10. To Determine the Height of a Vertical Object Standing on a Horizontal Plane.-Measure from the foot of the object any convenient horizon B FIG. 12 tal distance AB, Fig. 12; at the point A, take the angle of elevation BAC. Then, as B is known to be a right angle, two angles and the included side of a triangle are obtained. Assuming that the line AB is 300 ft. and the angle BAC 40°, the angle C=180°- (90°+40°)=50°. Then, sin C: AB=sin A: BC, or .766044: 300.642788 : BC, or BC=251.73+ ft. Or, by logarithms: Log 300 2.477121 Log sin 40°= 9.808067 12.285188 Log sin 50°= 9.884254 2.400934 or log of 251.73+ ft. 11. To Find the Distance of a Vertical Object Whose Height is Known. At a point A, Fig. 13, take the angle of elevation to the top of the object. Knowing that the angle B is a right angle, the angles B and A and the side BC are known. Assuming that the side BC=200 ft. and the angle A = 30°, a triangle is formed as follows: Angle A 30°, B=90°, C= 60°, and the side BC=200 ft. Then sin A: BC= sin C: AB, or .5: 200.866025: AB, or AB=346.41 ft. By logarithms: Log 200 2.301030 12.238561 Log sin 30° = 9.698970 2.539591 or log of 346.41 ft. D FIG. 13 12. To Find the Height of a Vertical Object Standing Upon an Inclined Plane. Measure any convenient distance DC, Fig. 14, on a line from the foot of the object, and, at the point D, measure the angles of elevation EDA and D B EDB, and the angle CDB; also at C measure the angle BCD. In the triangle BDC, the side BD may be calculated for the angles at D and C and the side CD are known. Then, in the right-angled triangle BED the sides BE and ED may be calculated, as the side BD and the angle at D are known. Next, in the right-angled triangle AED, the side AE may be calculated, for the are known. Finalmay be found by subfrom the length BE. FIG. 14 side ED and the angle at D ly, the height of the object stracting the length AE B FIG. 15 13. To Find the 4 Height of an Inaccessible Object Above a Horizontal Plane.-First Method.-Measure any convenient horizontal line AB, Fig. 15, directly toward the object, and take the angles of elevation at A and B. In the triangle ABC, the side AC may be calculated as the angles at A and B and the side AB are known. Then, in the right-angled triangle CDA, the side CD, which is the height of the object, may be calculated as the angle at A and the side AC are known. Second Method. If it is not convenient to measure a horizontal base line toward the object, measure any line AB, Fig. 16, and also measure the horizontal angles BAD, ABD, and the angle of elevation DBC. Then, by means of the two triangles ABD and CBD, the height CD can be found. Then, with the line AB and the angles BAD and ABD known, the third angle is readily found, and the side BD can be found. Then, in the triangle BDC, the angle B is known, by measurement, D=90°, and the side BD is known. Then, the side CD, or the vertical height, can be found by preceding methods. MECHANICS ELEMENTS OF MECHANICS GENERAL LAW All machinery, however complicated, is merely a combination of six elementary forms, viz.: the lever, the wheel and axle, the pulley, the inclined plane the wedge, and the screw; and these six can be still further reduced to the lever and the inclined plane. They are termed mechanical powers, but they do not produce force; they are only methods of applying and directing it. The law of all mechanics is: Law. The power multiplied by the distance through which it moves is equal to the weight multiplied by the distance through which it moves. Thus, 20 lb. of power moving through 5 ft. -100 lb. of weight moving through 1 ft. In the following discussion friction is not considered. LEVERS There are three classes of levers: (1) power at one end, weight at the other, and fulcrum between, as shown in Fig. 1; (2) power at one end, fulcrum at W FIG. 1 FIG. 2 FIG. 3 the other, and weight between, as shown in Fig. 2; (3) weight at one end, fulcrum at the other, and power between, as shown in Fig. 3. The handle of a blacksmith's bellows is a lever of the first class; the hand is the power and the bellows the weight, with the pivot between as the fulcrum. A crowbar used for prying down top rock is a lever of the second class; the hand is the power, the rock to be barred down the weight, and the point in the roof against which the bar presses is the fulcrum. The treadle of a grindstone is a lever of the third class; the foot is the power, the hinge at the back of the foot is the fulcrum, and the moving of the machinery is the weight. A lever is in equilibrium when the arms balance each other. The distances through which the power and the weight move depend on the comparative length of the arms. Let P= power; W = weight L-power's distance from fulcrum C; Arranging these terms according to the law of mechanics, PL WI, or P: W=1: L In first- and second-class levers, as ordinarily used, power is gained and time is lost; in the third class, power is lost and time is gained. EXAMPLE. Having a weight of 2,000 lb. to lift with a lever, the short end of which is 2 ft. from the fulcrum and the long end 10 ft., how much power will be required? SOLUTION.