25. A plane is inclined at an angle ẞ to the horizon; a particle is projected from a point in the plane at an inclination a to the horizon, with the velocity u, and the particle rebounds from the plane: find the time of describing n parabolic arcs.

26. In the preceding Example find the condition which must hold in order that after describing n parabolic arcs the particle should be again at the starting point.

27. A particle is projected with a given velocity at a given inclination to the horizon from a point in an inclined plane: find the whole time which elapses before the particle ceases to hop.

28. In the preceding Example find the condition which must hold in order that the particle may cease to hop just as it is again at the starting point.

29. In Example 25 find the cotangent of the inclination to the plane of the direction of motion of the particle at the beginning of the nth arc.

30. Shew that the time of descent to the lowest point of a very small circular arc is to the time of descent down its chord as the circumference of a circle is to four times its diameter.

III. 2. By Art. 34, forces 1, 1, 1 are in equilibrium
and may be omitted; thus the resultant is equivalent to
that of forces 1 and 2 at an angle of 120o. 4. See Art. 37.
5. 15 lbs., 20 lbs. 6. 4 lbs. 7. Let OA and OB denote
the equal forces, OD their resultant; produce AO to C so
that ŎC=20A; and let OE be the resultant of OB and
OC: then it is given that OE=OD. The resultant of OE
and OD is equivalent to that of twice OB and half OC,
and is therefore equal to OE. 8. It follows from 7 that
the angle EOD = 120°; and the straight line which bisects
the angle EOD must make with OB an angle equal to
EOC=the angle OED=the angle ODE. 12. The re-
sultant is 2√2 lbs., and it is parallel to a side of the
square. 13. The resultant coincides in direction with the
straight line from the point to the intersection of the
diagonals of the rectangle, and is equal to twice that straight
line. 14. Use the polygon of forces. 15. Use Ex. 14:
if n be the number of equal parts the resultant is repre-
sented by (n-1) times the radius.

IV. 1. The resultant is 9/2 lbs.: it is parallel to the

2

diagonal AC; and it crosses AD at the distance AD

from A. 2. 38 lbs. and 114 lbs.
the heavier weight. 8.
a is a side of the square.

9

3. 13 inches from

Pa~ Qb.

9.

Pa 2/2, where

V. 7. Take moments round H: thus we find that KI is parallel to BC. 8. Take moments round an end of one force: thus we find that the other two forces are bisected at 0.

VI. 1. (20) lbs. 5. The angle ACB is given; and since P, Q, and R are given, the angles which the direction of R makes with AC and CB are given. 6. See Art. 38, and Euclid, 11. 21, 22. S. The point must be at the intersection of the straight lines which join the middle points of opposite sides. 9. The forces 1 and

3 are at right angles; the forces 2 and 1 at 120o. 11. Let CD be the resultant of CA and CB. Let A come to a. Take Dd equal and parallel to Aa; then að is equal and parallel to CB. Thus Cd is the resultant of Ca and CB.

5 inches from the end at which the force of 4 lbs. acts. 4. At a distance from the centre of the hexagon equal to one-fifth of a side. 5. At the point at which the force

of 8 lbs. acts.

6. At the distance

1 n- -1

of the radius

7. 6 inches from the end.

12. The

from the centre.
force at the point A must be Q+R−P; and so on.

IX. 2. One foot from the end. 3. Suppose the straight line parallel to BC; let D be the middle point of BC: the

7

centre of gravity is on AD at the distance AD from A.

4. At a distance from the centre of the larger circle equal to one-sixth of the radius. 5. Equal forces.

tance from the base of the triangle equal to

altitude of the triangle. 17. At a distance from the

5

of the

18

base of the triangle equal to

3
8+2/3

of the base.

18. Put the rods so that the points in contact may be

of a foot from the middle of each, towards the 1lb. of the lower rod, and towards the 9 lbs. of the upper rod.

7

19. At of the whole length from the end. 20. 8,

18

81, 3 lbs.

X. 1. Three quarters of the square. 2. A straight line parallel to the base. 3. The centre of the spherical surface. 6. One is double the other. 10. Twelve inches. 11. The distance of the point from one end of the side must be twice its distance from the other end.

5

XI. 1. 1 to 3. 2. 52 inches from the end.
4. Two feet from the end.

7 lbs.

lbs.

lbs.,

5.

3. 1 Two inches.

6. 9 lbs., 6 lbs.: ratio that of 2 to 3. 7. 3 lbs. 8. 5 lbs., 9. The forces are 3 lbs. and 12 lbs. 10. 12 lbs., 11. One is double the other. 12. 2, 4 feet.

22 lbs.

5

√(12) lbs. 17. Pat a distance 14 feet from the fulcrum.

3

19. One inch from A; 10lbs. 20. 4 inches from the fulcrum.

XII. 2. 18 ounces. 3. 40 lbs. 4. He gets 15 ounces for 3s. 9d.; which is at the rate of 4s. per lb. 6. Seven inches from the point of suspension. 9. The point D, from which the graduations begin, is brought nearer to the point of suspension C. 10. The point D is taken further from C. 11. 2 feet from the end at which 10 lbs. is suspended; 60 lbs. 13. Pressure on Cis

half the weight of the rod, on D is rod. 14. 8 inches from the other

21. At

18. 20 lbs.

3

14

20. Two feet from the other end.

of an inch from the end of the lead bar. 22. 30 lbs.

5

24. At a distance of of the lever from the end where

the greater force acts.

12

XIII. 1. 56 inches. 2. 6 lbs. 3. The radius of the Wheel must be ten times the radius of the Axle. 4. 16 ounces. 5. The weight of 6 lbs. The prop must support lbs., leaving 5ğ lbs. on the Wheel to balance the

3 lbs. on the Axle. The pressure on the fixed supports is 85 lbs. 6. 15 cwt. 7. 18 inches; 3 inches. 8. 108 lbs. 9. The string which is nailed to the wheel hangs vertically so that its direction just touches the Axle. creased; see Algebra for Beginners, Art. 353.

XIV. 1. 3 lbs. 2. One lb. third of his weight. 4. As 12 to 1.

10. In

3. A force equal to a

5. 16 cwt. 6. 6.

7. The Weight will overcome the Power.

9

8. of his

8

weight, supposing him to pull upwards, as in Art. 196; but

7

8

of his weight if the Power end of the string passes over

10. One lb.

a fixed pully so that he pulls downwards. 9. 3 lbs. 11. W=P. 12. W= =w. 14. 3 lbs. 17. The Power will overcome the

15. 16.

Weight.

16. 6.

18. 7 cwt.

20. Three times the Power.

XV. 1. The perpendicular from the right angle on the length. 2. 7 lbs.