forces P and Q from the axis of rotation, the products P·p and Q·q are the static moments of the forces P and Q about the axis of rotation.
It follows that P and Q are in equilibrium, that is, they must rotate in opposite directions and have equal static moments with respect to the axis of rotation. Pp has the same value as Qq, The forces P and Q rotate in opposite directions, the numbers P and Q have therefore opposite signs, p and q are two distances, expressed by two numbers without leading signs. P·p and Q·q are therefore two numbers, which are equal in magnitude and have opposite signs, so that their sum is 0. The sum of quantities with different leading signs is called an algebraic sum. The condition P·p = Q·q or P·p-Q·q= 0 therefore means: P and Q are in equilibrium, if the algebraic sum of their static moments about the axis of rotation equals zero.
If arbitrarily many forces P1P2P3 · · · and Q1Q2Q3 · · · act at a section (Fig. 51) and the body remains at rest and the sense of rotation of the forces P · · · at the distances p · · · from the axis of rotation is opposite to that of the forces Q · · · acting at the distances q · · ·, it can be proved that there is equilibrium, if P1p1 + P2p2 · · · = Q1q1 + Q2q2, that is, if the algebraic sum of the static moments of all forces vanishes.
It follows from P·p = Q·q that P/Q = p/q and, depending on the way of writing this equation in the form n·Q/n·P = p/q or in the form Q/P = n·p/n·q or in the form
nP·(p/n)=(P/n)·np=nQ·(q/n)=(Q/n)·nq,
that:1. as long as the
distances p and q remain unchanged, if the body
is to remain at rest, you can change the forces only in the way
that, if one is doubled,
tripled, etc., then also the other one must be doubled, tripled,
etc.;
2. as long as the magnitudes of the forces P
and Q remain unchanged, if the body is to remain at
rest, the perpendicular distances of the forces from the axis of
rotation can only be changed in the way that if one is doubled,
tripled, etc.,also the others must be doubled, tripled, etc.;
3. every force P and Q can
be replaced by its n-th part at n
times its distance from the axis of rotation, whence the equation
P/Q = p/q can be stated more briefly as follows: Equilibrium occurs if the forces rotate in opposite directions and their magnitude are
inversely related to their distances from the axis of rotation.
If you draw circles about C with radii p and q and tangential lines, which represent the forces P and Q, at any points of the corresponding circles (Fig. 53), the distance from the axis nor the quantity nor the sense of rotation of the forces change, that is, neither their state of equilibrium. In this manner, you can place them in many different relative positions, for example, direct them in such a way that no longer (as up till now) C lies in between them. A disk, which can rotate in its plane about a vertical axis, becomes therefore a convenient arrangement for changing the direction of a force; for example, in Fig. 54, the disk changes the vertically downward directed force of the weight W into a horizontally directed force.
Equally directed, parallel forces acting on a rotatable body
Also parallel forces can have a resultant; this is remarkable, because apparently parallel forces to which correspond parallel lines do not have a common point of attack at which they can be composed into a resultant.
Let A and B be two points of a rigid body (Fig. 55) linked by a rigid line AB, P and Q two parallel forces, AP and BQ lie in the same plane - the plane of the drawing. The state of motion of A and B remains the same even if you let the forces P' and Q' act at A and B, the direction of which is the rigid line AB and which have the same magnitude (P' = Q'), but opposite direction. In this way, P and Q are replaced by R and R'. You can displace R and R', since they are not parallel and lie in the same plane, along their lines of action until their points of attack coincide at C. CD and CE then represent the forces R and R' and each of them can be decomposed into the same components, from which they originated, that is, R into P' and P (= CG), R' into Q' and Q (= CF). The forces PO' and Q' balance each other, because they have the same magnitude and opposite direction. P and Q are in the same direction and form the sum P + Q (= CG + CF). The forces R and R' are thus replaced by a single force of magnitude P + Q, and since R and R' have been replaced by P and Q, these two forces are replaced by P + Q. The design shows that CG is parallel to P and Q. Here too you can shift the resultant P + Q to H.