-Applying the formula L: 1=W: P, 10: 2=2,000: P: whence P= (2,000X2) ÷ 10=400 lb. The compound lever, Fig. 4, consists of several levers so constructed that the short arm of the first acts on the long arm of the second, and so on to the last. If the distance from A to the fulcrum is four times the distance from the E FIG. 4 fulcrum to B, then a power of 5 lb. at A will lift 20 lb. at B. If the arms of the second lever are of the same comparative length, the 20-lb. power obtained at B will exert a pressure of 80 lb. on E; and if the third lever has the same comparative lengths, this 80 lb. at E will lift 320 lb. at G. Thus, a power of 5 lb. at A will balance a weight of 320 lb. at G. But, in order to raise the weight 1 ft., the power must pass through 320÷5-64 ft. WHEEL AND AXLE The wheel and axle, Fig. 5, is a modification of the lever. The ordinary windlass is a common form. The power is applied to the handle, the bucket is the weight, and the axis of the windlass is the fulcrum. The long arm of the lever is the handle, and the short arm is the radius of the axle. Thus, F is the fulcrum, Fc the long arm, and Fb the short arm. The wheel and axle has the advantage that it is a kind of perpetual lever. It is not necessary to prop up the weight and readjust the lever, but both arms work continuously. By turning the handle or wheel around once, the rope will be wound once around the axle, and the weight will be lifted that distance. Applying the law of mechanics, power X circumference of wheel weight X circumference of axle; or, as the circumferences of circles are proportional to their radii, P: W=r: R; whence PR= Wr. Therefore, = FIG. 5 A train, Fig. 6, consists of a series of wheels and axles that act on one another on the principle of a compound lever. The driver is the wheel to which power is applied. The driven wheel or follower, is the one that receives motion from the driver. The pinion is the small gear-wheel on the axle. W the power. If the diameter of the wheel A is 16 in., and of the pinion B 4 in., a pull of 1 lb. applied at P will exert a force of 4 lb. on the wheel C; if the diameter of C is 6 in., and of D 3 in., a force of 4 lb. on C will exert a force of 8 lb. on E. If E is 16 in. in diameter, and F 4 in., a force of 8 lb. on E will raise a weight of 32 lb. on F. In order, however, to lift this amount, according to the principle already named, the weight will only pass through one thirty-second of the distance of FIG. 7 Thus, power is gained and speed lost. To reverse this, power may be applied to the axle, and, with a correspondingly heavy power, speed gained. Referring to Fig. 7, applying the law of mechanics, P= Wrr'r" PRR'R" W= in which n: n"=r'r" : RR' v: v' =rr'r'' : RR'R" n, n', n"= number of revolutions; v, v velocity or speed of rotation; etc. radii of pinions; R. R', 'R", etc. = radii of wheels. INCLINED PLANE In Fig. 8, the power must descend a distance equal to AC in order to elevate the weight to the height BC; hence, PXlength of inclined plane = WX height of inclined plane, or P: W = height of inclined plane : length of inclined plane; or, P= Wh W= Pl To Find Weight Required to Balance Any Weight on Any Inclined Plane.-Multiply the given weight by the sine of the angle of inclination. Thus, to find the weight required to balance a loaded car weighing 2,000 lb. on a plane pitching 18°, multiply 2,000 by the sine of 18°, or 2,000X.309017=618.034 lb. W FIG. 8 Or, if the length of the plane and the vertical height are given, multiply the load by the quotient of the vertical height divided by the length. Thus, if a plane between two levels is 300 ft. long and rises 92.7 ft., and the load is 92.7 2,000 lb., the balancing weight is found as follows: 2,000X =618 lb. 300 Case 1.-To find the horsepower required to hoist a given load up an inclined plane in a given time, use the formula (Load, in lb.