on his head will be enormous: thus a hat, although very light, would, if exceedingly tall, crack his head.

(6) A particle is placed on a square table at distances c1, c2, c, c, from the corners, and to it are attached strings passing over smooth pulleys at the corners and supporting weights P1, P1⁄2, P ̧, P ̧; to prove that, if there is equilibrium,

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Let O, fig (29), be the position of the particle; C the centre of the square  ̧Â: draw CM, OM, parallel to  ̧‚ ‚ to intersect in M: let CM= x, OM=y. Let

1 2 3

and let 2a represent the length of a side of the square.

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27

For the equilibrium of the particle, we have, resolving forces parallel and perpendicularly to CM,

By symmetry it is evident that we must also have

(7) A weight W is supported on an inclined plane AB,

W

fig. (30), by two forces, each equal to, one of which acts

n

parallel to the base AC: to determine the conditions of equilibrium.

Let a be the inclination of the plane, R the reaction, and e the inclination of the other supporting force to the plane.

Resolving forces at right angles to AB, we have

This result shews that the equilibrium is impossible if a be

-1

greater than 2 tan11. If a be less than 2 tan ̄1

-1

n

(3) gives us two equal values of e with opposite signs: thus the force, of which the direction is not given, may act as indicated in the diagram or at an equal inclination below the inclined plane.

Again, the equations (1) and (2) may be written in the

squaring these two equations and then adding, we get

which determines the magnitude of the reaction of the inclined plane against the weight W.

The values of R here given will be both possible and

sin a

positive if cos a +

n

be greater than 1, that is, if

sin a> n (1 cos a),

(8) Prove that the forces 3P√2, 4P, 5P, 7P, 2P, acting in one plane upon a point, the directions of the last four making angles of 45°, 135°, 225°, 315°, respectively with that of the first, will produce equilibrium.

(9) A weight of 10 pounds is placed on an inclined plane, the angle of which is 30°: to find the magnitude of a horizontal force which will sustain the weight.

(10) Two forces P, Q, acting respectively parallel to the base and length of an inclined plane, will each of them singly sustain upon it a particle of weight W: prove that

(11) A weight W is supported on an inclined plane AB, fig. (31), by three forces, each equal to P, one acting vertically upwards, another parallel to the horizontal line AC, and the third along AB: to find the inclination of the plane.

The required inclination is equal to

(12) If P be the force which, acting along an inclined plane, will support a weight W, R being the pressure on the plane; prove that, P' being the force which, acting horizontally, will support the same weight,

(13) A power P supports a weight W on a plane, the inclination of which is a; prove that, if the pressure on the plane be

equal to P, the angle which the direction of the power makes with the plane is equal to

(14) Two heavy particles, P and Q, are connected together by a fine thread passing over a smooth pully at C: P rests on a smooth inclined plane AB, and Q hangs freely: prove that, a denoting the inclination of the plane to the horizon, R the pressure, and the angle CPB,

(15) A weight W is supported on an inclined plane by a certain force acting at an angle e to the plane, the inclination of the plane being a: the same weight is sustained on the same plane by the same force acting at an angle -e: prove that in each case the pressure on the plane is equal to

W (cos a - sin a).

(16) A weight of 6 lbs. is placed on an inclined plane, the height of which is 3 feet and base 4 feet, and is attached by a string to an equal weight hanging over the top of the plane: find how much must be added to the weight on the plane that there may be equilibrium, and determine the pressure on the plane.

The amount which must be added is 4 lbs., and the pressure on the plane is 8 lbs.

(17) If two planes have the same height, and if two weights balance on them by means of a string which passes over the common vertex, prove that the weights will be to one another as the lengths of the planes.

(18) Two weights of 3 pounds and 5 pounds rest on a double-inclined plane, being connected together by a fine string which passes over the common summit of the two planes: supposing the inclination of the plane, on which the former weight rests, to be 30o, find the inclination of the other plane.

W. S.

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