If this force, which acts parallel to P and Q and passes through H, is really the resultant of the two forces, then the action of P and Q must be balanced, if you let at at H act a force (P + Q), which is parallel to P and Q, but acts in the opposite direction (Fig. 56). Indeed this is so:
The actions of P and Q cause: 1. the rigid line AB to move in its plane of action as a whole. Clearly, this stops the force (P + Q), which has the same magnitude and opposite direction of the force tending to move the point H in the direction, in which act P and Q; hence the point H remains at rest and therefore the line cannot move as a whole.
The action of P and Q is 2. that the line tends to turn in the plane of the drawing about the point H, that is, it tends to turn the line AB in the plane, in which they act, about an axis, which is perpendicular to the plane at H. But H subdivides the line AB into sections p' and q', and Fig. 55 allows to prove (on the one hand, from the similarity of the triangles CFE and CHB, on the other hand, from the equality of P' and Q' and the equality of P' and Q') that p'/q' = Q/P. Furthermore, in the two right-angled triangles of Fig. 56: p'/q'=p/q, whence Q/P = p/q and therefore Qq = Pp.
The lengths p and q measure the vertical distances of the forces P and Q from the axis of rotation. hence the products Pp and Qq are the static moments of P and Q with respect to the axis of rotation, and, since they are equal and the forces P and Q turn in opposite directions, there occurs no rotation. The force (P + Q), acting at H in Fig. 56 balances the action of P at A and Q at B. The force P + Q, which acts at H and is parallel to P and Q in the same direction, is therefore really the resultant of P and Q. It is called the resultant of parallel forces and its point of attack is called the central point of the parallel forces.
Parallel, oppositely directed (anti-parallel) forces. Couple
If P and Q are parallel, but act in opposite directions, you discover by a construction similar to that of Fig. 55 - assuming that they have not the same magnitude - that they too have a central point and a resultant, that their resultant equals the difference of P and Q, is parallel to P and Q and acts in the direction of the larger one. The central point of the forces is again located so that, if you place an axis, fixed in space, vertically to the plane of the drawing, the entire system remains at rest.
However, if the forces P and Q act in opposite directions and have the same magnitude, they do not have a resultant and therefore no central point, and it is impossible to balance them by a single force. Such a combination of equally large, parallel, oppositely directed forces is called a couple, their perpendicular distance p the arm of the couple, the product P·p the magnitude of the couple (Louis Poinsot 1777-1859 1804). The horses in Fig. 58 create a couple.
Parallel displacement of a force
There is no need to go into details regarding couples, but the following will become important in the sequel: The state of motion of a body (Fig. 59) under the action of the force P, applied at A, remains unchanged, if you apply anywhere at a point B equal, parallel forces P1 and P2.Then P and P1 form a couple with the arm p and P1 is obviously the force, the point of application of which is displaced from A to B.
This means, that you can remove a force at a point A (Fig. 59) from A and shift it parallel to itself to B, if at the same time you apply a couple PP2 to the body, as follows from these considerations. The couple PP2 tends to rotate the roller (Fig. 60) in the vertical plane, because its axle is fixed in the bearings.
Parallel displacement of a force is another aspect of the frequently referred to displacement of a force along its line of action (Fig. 48). For certain problems of Mechanics, the right to move forces to other points of application is important. You can undertake with all forces the same operations. The final result will be that all forces, which are applied at arbitrary points of a body, are applied at a single point, that is, are combined into a single resultant, so that then the same number of couples are applied, as initially there were single forces. The couples can be joined into a single resultant couple (the proof will be omitted here). In the end, only a single force is applied, which tends to move the body as a whole, as well as a single couple, which tends to rotate it. In order to keep the body nevertheless at rest, you must balance the force by an equally large force and the couple by an equally large couple. This couple must have the same moment like that couple, but rotate in the opposite direction.
The magnitude (P + Q) of the resultant force depends only on that of P and Q, the position of its point of application, as is shown by p/q = Q/P, only on the ratio of P and Q. Hence the magnitude of the resultant force and the position of H on AB (Fig. 56) does not change, if you turn P and Q into P1 and Q1 or P2 and Q2 (Fig. 61), as long as they remain equally large and parallel. However, the direction of the resultant force changes then, because it must be parallel to the parallel forces, but not its magnitude nor the location of its intersection with the line AB.
