(10) If P just supports W on a rough inclined plane, and the plane is depressed through an angle the tangent of which is the coefficient of friction; prove that, if friction suddenly ceases, there will still be equilibrium, P's direction remaining unaltered.

(11) A rough right-angled triangle stands on its hypotenuse, and equal particles, connected by a fine string passing round a small smooth pully at the vertex, so rest on the sides that one is but just supported: after the string is cut, the other is but just supported; find the angles at the hypotenuse.

(12) Two weights W, W', of different materials, connected by a fine string which passes over a smooth peg, are placed on a rough inclined plane passing through the peg, the inclination of the plane being a: the two parts of the string are respectively perpendicular and inclined at an angle ẞ to the intersection of the inclined plane with a horizontal plane: the weights are in equilibrium, the former being only just supported: prove that, μ, μ', being the respective coefficients of friction between the weights and the inclined plane,

W2 (tan a — μ)2 – 2 WW'tan a sin ß (tan a — μ)

+ W12 (tan2 a — μ'2) = 0.

CHAPTER V.

EQUILIBRIUM OF A BODY MOVEABLE ABOUT A FIXED AXIS.

IF two forces act upon an imponderable lever, and produce equilibrium, their resultant must pass through the fulcrum, and, conversely, if their resultant passes through the fulcrum, there will be equilibrium.

The above condition of equilibrium may be replaced by the following one viz. that the moments of the forces about the fulcrum shall be equal and opposite.

If more than two forces act on the lever and produce equilibrium, the resultant of all the forces must pass through the fulcrum, and, conversely, if their resultant passes through the fulcrum, there will be equilibrium.

The above condition may be replaced by the following one: viz. that the sum of the moments of all the forces which tend to twist the lever in one direction about the fulcrum shall be equal to the sum of the moments of all the forces which tend to twist the lever about the fulcrum in the opposite direction.

SECT. 1. Body acted on by two forces.

(1). If AB, fig. (38), be a rigid imponderable rod, moveable about A, and be in equilibrium under the action of any two forces represented by BF, BC; to shew that, if FB be produced to a point D such that BD= BF, and C, D, be joined, CD is parallel to BA.

Join CD: then the resultant of the two forces BF, BC, that is, of DB, BC, will be a force, equal and parallel to DC, acting on the point B. Now a single force acting on B cannot keep AB

at rest about the fulcrum A unless its direction coincides with AB: the line DC must therefore be parallel to AB.

(2) A shop-keeper has in his shop correct weights but a false balance, that is, one having one arm longer than the other: supposing that he serves out to each of two customers articles weighing, as indicated by his balance, W pounds, using for the commodities first one scale and then the other; to find whether he gains or loses by the peculiarity of his balance.

Let a, b, be the lengths of the arms of the balance, and let P, Q, be the actual weights of the two commodities, placed at the ends of the former and latter arm respectively.

The shop-keeper therefore loses by his sale to the two customers a weight of substance equal to

(a - b)2
W.
ab

(3) A uniform bent rod ABC, fig. (39), moveable about its lower end A, which is fixed, is in equilibrium: to find the inclination of AB to the horizon, ABC being a right angle.

Let AB=2a, BC=2b; then the weights of AB, BC, may be represented by λa, b, respectively. Let BAX=0, AX being a horizontal line.

L

Conceive a beetle to crawl from A to the middle point of AB: it will have arrived at the centre of gravity of AB, having travelled horizontally leftwards through a space a cos 0; thus the moment of AB about A will be λa. a cos 0. Again, conceive a beetle to crawl along the rod from A to the middle point of BC, that is, to the centre of gravity of BC: in going from A to B it will have travelled horizontally leftwards through a space 6

W. S.

2a cos 0, and, in going from B to the middle point of BC, it will have travelled horizontally rightwards through a space b sin 0: thus altogether the beetle will have travelled horizontally leftwards through a space 2a cos 0 -b sin 0, and accordingly the moment of BC's weight about A is equal to

Mb (2a cos 0-b sin 0).

Thus, for the equilibrium of the rod we have, taking moments about A,

λa. a cos 0 +λb (2a cos 0 -b sin 0) = 0,

(4) Two forces P and Q act at the ends A and B, respectively, of a weightless straight lever AB: to find the position of the fulcrum that equilibrium may be maintained, the inclinations of P's and Q's directions to AB being a and ẞ respectively.

Let AB=c; and let x, y, be the distances of the fulcrum from A, B, respectively. Then, taking moments about the fulcrum, we get

(5)

Pc sin a

y=Psin a+Q sin B

A rod AB, fig. (40), moveable about a smooth hinge at A, is attached, by means of a fine string passing over a fixed pully C, to a weight P: to find the position of the rod when in equilibrium.

Let the horizontal line through A be intersected by CP in D and by CB, produced, in E. Let W= the weight of AB, AB=2a, AD=h, CD=k, ▲ BAE= 0, ▲ AEB = 0.

L

Taking moments about A, for the equilibrium of AB, we

Again, from the geometry it is plain that D's distance from BC is equal to either k cos p, or to h sin 4 + 2a sin (0 +$): hence k cos = h sin 4 + 2a sin (0 + 4)

........

(2).

The equations (1) and (2) determine the magnitudes of the angles and 4, and thus define the position of AB.

(6) A uniform rod CD, fig. (41), moveable about a smooth hinge C, presses against a given inclined plane AB: to find the inclination of CD to the horizon in order that the pressure exerted by it on the plane may be equal to half its own weight.

Let R be the reaction of the plane against the rod, and W the weight of the rod. Let a = the inclination of the plane and of the rod to the horizon.

Then, for the equilibrium of the rod, taking moments about C, we have

R. CD cos (a — 0) = W. CG cos 0:

but, the rod being uniform, CGCD, and, by the hypothesis,

(7) When a boy of weight B ascends to a point b of a ladder AB, fig. (42), the lower end of which is fixed while the higher rests against a smooth vertical wall CD, its pressure against the wall is the same as when a man of weight M ascends to a point m: to prove that

B: M: Am : Ab.

Let G be the centre of gravity of the ladder and W its weight let R denote the reaction of the wall against the ladder, in both cases. Let BAC-a.