In the right-hand figure W+w=5P cos a + 2x. The pressure at the fixed point K is 2P cos a, and at the fixed point L is 4P cos a+x.

208. It will be seen that in every system of Pullies we find the condition of equilibrium by beginning at one end of the system and determining in order the tensions of all the strings of the system. We have always begun with the Power end, except in Art. 196; and in that Article we might also have begun with the Power end.

EXAMPLES. XIV.

1. In a single moveable pully if the weight of the pully be 2 lbs., find the force required to raise a weight of 4 lbs.

2. If there be two strings at right angles to each other and a single moveable pully, find the force which will support a weight of 2 lbs.

3. A man stands in a scale attached to a moveable pully, and a rope having one end fixed passes under the pully, and then over a fixed pully: find with what force the man must hold down the free end in order to support himself, the strings being parallel.

4. If on a wheel and axle the mechanical advantage be six times as great as on a single moveable pully, compare the radii of the wheel and the axle.

The following ten Examples relate to the First System of Pullies; see Art. 196:

5. If n=6, and P=28 lbs., find W.

6.

If W=4 lbs., and P=1 ounce, find n.

7. If n=3, find the consequence of adding one ounce to P, and ten ounces to W.

8. If a man support a weight equal to his own, and there are three pullies, find his pressure on the floor on which he stands.

9. If there are three pullies, each weighing one lb., find the Power which will support a Weight of 17 lbs.

10. If there are three pullies of equal weight, find the weight of each in order that a Weight of 56 lbs. attached to the lowest pully may be supported by a Power of 7 lbs. 14 ounces.

11. If the weight of each pully is P, find W and the tension of each string.

12. If there be three pullies each of weight w, and W=P, find W.

13. If there are two pullies each of weight 4w, and the Power be 3w, shew that no Weight can be supported by the system.

14. In a system of three pullies if a weight of 5 lbs. is attached to the lowest, 4 lbs. to the next, and 3 lbs. to the next, find the Power required for equilibrium.

The following six Examples relate to the Second System of Pullies; see Art. 198:

15. Find the number of strings at the lower block in order that a Power of 4 ounces may support a Weight of 4 lbs.

16. Find the number of pullies at the lower block if P 12 stone and W=18 cwt.

=

17. If there are four strings at the lower block, find the consequence of adding one ounce to P, and three ounces to W.

18. If there are six strings at the lower block, find the greatest Weight which a man weighing 10 stone can possibly raise.

19. A man supports a Weight equal to half his own weight; if there are seven strings at the lower block, find his pressure on the floor on which he stands.

20. Find what Weight can be supported if there are three pullies at the lower block, the string being fastened to the upper block, and the weight of the lower block being

three times the Power.

XV. The Inclined Plane.

209. An Inclined Plane in Mechanics is a rigid plane inclined to the horizon.

B

When an Inclined Plane is used as a Mechanical Power the straight lines in which the Power and the Weight act are supposed to lie in a vertical Plane perpendicular to the intersection of the plane with the horizon. Thus the Inclined Plane is represented by a right-angled triangle, such as ACB; the horizontal side AC is called the base; the vertical side CB is called the height; and the hypotenuse AB is called the length, The angle BAC is the inclination of the Plane to the horizon.

The Plane is supposed perfectly rigid, and, unless the contrary be stated, it is supposed to be perfectly smooth; so that the Plane is assumed to be capable of supporting any amount of pressure which is exerted against it in a perpendicular direction.

210. If a Weight be supported on an Inclined Plane at a point L, and LM be drawn in the direction of the Power, and LN at right angles to the Plane, so as to meet a vertical line at M and N, the Power is to the Weight as LM is to MN.

Let BAC be the Inclined Plane. Let a heavy body whose weight is W be placed on it at any point L, and be kept at rest by a Power, P, acting in the direction LM. Let MN be drawn vertically downwards, and LN at right angles to the Plane.

The body at L is acted on by three forces; the Power P in the direction LM, its own Weight in a direction parallel to MN, and the resistance of the Plane in the direction NL.

L

MA

N

B

Hence, by Art. 36, since there is equilibrium the sides of the triangle LMN are respectively proportional to the forces. Therefore

Let R denote the resistance of the Plane; then

We may write these results thus:

P: WR LM: MN: NL.

It is usual to consider separately two special cases of the general proposition; and this we shall do in the next two Articles.

211. When there is equilibrium on the Inclined Plane, and the Power aets along the plane, the Power is to the Weight as the height of the Plane is to the length.

Let W denote the Weight of a heavy body, and P the Power. From any point Z in the Plane, draw LN at right angles to the Plane, meeting the base at N; and draw NM vertical, meeting the plane at M.

M

B

Then the sides of the triangle LMN are parallel to the directions of the forces which keep the heavy body at rest; therefore, by Art. 36,

P LM
WMN

But the triangle LMN is equiangular to the triangle CBA; for the angle LMN is equal to the angle ABC, by Euclid, 1. 29; the right angle ÑLM is equal to the right angle ACB; and therefore the third angle MNL is equal to the third angle BAC.

Let R denote the resistance of the plane; then

R AC

W AB

We may write these results thus:

212.

P: WR: CB : BA : AC.

When there is equilibrium on the Inclined Plane, and the Power acts horizontally, the Power is to the Weight as the height of the Plane is to the base.

Let W denote the Weight of a heavy body, and P the Power. From any point L in the Plane draw LN at right angles to the Plane, meeting the base at N, and NM vertical meeting at M the horizontal straight line drawn through L.

A

L

M

B

N

C

Then the sides of the triangle LMN are parallel to the directions of the forces which keep the heavy body in equilibrium; therefore, by Art. 36,

But the triangle LMN is equiangular to the triangle BAC. For the angle BLN, being a right angle, is equal to the sum of the two angles BAC and ABC; and BLM is equal to BAC, by Euclid, 1. 29: therefore MLN is equal to ABC. And the right angles LMN and BCA are equal. Therefore the third angle LNM is equal to the third angle BAC.