In other words: The parallel forces P and Q have one resultant of unique magnitude and only one central point. If yet a third parallel force S is applied, then the forces P + Q and S have a resultant P+Q+ S, its point of attack is determined by the condition, that it must lie on the rigid line and its distance from H to that of S must be in the inverse proportion to that of P + Q and S. However many forces are applied to the line, they can be replaced by a resultant force. In order to balance them, they can be replaced by a resultant force, the magnitude of which equals the sum of the individual forces, is parallel to the forces and has the opposite direction.
The same results, which are valid for two separate points and then for arbitrarily many points, can be extended to arbitrarily many mass points, which lie in any positions with respect to each other; they must only be linked to a rigid body.
Centre of gravity as central point of parallel forces
You can conceive the weight of a rigid body as the resultant force of a multitude of parallel forces. Each of its mass particles m1, m2, m3, ··· is attracted by gravity to the centre of Earth, the dimensions of the body in relation to its distance from the centre of Earth are negligibly small, whence the directions of the forces m1g, m2g, m3g, ··· can be viewed to be parallel. The action of all these forces is replaced by the single force m1g + m2g + m3g _ ··· = g(m1 + m2 + m3 + ···). The expression in brackets is the mass M of the body, that is, the resultant force Mg equals the weight of the mass. The action of gravity on the body manifests itself just as if its mass were concentrated in a single material point with mass M, the centre of gravity, and only this point were attracted by gravity. The centre of gravity is also referred to as centre of mass.
We know the magnitude and direction of the resultant force , called weight. Where lies its point of attack, the centre of gravity (C.G.)? For certain shapes of bodies, the form and mass distribution has a formula and you can compute the coordinates of the C.G. But this point can also be determined experimentally in accordance with the earlier concept, that all the mass of the body is concentrated at it. If you suspend the body by a string (Fig. 62), it will always assume a position, in which the centre of gravity lies on the vertical straight line, formed by the string.
If you suspend it consecutively in two different positions, then both times the centre of gravity must lie on the line which continues the string and passes through the body. But this can only happen if both lines lie in the same plane. and intersect each other. Their point of intersection is identical with the centre of gravity.
Conservation of the centre of gravity.
The concept of the centre of gravity was the result of collection of all mass particles m, representing the body. If you bisect the body by a plane into the two parts A1 and A2, then each of these two parts has a centre of gravity. But the centre of gravity of A1 and A2 is again that of the entire rigid body. The same considerations holds also, if you subdivide a body by planes into arbitrarily many particles: Their common centre of gravity is that of the rigid body out of which they were formed. Now imagine that the body decomposes as a result of an internal cause ( not an externally acting force!) into individual particles, that is, it explodes like a grenade: Nevertheless, the centre of gravity of this multitude of particles remains the same, it moves as long as none of the particles is affected by external forces as it would have moved, if the body had not exploded. In simple words: The internal forces of the system do not affect the state of motion of the centre of gravity. This fact is referred to as conservation of the centre of gravity (Fig. 63)..
It is readily seen from a calculation that two masses cannot displace by interaction their common centre of gravity. Let two masses 2m and m be located at a and b, their common centre of gravity be S and bS = 2aS and the masses interact, that is attract or repulse each other, then the accelerations which they give each other are inversely proportional to the masses. Thus, when 2m has traveled ad, m travels bc = 2ad. S remains the centre of gravity, since bc = 2ad. If you deal with several masses in space, you will recognize how pairs of such masses cannot displace their centre of gravity, that is, that the centre of gravity of the entire system of masses cannot be displaced by the interaction of the masses. Thus, the motion of the C.G. of a system is only determined by the external forces, the internal forces only cause actions which cancel each other.
You obtain the acceleration of the C.G. in a given direction by summing all forces in that direction and dividing by the total mass. Its centre of gravity moves as if all mass were joined at it and all forces acted on it. Neither a mass without external forces nor its centre of gravity is accelerated .
