290. Let 1, 2

r, be the radii of the several axles, R1, those of the corresponding wheels; let also W, P1, P2 P1, P be the weights applied to the axles sus

R1, R2, R3, .....

tained by the several powers P1, P2 P applied to the corre

sponding wheels. Then, in case of an equilibrium, we have

1

.. P: W:: r1 72 73 ............2: R1 R the relation between the power and weight generally.

R

and n may be found by the solution of a quadratic, and by refer

ence to the tables.

291..

When the bar is in equilibrium, its centre of gravity

and point of suspension are in the same vertical line.

Hence the inclination of the line joining the point of suspension and centre of gravity to the bar, is the inclination also of the bar to the vertical; which determines its position.

292.

The strength (s) of the beam at any point is measured by the area of its lateral section x distance of its centre gravity from its under surface,. Hence, x and y being the vertical and horizontal sides of the rectangular section of the beam required, and r the radius of the base of the cylinder, we have

293.

When the paraboloid is at rest, the straight line joining its centre of gravity and point of contact with the plane on which it rests, is vertical.

Hence the equation to the generating parabola being

a the length of the axis, and the inclination of the axis to the

vertical, g the distance of the centre of gravity from the vertex

294. Let the weight (W) of the lever be supposed to be collected in its middle point, its length = a, and the distance of the fulcrum from the end to which A is attached = x. Then, A being the greater weight, we have

295.

The beam being in equilibrium, the directions of the forces, viz., the reactions of its extremes acting planes, and the weight of the beam at its centre of gravity acting vertically, must conspire to the same point.

Hence, supposing 0, ' the inclinations of the planes to the horizon, that of the beam of the vertical, and a, a' its parts as divided by the centre of gravity, we readily obtain

whence it is very easy to determine the actual position of equilibrium of the beam.

296.

Let Q the weight of each pulley, and suppose P1 P2 P the several powers requisite to sustain each pulley,

.....

+ +

Now the weight of the pulleys not being considered, we have

297.

Whatever may be the number of tacks, it is evident, that in the case of an equilibrium the sum of the forces estimated in directions parallel to any given direction must = 0; for otherwise motion would ensue. Hence, in the problem, the vertical tendency of the weights, which is measured by their sum, must be counteracted by the sum of the vertical reactions (in the opposite direction) of the tacks. .. sum of the weights = (sum of reactions) sum of the vertical pressures. For another proof, see Whewell's Mech. p. 48.

298. If the straight line joining the centre of gravity of the A and one of its angles be produced through the middle point

of the opposite side until it is doubled, the whole line will be the diagonal of the parallelogram whose sides are the other two lines joining the centre of gravity, and the remaining angles. Hence the truth of the proposition is manifest.

299.

Let w, w'; g, g', be the weights, and distances of the centres of gravity, from the fulcrum of the greater and less arms of the steelyard, respectively. Also let the weights W, W', W", &c., attached to the extremity of the less arm (whose length supposem), be balanced by P, P', P', &c., placed on the greater arm at the distances d, d', d", &c., from the fulcrum. Then Wm+w'g' = Pd+wg

which intervals will evidently be constant, provided the weights W, W, &c., have a common difference, or are in arithmetical

300.

If a be the tension of the catenary at its lowest point (which being measured along the tangent is then wholly horizontal) x, y the vertical and horizontal co-ordinates originating in that point, and s the length of the corresponding arc; then it is known that

s22ax + x2

Hence, putting length of the chain

(1)

2n, and the distance be

tween the tacks, which are supposed in the same horizontal line

= 2m, we have,