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from Ex. 2, art. 656, Marrat's Mechanics, that y fr2+2rx+3x)

From these equapo? tra + x? tions, I find (by the rule of double position) x = 1,081 ft. nearly; and hence the solidity of the part EQ R to be cut off is = 0.97 feet, as required. * Answers were also given by Messrs. Rylando,"

Tomlinson, and Maar. 18. Qu. (73) Answered by Mr. J. Butterworth, Haggate

- Let A P and BP be two tangents in. - - :D tersecting in P, and draw the radii O A, O B, and join OP; then by the question AP: + B P2 is given, and AO and O B are given-; therefore, A P + A 02 + BP2 + O Bʻ = 2 O P is given; hence O P is given; and O is given, therefore the locus of P will be a circle concentric with the given circles. This is only a particular case of question 11.

In the same manner it was answered by Messrs. Baines, Brooke, Eyres, Gawthorp, Hine, Maffett, Nesbit, Putsey, Whitley, Winward, and Vayheeg.

19. Qu. (74) Answered by Messrs. Gawthorp, Nesbit, :- . . 'Rylando, and Whitley.

Retaping the notation made use of in art. 786, Mar.. rat’s Mechanics, we have by trig cin (1-x2)::1: 21 (1xo), hencea V (1 — ) – A + Ams is

.. n the friction on the inclined plane; which, by the ques.

tion, is to be the least possible ; therefore its fluxion * 2 rejicht maps or, (art. 773, ibid. when m = 1-and n= 3) moto. This result is different from that given at art. 19, p. 191, Young's Analysis, but that is,. in all probability, a press error.

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20. Qu. (75) Answered by Rylando. . Composition. From the wind D centre 0 of the given circle A E B, demit O D perpendicular to the line D P given , in position, and from the point D draw the tangents DE, DE; then through the points E, E, of contact draw the indefinite right line G E C, which will be that required. ,

Demonstration. From any point P in PD draw the tangents P BG, PCA meeting GEC in the points GC; it is to be proved that B G is equal CA. Draw the other lines as per figure, then if O P meet A B in I, and OD meet EF in F, we shall have by the wellknown properties of the figure O F .XOD= E O’ and OIX OPA O2 = 0 E2; therefore O EXODOLXOP; or 01:0E::OD:OP; thus it appears that the point F is in the right line AB; therefore 2 AFC=BFG, but the angles G BO, GFO are right angles, and therefore the points G, B, F, 0, are in a circle, wherefore 2 BFG= LBOG. In the same way exactly it may be shown that the angle AFC= LA O C, therefore a BOG=LAOC, and because the right angled triangles O BG, O AC have the BOG = L AOC and O B=AO, they are, therefore, equal in all respect, and consequently B G is equal to AC. 2. E. D.

Again, by Mr. Butterworth. . pt Construction. From the cen- o g sth tre 0 of the given circle, demit · QC, perp. to AB, the line

given in position; and from C /
draw the tangents CF, CL, G
and through the points F, L,
draw the indefinite line DE,
and it will be that required.
1. Demonstration;' From any R
point B, in A B, draw the tan
gents BH, BI, meeting DE 12
in D, and E; and take CA A

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equal to C B, and draw the tangents A G, AK; also let the tangent AG meet the tangent BH in D'; draw the radii O G, OH, and join O Do; also, join GK, and HI, and it is well-known that they will intersect each other, and the lines D E, and 0 C in the same point P. Then, because AC and C B are equal, the arcs HG and K I will be equal, therefore the angle HPG = HOG, consequently the circumference of a circle will pass through the points G, P, H, D'; and the angle GPD=ĚPI = HPD, 'therefore, the angle HPD =HO D'; consequently the points D, D' coincide in D. Again, the tangents DH, DG, EK, and E I, are evidently all equal, therefore HD=EI, 2. E. D. .

