60. In Ex. 76 show that at the middle point the shearing force is 0, and the bending moment wa.

61. In the last Question, if w were concentrated at the middle point of the rod, show (Ex. 72) that the shearing force is w and the bending moment wa.

62. In Ex. 76, suppose w to equal 2000 lbs., a to be 10 ft. long; show that the bending moments at the middle point and at points distant 5 ft. from each end are respectively 5000 and 3750.

63. Four planks equal in all respects are placed flat on one another, and rest with their ends on two trestles, it is found that a weight w bearing on their middle point will be sufficient to break them; if the four planks are put together so as to form a hollow box open at both ends, it will be found that they will now support w; account for this.

Art. 67.)

(See

64. Let the rod in Ex. 76 be loaded not only by its own weight (w), but also by a weight (w,) acting at a point c (fig. 75); if b and x denote BC and BD, show that the bending moment at D equals

65. In the last Question let w, w1, 2a, b equal 1000 lbs., 400 lbs., 20 ft., 15 ft.; show that the bending moment at a point between в and c, and at a distance x from в, is 600x—25x2, and that it has its greatest value at a point distant 12 ft. from B.

N.B.-The Student will observe that the references to Examples are to those that occur as parts of the several chapters, and are marked thusEx. 75, Ex. 101, &c. The references to Questions are to those added to the ends of the chapters under the heading of Questions.

CHAPTER IV.

WORK.

68. Definition and measure of work.-When the point of application of a force moves wholly, or partly, in the direction of the force, work is said to be done by the force. Suppose a force of P units to act on a point, and suppose the point to move in the direction of the force, and to describe a distance of p units; the work done by the force while its point of application describes this distance is measured by the product p p. A force of one unit does a unit of work when its point of application moves through a unit of distance in the direction of the force; consequently the product Pp is the number of units of work done by P. When force is measured in pounds and space in feet it is sometimes convenient to call the unit of work a "foot-pound.'1 Thus if a carpenter urges forward a plane through 3 ft. with a force of 12 lbs., he does 36 footpounds of work; or, if a weight of 7 lbs. descends through 10 ft., gravity does 70 foot-pounds of work.

If in virtue of the action of other forces the point moves in a direction opposite to that of the force, the force resists the motion of the point, and work is expended in overcoming that resistance, or the work is said to be done against the force. If the force is one of p lbs. and the point moves through p ft. in a direction

When there is occasion to draw attention to the fact that distance is assumed to be measured in feet and force in pounds, it is best to use the term foot-pound; under other circumstances it is, perhaps, best to use the more general term 'unit of work.'

WORK OF RAISING A SYSTEM OF WEIGHTS. 91

opposite to that of the action of the force, P p units of work will be expended in overcoming the resistance of the force. Thus, if a weight of 10 lbs. is raised to a height of 5 ft., 50 foot-pounds of work must have been expended in overcoming the resistance of gravity, i.e. 50 foot-pounds of work have been done against gravity.

It is not necessary that the point should move in a straight line. The above statements are true without any modification, when the point moves in a curved line, provided the direction of the force is always tangential to the curve. Thus, if a man presses with a force of 30 lbs. at right angles to the arm of a capstan, and if, in one turn of the capstan, the point at which he pushes describes a circumference of 60 ft., he does 30 × 60, or 1800 footpounds of work.

Ex. 82.-If the area of the piston of a steam-engine is 5000 sq. in., and the mean pressure of the steam is 12 lbs. per sq. in., the whole force will be 60,000 lbs. ; and the work done by the steam in one stroke of 8 ft. will be 480,000 foot-pounds.

Ex. 83.-If two weights of 150 and 200 lbs. are raised through heights of 80 and 120 ft. respectively, the whole work done must be 150 x 80+ 200 × 120, i. e. 36,000 foot-pounds. It is plain that this is the same number of units as must be done if 350 lbs. were raised through 1029 ft. Now the student will easily prove that when the weights are raised as in the Question, their centre of gravity will be raised through 1029 ft. So we see that the number of units of work done in raising the weights separately is the same as the number that would be done if the whole weight were raised through the same height as that through which the centre of gravity is raised.

