# Page:EB1911 - Volume 17.djvu/1028

APPLIED DYNAMICS]
1009
MECHANICS

Z

o

c

x

8 b

A ' D |>

FIG. 123. FIG. 124.

the solution is based is that the only motion which B can have relatively to an axis through C Hxed to the link CD is one of turning about C. Choose any pole O (fig. 124). From this pole set out Oc to represent the velocit of the point C. The direction of this must be at right angles to the line CD, because this is the only direction possible to the point C. If the link BC moves without turning, Oc will also represent the velocity of the point B; but, if the link is turning, B can only move about the axis C, and its direction of motion is there lore at right angles to the line CB. Hence set out the possible direction of B's motion in the velocity diagram, namely cbl, at right angles to CB. But the point B must also move at right angles to AB in the case under consideration. Hence draw a line through O in the velocity diagram at right angles to AB to cut cbi in b. Then Ob is the velocity of the point b in magnitude and direction, and cb is the tangential velocity of B relatively to C. Moreover, whatever be the actual magnitudes of the velocities, the instantaneous velocity ratio of the points C and B is given by the ratio Oc/Ob. A most important property of the diagram (fis. 123 and 124) is the following z If points X and x are taken dividing the link BC and the tangential velocity cb, sothat cx: xb=CX:XB, then Ox represents the velocity of the point X in magnitude and direction. The line cb has been called the velocity image of the rod, since it may be looked upon as a scale drawing of the rod turned through 90° from the actual rod. Or, put in another way, if the link CB is drawn to scale on the new length cb in the velocity diagram (fig. 124). then a vector drawn from O to any point on the new drawing of the rod will represent the velocity of that point of the actual rod in magnitude and direction. It will be understood that there is a new velocity diagram for every new configuration of the mechanism, and that in each new diagram the image of the rod will be different in scale. Following the method indicated above for a kinematic chain in general, there will be obtained a velocity diagram similar to that of fig. 12 for each configuration of the mechanism, a diagram in which the velocity of the several points in the chain utilized for drawing the diagram will appear to the same scale, all radiating from the pole O. The lines joining the ends of these several velocities are the several tangential velocities, each being the velocity image of a link in the chain. These several images are not to the same scale, so that although the images may be considered to form collectively an image of the chain itself, the several members of this chain-image are to different scales in any one velocity diagram, and thus the chain image is distorted from the actual proportions of the mechanism which it represents.

§ 82.* Acceleration Diagram. Acceleration Image.-Although it is possible to obtain the acceleration of points in a kinematic chain with one link fixed by methods which utilize the instantaneous centres of the chain, the vector method more readily lends itself to this purpose. It should be understood that the Instantaneous centre considered in the preceding paragraphs is available only for estimating relative velocities; it cannot be used in a similar manner l

9

for questions regarding acceleration. That is to say, although the instantaneous centre is a centre of no velocity for the instant, it is not a centre of no acceleration, and in fact the centre of no acceleration is in general a quite different point. The general principle on which the method of drawing an acceleration diagram depends is that if a link CB (fig. 125) have plane motion and the acceleration of any point C be given in magnitude

and direction, the acceleration of any

other point B is the vector sum of

the acceleration of C, the radial B

acceleration of B about C and the

tangential acceleration of B about C.

Let A be any origin, and let Ac

represent the acceleration of the

point C, ct the radial acceleration of

B about C which must be in a direction

parallel to BC, and tb the tangential

which must of course be at right

angles to ct; then the vector sum of

X c

A

b

r

C

FIG. 125.

CHAPTER II. ON APPLIED DYNAMICS

§ 83. Laws of Motion.—The action of a machine in transmitting force and motion simultaneously, or performing work, is governed, in common with the phenomena of moving bodies in general, by two “ laws of motion.”

Division I. Balanced Forces in Machines of Uniform Velocity. § 84. Application of Force to Mechanism.-Forces are applied in units of weight; and the unit most commonly employed in Britain is the pound avoirdupois. The action of a force applied to a body is always in reality distributed over some definite space, either a volume of three dimensions or a surface of two. An example of a force distributed throughout a volume is the weight of the body itself, which acts on every particle, however small. The pressure exerted between two bodies at their surface of contact, or between the two parts of one body on either side of an ideal surface of separation, is an example of a force distributed over a surface. The mode of distribution of a force applied to a solid body requires to be considered when its stiffness and strength are treated of; but, in questions respecting the action of a force upon a rigid body considered as a who e, the resultant of the distributed force, determined according to the principles of statics, and considered as acting in a single line and applied at a single point, may, for the occasion, be substituted for the force as really distributed. Thus, the weight of each separate piece in a machine is treated as acting wholly at its centre of gravity, and each pressure applied to it as acting at a point called the lcegtre of pressure of the surface to which the pressure is really a Ie .

pg 85. Forces applied to Mechanism Classed.-If 6 be the obliquity of a force F applied to a piece of a machin¢=%that is, the angle made by the direction of the force with the direction of motion of its point of application-then by the principles of statics, F may be resolved into two rectangular components, viz.:-

Along the direction of motion, P=-'F cos 0 () Across the direction of motion, Q =F sin 0 49