Again, by Mr. Whitley. From O, the centre of the given circle BG AF, demit on PE the straight line given in position, the perp. OE: draw the tangents EF, EG;, then. through the points of con- L tact G,F,draw the indefinite P right line CD, which will be the line required. * Demonstration. From any point P in PE, draw the tangents PA, PB, meeting Č Ď in the points D, C; then will AD be equal to B C. For drawing the lines as in the figure, let A B nieet O Pin I, and ČD meet OE in S; then the triangles OS F, OFE will be similar, as will also the triangles OIB,OB P; therefore OS:OF:: OF::0E, and 01:OB=OF::OB=OF:0P; hence by equality OI:OS::0E: OP; thus it evidently appears that the point S is in the straight line A B. Again: since the angles OSD, O AD are right angles, a circle, might pass through the points 0, S, A, D, and for the same reason a circle might pass through the points O, S, C, B; therefore the angle ODA is equal to the angle OSB, that is, equal to the angle OC B in the same segment; hence, since in the triangles OʻAD, O BC, the angle O D A is equal to the angle O C B, the angle O A D equal to O B C, each being a right angle, and the side o A equal to the side o B the triangles

then -7076

are identical; and therefore the segment A D is equal to BC. 2. E. D.

It was also answered by Mr. Gawthorp. Mr. Jesse Winward is requested to send for any book, the price of which does not exceed Half á Guinea.

* NEW MATHEMATICAL QUESTIONS

TO BE ANSWERED IN NO. IX.

1. Qu. (97) By Mr. Rt. Maar. · Extract the .6' root of .0005. '

. 2. Qu.(98) By the same. Find the value of 5x17m 6 places of decimals.

* 3. Qu. (99) By Mr. J. Baines, jun. At what time of the day, on the 9th of Dec. 1812, in lat. 59°, 42' N. will the sun's alt. above the horizon be equal to his azimuth from the south?

* 4. Qu. (100) By the same, The square of any prime number of the form 31n + 1, is of the form 3 a + b?; where b is a prime, and o a composite number: required proof.

5. Qu. (101) By Mr, W. Putsey, Pickering, It is proposed to divide the beam of a steelyard, or to find the points of division where the weights of 1, 2, 3, 4, &c. tbs, on one side, will just balance a constant weight of 95 lbs. at the distance of two inches on the other side of the centre of suspension; the weight of the beam being 10 lbs. and its whole length 36 inches. (Hutton's Course.) .. - 6. Qu. (102) By Mr. Nesbit, Farnley. .

There is a field in form of a trapezoid, the parallel ends of which, perp. to one side (the base) are 9 and 15 chains, and the base 40 chains; divide it equally among three persons, by fences, parallel to the perpendicular sides. Les 7. Qu. (103) By Mr. Rt. Maar.

Every thing being as in question 8 (63), to determine when and where the bodies will both be in a right line parallel to the horizon.

'. 8. Qui (104) By Mr. Gawthorp. ** Determine a point in the side of a cone, the perp. alt. and base of which are respectively 3 and 2 feet, to which if a string be fastened, and the cone suspended by it, the axis of the cone may make an angle of 25° 15' with a plane parallel to the horizon..

9. Qu. (105) By Mr. Nesbit. If the greatest altitude of the piston of a common sucking pump above the surface of the water be 18, and the least alt. 15 feet ; how high will the water rise in the pump by 3 strokes of the piston, admitting the height of a column of water, equivalent to the pressure of the atmosphere, to be 32 feet?' . 10. Qu. (106) By Mr. J. Winward, R. L. Militia,

. . Plymouth., · Determine the area of the curve, whose equation is y = (as + x^)i + (a2 + ax + x)}. . 11. Qu. (107) By Mr. J. Baines, jun. Horbury Bridge.

Reduce a given triangle to a trapezium, having the same area and perimeter. .

12. Qu. (108) By Mr. J. Whitley, Masbro'. Determine a triangle whose three sides shall be integers, in arithmetical progression, such, that the area and the diameters of the inscribed and circumscribed circles, shall be all integers. i n pr .

13. Qu. (109) By the same. Let ABC be a plane triangle, and let A F, BE, be tangents to the circumscribing circle at A,B; draw CEF perp. to A B, cutting A F, BE, in F, E; the square of AC is to the square of BC, as CF to CE; required a demonstration. i' en .... ....*14. Qui (110) By the same. ..!

Find p and g in general terms, so that p? t'q + 2 and p? 92 - 1' may be squares.

- 15. Qu. (111) By the same... Find expressions for the principal diameters of the maximum ellipse inscribed in a given sector of a circle ;

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