Page images
PDF
EPUB

strokes and draw another. Then without moving & push the slides back and draw a third line close to the first; take a hundred strokes and draw a fourth line near the second. Measure the interval between the first and third, and the second and fourth. They should be equal, but if not, the difference divided by an hundred gives the average difference in length of a stroke the second time, compared with the first.

Instead of a pencil, a pen may be used to draw the lines, or a graver, if a metallic scale is desired. The finest scales are ruled on glass by a diamond. Instead of using the natural edge of the gem, as when cutting glass, an engraver's diamond should be employed, which is ground with a conical point; the direction in which it should be held, and the proper pressure, being obtained by trial. Scales may also be etched by covering the surface with a thin coating of wax or varnish, and the lines marked with a graver. If metallic, it is then subjected to the action of nitric acid; if of glass, to the fumes of fluorhydric acid. It is possible that the new method of cutting glass by a sand-blast may prove applicable to this purpose with a great saving of time and trouble.

MECHANICS OF SOLIDS.

23. COMPOSITION OF FORCES.

Apparatus. Two pulleys A and B, Fig. 19, are attached to a board which is hung vertically against a wall. Two threads pass over them, and a third C, is fastened to their ends at D. Three forces may now be applied by attaching weights to the ends of the cords. The weights of an Atwood's machine are of a convenient form, but links of a chain, picture hooks, cents, or any objects of nearly equal weight may be used. Small beads are attached to the three threads at distances of just a decimetre from D.

D

M

GG

[ocr errors]

C

B

Experiment. Attach weights 2, 3 and 4 to the three cords, and let D assume its position of equilibrium. Owing to friction it will remain at rest in various neighboring positions, their centre being the true one. Now measure the distance of each bead from the other two with a millimetre scale, and obtain the angle directly from a table of chords. If these are not at hand, dividing the distance. by two, gives the natural sine of one half the required angle. By the law of the parallelogram of forces, the latter are proportional to the sides of a triangle having the directions of the forces. But these sides are proportional to the sines of the opposite angles, hence the sines of the angles included between the threads should be proportional to the forces or weights applied. Divide the two larger-forces by the smaller, and do the same with the sines of the angles, and see if the ratios are the same. The angles themselves should first be tested by taking their sum, which should equal 360°. If either angle is nearly 180°, it cannot be accurately measured in this way, but must be found by subtracting the sum of the other two from 360°, or measuring one side from

Fig. 19.

the prolongation of the other. It is well to draw the forces from the measurement, and see if a geometrical construction gives the same result as that obtained by calculation. Repeat with forces in several other ratios, as 3, 4, 5; 2, 2, 3; 3, 5, 7; taking care in all cases to include in the weights the supports on which they rest.

24. MOMENTS.

Apparatus. A board AB, Fig. 20, is supported at its centre of gravity on the pin C. It should revolve freely, and come to rest in all positions equally. Two forces may be applied to it by the weights D and E, attached by threads to the pins F and G. Their magnitudes may be varied from 1 to 10 by different weights, and their points of application by using different pins, as H, I and J. To measure their perpendicular distances from the pin C, a wooden right-angled triangle or square is provided, one edge of which is divided into millimetres, or tenths of an inch.

M
A

I J
G H

K

B

F

Experiment. Various laws of forces may be proved with this apparatus. 1st. When a single force acts on a body AB fixed at one point, as C, there will be equilibrium only when it passes through this point. Remove FD and attach a weight E to G. It will be found that the body will remain at rest only when the point & is in line with M and C. 2d. A force produces the same effect if applied at any point along the line in which it tends to move the body. Apply the two weights D and E, which tend to turn the board in opposite directions. Make their ratio such that MG shall be in line with G, H, J. Now transfer the end of the thread from G to H, I and Jin turn, when it will be found that the position of the board will be unchanged. It should be noticed, however, that in the last case the board is in unstable equilibrium, since FJ falls beyond the point of support C. 3d. The moment of a force, or its tendency to make a body revolve, is proportional to the product of its magnitude by its perpendicular distance from the point of support. Make D equal 2, and attach it to K, so that the thread rests over the edge of the board, which is the arc of a circle with centre at C, and radius .6. Its tendency to make the board revolve is therefore the same, what

