If the pressure exerted by the force is variable, we may conceive the path described by the point of application to be divided into equal parts, so small that, for each part, the pressure may be regarded as constant. If we denote the length of one of these equal parts, by p, and the force exerted whilst describing this path, by P, we shall have for the corresponding quantity of work, Pp, and for the entire quantity of work denoted by Q, we shall have the sum of these elementary quantities of work; or, since p is the same for each,

The quotient obtained by dividing the entire quantity of work by the entire path, is called the mean pressure, or the mean resistance, and is evidently the force which, acting uniformly through the same path, would accomplish the same work.

Work, when the power acts obliquely to the path. 144. Let PD represent the force, and AB the path which the body D is constrained to

8

Fig. 126.

D

B

follow. Denote the angle PDs by a, P and suppose Pto be resolved into two components, one perpendicular, and the other parallel to AB. We shall have, for the former, Psina, and, for the latter, Pcosa. The former can produce no work, since, from the nature of the case, the point cannot move in the direction of the normal; hence, the latter is the only component which works. Let SD be the space through which the body is moved in any time whatever. If we denote the pressure exerted in the direction of PD, by P, and the quantity of work, by Q, we shall have,

Let fall the perpendicular ss' from s, on the direction of the

force P. From the right-angled triangle Dss', we shall

have,

SD x cosa s'D.

Substituting this in the preceding equation, we get,

Q Px s'D.

That is, the quantity of work of a force acting obliquely to the path along which the point of application is constrained to move, is equal to the intensity of the force multiplied by the projection of the path upon the direction of the force. We have supposed the intensity of the force P, to be expressed in pounds, or units of mass.

If we take the distance sD, infinitely small, s'D will be the virtual velocity of D, and the expression for the quantity of work of P will be its virtual moment (Art. 38). Hence we say that the elementary quantity of work of a force is equal to its virtual moment, and, from the principle of virtual moments, we conclude that the algebraic sum of the elementary quantities of work of any number of forces applied at the same point, is equal to the elementary quantity of work of their resultant. What is true for the elementary quantities of work at any instant, must be equally true at any other instant. Hence, the algebraic sum of all the elementary quantities of work of the components in any time. whatever, is equal to the algebraic sum of the elementary quantities of work of their resultant for the same time; that is, the work of the components for any time, is equal to the work of their resultant for the same time. This principle would hardly seem to require demonstration, for, from the very definition of a resultant, it would seem to be true of necessity. If the forces are in equilibrium, the entire quantity of work will be equal to 0.

This principle finds an important application, in computing the quantity of work required to raise the material for a wall or building; for raising the material from a shaft; for raising water from one reservoir to another; and a great

variety of similar operations. In this connection, the principle may be enunciated as follows: The algebraic sum of the quantities of work required to raise the parts of a system through any vertical spaces, is equal to the quantity of work required to move the whole system over a vertical space equal to that described by the centre of gravity of the system.

It also follows, from the same principle, that, if all the pieces of a machine which moves without friction be in equilibrium in all positions, under the action of weights suspended from different parts of the machine, the centre of gravity of the system will neither ascend nor descend whilst the machine is in motion.

Work, when a body is constrained to move upon a curve.

145. Let AB represent the curve, and suppose that the force is so taken that its line of direction shall

P

Fig. 127.

always pass through a point P. Divide the curve into elements so small that each may be taken as a straight line, and, with P as a centre, and the distances from P to the points of division as radii, describe arcs of circles. -Then, denoting the force supposed constant, by P, we shall have (from Art. 144) the elementary quantity of work performed whilst the point is moving over aa', equal to P x ac, or P x bb'. In like manner, the quantity of work performed whilst the point is describing a'a" will be equal to P x b'b", and so on. Hence, by summation, we shall find the entire quantity of work performed in moving the body from B to A will be equal to P x BB'. If now we suppose the curve AB to lie in a vertical plane, and the force to be the force of gravity, the point P may be regarded as infinitely distant, the lines Pa, Pa' &c., will become vertical, and the lines a'b', a"b", will be horizontal. We may, therefore, enunciate the following principle: The quantity of work of the weight of a body

in descending a curve, is equal to the quantity of work of the same weight in descending vertically through the same height. This principle is immediately connected with the discussion in Art. 74.

If a body in a stable position, as a pyramid resting on its base, be overturned by any extraneous force, the quantity of work will be equal to the weight of the body, multiplied by the vertical height to which the centre of gravity must be raised before reaching its highest point. This product might be taken as the measure of the stability of a body.

EXAMPLES.

1. What amount of work is required to raise 500 lbs. to the height of 5 yards? Ans. 7500 units, or 7500 lbs. ft. 2. To what height can 2240 lbs. be raised by the expenditure of 5600 units of work? Ans. 2.5 ft.

3. What weight can be raised to the height of 25 feet by 224000 units of work? Ans. 8960 lbs.

4. What is the effective horse power of an engine which raises 80 cubic feet of water per minute from the depth of 360 feet, a cubic foot of water weighing 62 lbs.

Ans. 51.11 horse power.

5. What must be the effective horse power to raise the same quantity of water per minute, from a depth of 40 feet? Ans. 6. horse power.

6. How many tons of ore can be raised per hour from a mine 1800 feet deep, by an engine of 28 effective horse power, reckoning 2240 lbs. to the ton? Ans. 13 tons.

7. From what depth will an engine of 16 effective horse power raise 5 cwts. of coal per minute.

Ans. 943 feet, nearly.

8. In what time will an engine of 40 effective horse power raise 44000 cubic feet of water from a mine 360 feet deep, allowing 623 pounds to the cubic foot?

Ans. 12 h. 30 min.

9. Required the quantity of work necessary to raise the material for a rectangular granite wall 25 feet long, 24 feet thick, and 20 feet high, the weight of granite being 162 lbs. per cubic foot?

SOLUTION.

The weight of the wall is equal to

162 lbs. x 25 × 2.5 × 20 = 202500 lbs.

The height of the centre of gravity being 10 feet, the quantity of work is equal to

202500 X 102025000 lbs. ft. Ans.

10. How long would it take an engine of 4 effective horse power to raise the material for the wall in the last example? Ans. 15 minutes, nearly.

11. What quantity of work must be expended in drawing a chain from a shaft, the length of the chain being 450 feet, and its weight 40 lbs. to the foot? Ans. 4050000 lbs. ft.

12. A cylindrical well is 150 feet deep, and 10 feet in diameter. Supposing the well to be filled with water to the depth of 50 feet, how much work must be expended in raising it to the top, water being taken at 62.5 lbs. per cubic foot?

SOLUTION.

The weight of the water is equal to

TX 52 X 50 × 62.5 lbs. 245437.5 lbs.

The distance of the centre of gravity from the top is 125 feet. Hence, the required quantity of work is equal to

245437.5 lbs. X 125 ft. = 30679687.5 lbs. ft. Ans.

13. What quantity of work will be required to overturn a right cone, with a circular base, whose altitude is 12