+weight of hoisting rope, in lb.) X vertical height load is raised, in ft. 33,000X time of hoisting, in minutes EXAMPLE. Find the horsepower required to raise, in 3 min., a car weighing 1 T. and containing 1 T. of material up an inclined plane 1,000 ft. long and pitching 30°, if the rope weighs 1,500 16. SOLUTION.-The total load equals car+contents+rope = 2,000+2,000 +1,500=5,500 lb. The vertical height through which the load is hoisted equals 1,000X sin 30° = 1,000X.5=500 ft.; therefore, H. P.= 5,500 X 500 33,000X3 =27.7. Case II.-When the power acts parallel to the base, use the formula WXheight of inclined plane =PXlength of base. These rules are theoretically correct, but in practice an allowance of about 30% must be made for friction and contingencies. SCREW The screw consists of an inclined plane wound around a cylinder. The inclined plane forms the thread, and the cylinder, the body. It works in a nut that is fitted with reverse threads to move on the thread of the screw. The nut may run on the screw, or the screw in the nut. The power may be applied to either, as desired, by means of a wrench or a lever. When the power is applied at the end of a lever, it describes a circle of which the lever is the radius r. The distance through which the power passes is the circumference of the circle; and the height to which the weight is lifted at each revolution of the screw is the distance between two of the threads, called the pitch, p. Therefore, PX circumference of circle = WX pitch; or P: W p: 2πr; whence, Wp P= W= The power of the screw may be increased by lengthening the lever or by diminishing the distance between the threads. EXAMPLE. How great a weight can be raised by a force of 40 lb. applied at the end of a wrench 14 in. long, using a screw with 5 threads per in.? SOLUTION. Substituting in the formula WX=40×28×3.1416; whence W = 17,593 lb. WEDGE The wedge usually consists of two inclined planes placed back to back, as shown in Fig. 9. In theory, the same formula applies to the wedge as to the inclined plane, Case II. P: W=thickness of wedge : length of wedge. Friction, in the other mechanical powers, materially diminishes their efficiency; in this it is essential, since, without it, after each blow the wedge would fly back and the whole effect be lost. Again, in the others the power is applied as a steady force; in this it is a sudden blow, and is equal to the momentum of the hammer. FIG. 9 PULLEY The pulley is simply another form of the lever that turns about a fixed axis or fulcrum. With a single fixed pulley, shown in Fig. 10, there can be no gain of power or speed, as the force P must pull down as much as the weight W, and both move with the same velocity. It is simply a lever of the first class with equal arms, and is used to change the direction of the force. If v = velocity of W; v' = velocity of P; then P=W and v=v'. A form of the single pulley, where it moves with the weight, is shown in Fig. 11. In this, onehalf of the weight is sustained by the hook, and the other half by the power. As the power is only one-half the weight, it must move through twice the space; in other words, by taking twice the time, twice as much can be raised. Here power is gained and time lost; therefore, PW and v' = 2v. FIG. 10 FIG. 11 Combinations of Pulleys.-(1) In Fig. 12, the weight W is sustained by three cords, each of which is stretched by a tension equal to the P; hence, 1 lb. of power will balance 3 lb. of weight. (2) In Fig. 13, a power of 1 lb. will in the same manner sustain a weight of 4 lb., and must descend 4 in. to raise the weight 1 in. (3) Fig. 14 represents the ordinary tackle block used by mechanics, which can be calculated by the following general rule: Rule. In any combination of pulleys where one continuous rope is used, a load on the free end will balance a weight on the movable block as many times as great as the load on the free end as there are parts of the rope supporting the load, not counting the free end. (4) In the cord marked 1, Fig. 15, each part has a tension equal to the power P; and in the cord marked 2, each part has a tension equal to 2 P, and so on with the other cords. The sum of the tensions acting on the weight Wis 16; hence, W=16 P. If n=number of pulleys, W W=2"P 2PR Differential Pulley. In the differential pulley, shown in Fig. 16, W= In all pulley combinations, nearly one-half the effective force is lost by friction. |