DE

Fig. 20.

ever the position of the latter. Make E successively 1, 2, 3, 4, 5, 6, and measure the perpendicular distance of the thread to which it is attached in each case from C. This distance is measured by resting the triangle against the thread and measuring the distance of C'by its graduated edge. In each case the moment of E will be found to be the same, and equal to 2 × 6, the moment of D. 4th. When two forces hold the body in equilibrium their resultant must pass through the fixed point. Make D equal 2, and attach it to F, and E equal 3, applied at G. Lay a sheet of paper on the right hand portion of AB, making holes for F, C and J to pass. Draw on it with a ruler the direction of the two threads prolonged, and then removing it, construct their resultant geometrically by means of the parallelogram of forces. It will be found to pass through C. Repeat two or three times with different weights and points of application.

25. PARALLEL FORCES.

Apparatus. The apparatus used is shown in Fig. 21. AB is a straight rod about two feet long, with a paper scale divided into tenths of an inch attached to it. It is supported by a scale-beam CD with a counterpoise, so that it is freely balanced, and remains horizontal. Weights formed like those of a platform scale may be attached to it at any point, by riders, as at E, F and G. Taking each rider as unity, four sets of weights are required of magnitudes 10, 5, 2, 2, 1, .5, .2, .2, .1. Two other beams, like CD, should also be provided, to which these weights may be attached, as at E, so as to produce an upward force of any desired magnitude. All these scale-beams may be very roughly made, even a piece of wood supported at the centre by a cord, being sufficiently accurate. English beams of iron may, however, be obtained at a very low price.

H
E

C

B

Experiment. The resultant of any system of parallel forces lying in one plane may be found by this apparatus. Thus suppose we have a force of 15.7 acting upwards, and two of 8.3 and 1.4 acting downwards, and distant from the first 6.4 and 8.7 inches respectively. Produce the upward force by adding the weights 14.7 to E, and the

Fig. 21.

two downward forces by weights 7.3 and .4 (allowing 1 for each of the scale-pans) at F and G, setting them at the points of the beam marked 3.6 and 18.7. They are then at the proper distance from C, which is at 10 inches from the end. We now find that A goes down and B up; by placing the finger on the beam we see that it can be balanced only by applying a downward force to the right of C. Now place a rider in this position, and move it backwards and forwards, varying the weight on it until the beam is exactly balanced. The magnitude of this weight will be found to be 6, and its position 16.8, or 6.8 inches from C. The resultant of the three forces will be just equal and opposite to this. Had the force required to balance them acted upwards, we should have used one of the auxiliary scale-beams. To test the correctness of this result we compute the resultant thus: R 15.7 (8.3 + 1.4)=6, and taking moments around C we have 8.3 × 6.4 1.4 × 8.7 = 6 × x, or x = 6.8 the observed distances.

Determine the position and magnitude of the resultant in several similar cases, as for example the following, in which U means an upward, and D an downward force, and each is followed first by its magnitude, and then by the point on the bar at which it is to be applied.

1. D, 5.0, 4.3; U, 10.0, 10.0.

2. D, 2.6, 3.2; U, 7.8, 10.0.

3. D, 7.4, 3.7; U, 17.1, 10.0.

4. D, 11.1, 2.1; D, 6.5, 5.6; U, 2.3, 18.4.

5. D, 5.2, 1.9; U, 15.2, 10.0; D, 8.4, 12.6; U, 3.0, 18.1. Two equal parallel forces acting in opposite directions and not in the same line, form what is called a couple, and have no single resultant. Thus apply the two forces D, 12.0, 5.0, and U, 12.0, 10.0. No single force will balance the beam. Equilibrium is obtained only by a second couple having the same moment, and turning in the opposite direction; thus the moment being 12.0 × 5.0 60.0, the beam may be balanced by two forces of 10.0, each distant 6 inches from one another, placing the upward force to the left. Find in the same way some equivalent to D, 4.3, 7.6, and U, 4.3, 10.0, and notice that it makes no difference to what part of the beam the two forces are applied, provided their distance apart remains unchanged.

« PreviousContinue »