Let R denote the resistance of the Plane; then

We may I write these results thus:

P: WR :: BC : CA: AB.

213. We may obtain convenient expressions for the proportionate values of P, W, and R by the aid of Trigonometry.

Let the angle BAC-a, and the angle_MLB=B; therefore the angle MLN=90°+ß.

P

Then

P: W:R:: LM: MN: NL

:: sin LNM: sin MLN: sin NML

L

M

B

:: sin a : sin (90o + ß) : sin (90o — a— ß) A :: sin a cosẞ: cos (a +ẞ).

N

C

214. In the figure of the preceding Article the resistance of the Plane is represented by the straight line NL; that is, the resistance acts from N towards L. Thus if the body be placed on the Plane, and be in equilibrium, the straight line LN must be below the Plane; that is, the sum of the angles MLB and BAC must be less than a right angle.

215. The results of Art. 213 may also be obtained by the method of resolving the forces given in Art. 57; and thus we obtain a good example of the method, and assistance in remembering the results.

Resolve the forces along the plane: this gives

P cos B W sin a.

=

Resolve the forces at right angles to the plane: this

gives

P sin B+RW cos a.

Hence we deduce

RW cos a—·

W sin a sin B W cos (a+B)

cos B

cos B

216. Perhaps it may seem that an Inclined Plane can scarcely be called a Machine; it is not obvious that it can be usefully employed like the other Mechanical Powers. But we may observe that if we have to raise a body we may draw it up an Inclined Plane by means of a Power which is less than the Weight of the body.

EXAMPLES. XV.

In the following twelve Examples the Power is supposed to act along the Plane:

1. If the Weight be represented by the height of the Plane, shew what straight line represents the pressure on the Plane.

2. If W=12 lbs., and the height of the Plane be to its base as 3 is to 4, find P.

3. If W=10 lbs. and P=6 lbs., find R.

4. If P=R, find the inclination of the Plane, and the ratio of P to W.

5. If P is to R as 3 is to 4, express each of them in terms of W.

6. If P-9 lbs., find W when the height of the Plane is 3 inches, and the base 4 inches.

7. An Inclined Plane rises 3 feet 6 inches for every 5 feet of length: if W=200, find P.

8. If the length of the Plane be 32 inches, and the height 8 inches, find the mechanical advantage.

9. When a certain Inclined Plane ABC, whose length is AB, is placed on AC as base, a Power of 3 lbs. can support on it a Weight of 5 lbs.: find the Weight which the same Power could support if the Plane were placed on BC as base, so that AC is then the height of the Plane.

10. A railway train weighing 30 tons is drawn up an Inclined Plane of 1 foot in 60 by means of a rope and a stationary engine: find what number of lbs. at least the rope should be able to support.

11. A Weight of 20 lbs. is supported by a string fastened to a point in an Inclined Plane, and the string is only just strong enough to support a weight of 10 lbs.: the inclination of the Plane to the horizon being gradually increased, find when the string will break.

12. If it takes twice the Power to support a given Weight on an Inclined Plane ABC when placed on the side AC, that it does when the Plane is placed on the side BC, find the greatest Weight which a Power of one lb. can support on the Plane.

In the following four Examples the Power is supposed horizontal:

13. If W=12 lbs., and the base be to the length as 4 is to 5, find P.

14. If W-48 lbs., and the base be to the height as 24 is to 7, find P and R.

15. If R=2 lbs., and P=1lb., find W, and the inclination of the Plane.

16. If W=12 lbs., and P=9 lbs., find R.

17. If the Power which will support a Weight when acting along the Plane be half that which will do so acting horizontally, find the Inclination of the Plane.

18. If R be the pressure on the Plane when the Power acts horizontally, and S when it acts parallel to the Plane, shew that RS=W2

19. A Power P acting along a Plane can support W, and acting horizontally can support : shew that

20. A weight W would be supported by a Power P acting horizontally, or by a Power Q acting parallel to the Plane: shew that

21. Give a geometrical construction for determining the direction in which the Power must act when it is equal to the Weight; and shew that if S be the pressure on the Plane in this case, and R the pressure when the Power acts along the plane, S=2R.

22. The length of an Inclined Plane is 5 feet, and the height is 3 feet. Find into what two parts a weight of 104 lbs. must be divided, so that one part hanging over the top of the Plane may balance the other resting on the Plane.

23. The inclination of a Plane is 30°; a particle is placed at the middle point of the Plane, and is kept at rest by a string passing through a groove in the Plane, and attached to the opposite extremity of the base: shew that the tension of the string is equal to the weight of the particle.

24. A Weight of 20 lbs. is supported by a Power of 12 lbs. acting along the Plane: shew that if it were required to support the same Weight on the same Plane by a Power acting horizontally, the Power must be increased in the ratio of 5 to 4, while the pressure on the Plane will be increased in the ratio of 25 to 16.

XVI. The Wedge. The Screw.

217. The Wedge. The Wedge is a solid body in the form of a prism; see Euclid, Book xI. Definitions. In the Wedge two parallel faces are equal and similar triangles, and there are three other faces which are rectangles.

The wedge may be employed to separate bodies.

We may suppose the wedge urged forward by a force P acting on one of the rectangular faces, and urged backwards by two resistances and R arising from the bodies which the wedge is employed to separate. These forces may be

B

supposed to act in one plane perpendicular to the rectangular faces; and we shall assume that the wedge and the bodies are smooth, so that the force acting on each face is perpendicular to that face.

Let the triangle ABC represent a section of the wedge made by a plane perpendicular to its rectangular faces; and suppose the wedge kept in equilibrium by the forces P, Q, R perpendicular to AB, BC, CA respectively: then by Art. 37

P: QR AB: BC: CA.

If AC=BC the wedge is called an isosceles wedge; in this case QR, and P: R:: AB: CA.

Let the angle ACB be denoted by 2a, then when the wedge is isosceles AB=2AC sin a, and

PR: 2 sin a : 1, so that P=2R sin a.

218. There is very little value or interest in the preceding investigation, because the circumstances there supposed scarcely ever occur in practice. A nail is sometimes