69. Work of raising a system of weights.-The result exemplified in the last Example is perfectly general and may be stated thus:- -When two or more heavy points are raised through different heights, the work done equals the sum of the weights of the points, multiplied by the height through which their centre of gravity has been raised. Two remarks may be made on this principle. First, if the weights are the parts of a continuous body,

the work expended in lifting it is the product of the whole weight multiplied by the vertical height through which the centre of gravity is raised. This is equally true whether it is raised as a whole or in parts; e.g. the work done in lifting the earth when a well is sunk through a stratum of uniform density is the weight of the earth multiplied by half the depth of the well. Secondly, if a body moves in such a way that its centre of gravity neither rises nor falls, work is neither expended on gravity nor done by gravity. If work is expended on the motion at all, it is expended on friction or on some force other than gravity; e.g. when a train moves on horizontal rails, work is expended only on friction, resistance of the air, &c., none on the weight of the train.

Ex. 84.-A well-shaft is 300 ft. deep and 5 ft. in diameter; it is full of water; how many units of work must be expended in getting this water out of the well (i. e. irrespectively of any other water flowing in)?

The weight of the water is rx (2·5)2 × 300 × 1000-16 or 36,815 lbs. The centre of gravity is 150 ft. below the ground; consequently the work expended must be 368,155 × 150, or 55,223,250 foot-pounds.

FIG. 81.

P

70. Cases in which the point of application does not move along the line of action of the force.-Such a case arises when a weight falls obliquely. Thus, let A N be a horizontal line, and suppose a weight to fall from P to A along the curved line; draw the vertical line PN. The work done by gravity is the weight multiplied by P N. Since gravity acts on the body along a vertical line throughout the whole motion, PN is the distance through which gravity has acted, measured in the direction of the force. Other cases can be similarly treated, and thus we may say generally that the work done by a force equals the product of the force, and the distance through which its point of application moves, measured in the direction of the force. Of

CASES IN WHICH A FORCE DOES NO WORK. 93

course, if the weight is lifted from A to P, the work expended equals the weight multiplied by N P.'

Ex. 85.-A train weighing 100 tons is made to run up an incline a mile long of 1 vertical to 160 horizontal; the train is lifted through 5280÷160, or 33 ft. of vertical height; consequently 224,000 × 33, or 7,392,000 footpounds, must be expended.

71. Cases in which a force does no work.—It is very possible for a force to act on a moving body and to do no work. This is evident; for in order that a force may do work, its point of application must move wholly or partly in its direction. Suppose, then, that the point of application of the force is at rest, or that it moves in such a manner that no part of the motion is in the direction of the force; in either case the force does no work. Thus :

(a) Suppose a point to slide along a smooth horizontal plane; the only forces acting are its weight and the reaction of the plane; both act at right angles to the direction of the motion, so that no part of the motion is in the direction of the forces, and consequently neither force does work, nor has work expended on it. In actual cases of sliding there is, in addition to the above-named forces, a force of friction which acts horizontally in a direction opposite to the motion, and as long as motion lasts work must be expended in overcoming it.

(b) Consider the case of a wheel rolling along a road. If we suppose this to be the case of a perfectly hard circle rolling (without any sliding) on a perfectly hard straight line, it is plain that at each instant the circle touches the line at one point only, and that point is for the instant at rest.

1 If the space through which the point of application of the force moves is indefinitely small, the work done by the force is what is generally called its virtual moment, and may be reckoned positive; in a similar case, if work is expended on the force, this work is the virtual moment of the force, and is reckoned negative. Suppose now the space through which the point moves to be finite; divide the path into an indefinitely great number of parts, let P denote the force, and p any one portion of the path measured in p's direction, i.e. P's virtual velocity; and let the sum of all the products Pp be taken for all the parts of the path; this sum is the work done by P. It will be observed that this statement includes the cases in which P is variable, and in which the virtual moment of Pis positive in some part of the path and negative in others.