Page images
PDF
EPUB

DF is

same; for, in this case, D G is one-third of A G, one-third of B F, and D E is one-third of CE; therefore each prop sustains one-third of the weight.

If the weight be supported on more than two props, the problem appears to admit of innumerable solutions; but if the centre of gravity of W, as it rests on the plane, be the same with the centre of gravity of the figure made by joining the tops of the props by straight lines, the pressures on the props are all equal to one another.

Example 3.

A uniform beam of timber, 10 feet in length, being suspended at the distance of 8 feet from one of the ends, it required a weight of 2 cwt. attached to the other end to keep it parallel to the horizon; what is the weight of the beam?

Rule.-Multiply the weight which is suspended at the end by its distance from the fulcrum, and take twice this product for a dividend. Then subtract the square of the distance between the fulcrum and the end where the weight is suspended, from the square of the distance between the fulcrum and the other end, for a divisor. And if the above dividend be divided by this divisor, the quotient will give the weight of one foot or one inch in length, according as you take the length in feet or inches; and this quotient, multiplied by the whole length of the beam, will give the whole weight of the beam.

Here 2 feet is the distance between the fulcrum and the end where the weight is suspended, and 8 feet is the distance between the fulcrum and the other end.

[blocks in formation]

weight of one foot in length of the beam; therefore 141

=

× 10 1493 lbs. the whole weight of the beam.

Example 4.

Two men carrying a burden of 200 lbs. weight between them, hung on a pole, the ends of which rest on their shoulders; how much of this load is borne by each man, the weight hanging 6 inches from the middle, and the whole length of the pole being 4 feet?

Now 4 feet 48 inches, and the burden being hung on at 6 inches from the middle, it will be 18 inches from one of the men, and 30 inches from the other.

And by the note at page 17, we have 48: 200 :: 30 : 125 lbs. the weight borne by the man who is nearest to the burden.

And 48 200 :: 18: 75 lbs. the weight borne by the other man.

Example 5.

If the altitude of a cone be double the diameter of its base, what is the inclination of its axis with the horizon when it is just ready to fall over? (Fig. 49.)

By Art. 74, Cor. 1, when the cone is just ready to fall over, a perpendicular from the centre of gravity will fall on the extremity A of the base.

=

=

From the centre of gravity of a cone (page 46) we have BD 4 G D, and by the question B D = 2 AC = 4 A D; hence G D A D, and therefore the angles GAD and A G D are each 45°, or half a right angle. But the triangles A E G, GA D, and D A E, are all similar (Euclid, Book 6, Prop. 8); consequently, the angle A E G, which is the angle of inclination of the axis with the horizon, is 45°; and its complement, the angle D A E, which is the angle the base of the cone makes with the horizon, is also 45°.

Example 6.

Two inclined planes A B and B C have the same height, and upon these planes two weights keep each other in

equilibrio in the same manner as in Example 3, page 38: given the length of the planes 30 and 40 inches respectively, and the horizontal distance A C between the feet of the planes 50 inches; required their common height, and the ratio of the weights.

Rule.-As the base or longest side is to the sum of the other two sides, so is their difference to the difference of the segments of the base. And half the difference of the segments, added to half their sum, gives the greater segment; and half the difference of the segments, subtracted from half their sum, will give the lesser segment.

[blocks in formation]

30 = 70 :: 40.

[ocr errors]

=

difference of the segments of the base; and

the difference of the segments.

Therefore 25+ 7

18 is the lesser segment.

10: 14, the

[blocks in formation]

32, the greater segment; hence

[merged small][ocr errors][merged small][merged small][merged small][merged small]

And, by Example 3, pages 38 and 39, P: W:: BC:

[blocks in formation]

A cast iron beam, of uniform thickness, and in the form of a parabola, is supported upon two pins, one of which is fixed at the vertex, and the other at the middle of the base: required the pressure on each pin when the distance between the vertex and the middle of the base is 5 feet, the base 2 feet, and the weight of the beam 56 lbs.

Here, in this Example, the centre of gravity is not in the middle of the beam; and by Form. 4, page 46, the distance of the centre of gravity from the vertex of a parabola is of the axis.

Hence × 53 feet, the distance of the centre of gra vity from the vertex.

Now we must consider the whole weight of the beam to be collected in its centre of gravity; and by note, page 17, we have 5 : 56 :: 3 : 333 lbs. the pressure on the pin which is fixed in the middle of the base.

And 5 : 56 :: 2 : 223 lbs. the pressure on the pin which is fixed in the vertex.

STRENGTH AND STRESS OF MATERIALS.

81. A knowledge of the strength and stress of materials is of so much importance to the practical mechanic, that it demands his most serious attention; and he will find the works of Barlow and Tredgold* to be treasures of inestimable value, being fraught with every kind of useful information relating to these subjects.

Mr. Barlow shows that there are four distinct strains to which every hard body may be exposed, and which may be stated as follow::

1. A body may be pulled or torn asunder by a stretching force, applied in the direction of its fibres; as in the case of ropes, stretchers, king-posts, tie-beams, &c.

2. It may be broken across by a transverse strain, or by a force acting either perpendicularly or obliquely to its length; as in the case of levers, joists, &c.

* Barlow on the Strength and Stress of Timber, and Tredgold on the Strength of Cast Iron.

3. It may be crushed by a force acting in the direction. of its length; as in the case of pillars, posts, and trussbeams.

4. It may be twisted or wrenched by a force acting in a circular direction; as in the case of the axle of a wheel, the nail of a press, &c.

ON THE COHESIVE STRENGTH OF MATERIALS.

82. The force of cohesion may be defined to be that force by which the fibres or particles of a body resist separation, and is therefore proportional to the number of fibres in the body, or to the area of its section.

Mr. Emerson gives the load that may be safely borne by a square inch rod of each of the following:

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

He also gives the following practical rule, viz. That a cylinder, the diameter of which is d inches, loaded to onefourth of its absolute strength, will carry as follows:

[merged small][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]
« PreviousContinue »