FIG. 101. is supposed to be on the point of turning round the axis, and is opposed by the friction, it is required to determine the moment of the frictions with respect to the axis oc. A P in inches), C πρ 2 B It may be assumed that the inequalities of the surfaces will wear away, and that the pressure will be equally distributed; consequently if pis the radius of the pivot (say will be the pressure per square inch, and consequently will be the friction per square inch; 2 πρε hence if we consider a small ring enclosed between two circles, whose radii op and op are respectively rand r+dr, its area will ultimately equal 2wrdr, and the friction on it will equal 24oror. Now the friction at every point of this ring acts in a direction perpendicular to the radius at that point, and hence the sum of the moments of the frictions on this ring with respect to the axis will ultimately equal 2028; the same will be true of any other ring, and therefore we shall obtain the required moment if we divide the area into a great number of rings, and ascertain the limit of the sum of the moments of the frictions on each ring; this can be done as follows: Take DE=p and at right angles to it draw EF=p, perpendicularly to both draw EH = 240, complete the و rectangle EFGH, and complete the pyramid DEFGH; take DP=r and Pp=dr, and through P and p draw planes parallel to the base inclosing the lamina PRS, then it is plain by similar triangles that ps=r and PR=2Qr, con sequently the volume of the lamina is ultimately equal to 2uoradr, i. e. the moment of the friction on the ring is correctly represented by the volume of the lamina, and the same being true of any other lamina, we shall have the moment of the whole correctly represented by the volume of the pyramid,* i.e. the moment equals 4px px 2μα or moment of friction =ραμ. Ex. 392. If the screw rests on a hollow pivot whose internal and external radii are respectively p1 and p, show that the moment of the friction round the axis of the screw is given by the formula and show from this formula that when p1 is very nearly equal to p the friction is very nearly equal to ραμ. Ex. 393. In the screw when the friction on the end as well as the friction on the thread is taken into account then where p is the radius of the end on which the screw rests. * The student who understands the Integral Calculus will perceive that the above construction is equivalent to integrating the expression between the limits of r=0 and r=p. 2αμ redr ρ [Referring to Ex. 384 the equation deduced from the principle of moments will become Pa=rp1+rp2+rP3 + ... + ραμ] Ex. 394. It is required to compress a substance with a force of 10,000 lbs.; the screw with which this is done has a diameter of 3 in., and its thread makes 1 turn to the inch; the arm of the lever is 2 ft. long; determine the force P that would be required-(1) if all frictions were neglected; (2) if the friction between the thread and groove were taken into account; (3) taking also into account the friction on the end of the screw which is 1 in. in radius; the surfaces being iron on iron well greased. Ans. (1) 66.3 lbs. (2) 129.6 lbs. (3) 157.5 lbs. Ex. 395. An iron screw 4 in. in diameter communicates motion to a nut, the force is applied at the extremity of a lever 1 ft. long; the inclination of the thread of the screw is 6°; determine the relation between the force applied and the weight raised by the nut, taking into account the frictions between the thread and groove, and the end of the screw whose diameter is 3 in. the surfaces are cast iron---(1) when well greased, (2) when ungreased. Ans. (1) P = 0.0427 Q. (2) P=0.0583 Q. Ex. 396. If the angle of the screw were 12°, the diameter of the screw and of its end 4 in., and the lever by which it is turned 2 ft. long, the surfaces being of cast iron and ungreased, what weight will a force of 1 cwt. overcome? Ans. 2730 lbs. Ex. 397. Determine the force required in Ex. 394 if the surfaces are of Ans. 488 lbs. ungreased oak. [The fibres may be reckoned to rest endwise between the thread and the groove as well as between the end and the movable piece.] Ex. 398. Given Q the pressure to be produced by the screw, r the radius of the mean thread, R the length of the arm, h the pitch, μ the coefficient of friction between the thread and the groove, if the friction between the thread and the groove is the only one taken into account, show that the force to be applied at the end of the arm is given by the formula * 83. The Endless Screw. It is not very unusual to make a screw work with a toothed wheel; the arrangement of the pieces when this is done will be sufficiently understood by an inspection of the annexed diagram; the screw A B may * This is the formula given in General Morin's Aide-Mémoire, p. 309. be mounted in a frame, and turned by a winch; the teeth of the wheel (c) work with the worm of the screw, on turning which the wheel is caused to revolve; as the screw has no forward motion, it will never go out of action with the wheel, and is, on that account, termed an endless screw. The reader will find in Mr. Willis's Principles of A FIG. 103. B Mechanism * a discussion of the form that must be given to the teeth in order to secure equable working. When the machine is employed, it commonly happens that the screw drives the wheel; sometimes, however, the screw is driven by the wheel, as in the case of the fly of a musical box. In the former case it is easily shown that if p is the force at the end of the arm which turns the screw, and Q the force exerted by the screw on the wheel in a direction parallel to the axis, then the relation between pand Q is the same as that determined in Ex. 384. Ex. 399. If a force P acting on the thread of a screw in a direction parallel to its axis is on the point of driving a force acting along a tangent to its base, show that Q=Ptan (a-p) where a is the inclination of the thread of the screw at the working point, and the limiting angle of resistance between the driving and driven surfaces. Ex. 400.-If the action of an endless screw is reciprocal, i. e. if it will act whether wheel or worm is driver, show that the inclination of the thread of the screw must be greater than & and less than its complement. Ex. 401. An endless screw consists of a cylinder of cast iron the radius of whose base is 3 in.; the thread makes one turn in 4 in.; what is the greatest extent to which the thread can project if the tooth by which it is driven is of cast iron and is ungreased ? Ans. 0.98 in. Ex. 402. In the last Example, if the depth of the thread be 1 in. what is the least pitch with which the machine can work if the surfaces are greased? Ans. 2513 in. * Page 160. AP G FIG. 104. R H B 84. Friction of Guides. One or two instances of the friction of Guides have been given already (Ex. 373-6); the following case will still further illustrate the subject:-EF is a beam constrained to move in a vertical direction by the four guides A, B, C, D ; a projection Guat right angles to EF works with a tooth or cam, k, revolving on a wheel: by the action of the cam the beam is lifted and then allowed to fall by its own weight, thereby serving as a ☑ hammer. In the fundamental case the forces act as in the figure: and we treat the beam as a straight line, the guides as points, and represent AC by a, &H by b, нс by x.* K C F W D Ex. 403. In the above case show that p {a-26μ-(a-2x)μμ'}=aw Ex. 404. In the above case if μ'x>b show that the forces will not act exactly as shown in the figure, and that P(1-μμ') =w * For a fuller discussion of this case, see Traité de Mécanique appliq. aux Machines, par J. V. Poncelet, vol. i. p. 234-238. |