1. A man steps on to an elevator, which thereupon descends with a uniform acceleration of 10 feet per second per second. What sensation will he experience, and calculate its amount?

2. A mass of 20 lbs. is placed upon a horizontal plane which is made to descend with a uniform acceleration of 30 feet per second per second. Find the pressure on the plane.

3. A balloon is ascending vertically with a velocity which is increasing at the rate of 3 feet per second per second; find the apparent weight of one pound weighed in the balloon by means of a spring balance.

4. Three ropes are tied together, and a man pulls at each. If, when their efforts are in equilibrium, the angle between the first and second rope is 90°, and that between the first and third is 150°, what are the relative strengths of the men as regards pulling?

5. A particle is acted on by a force whose magnitude is unknown, but whose direction makes an angle of 60° with the horizon; the horizontal component of the force is known to be 1:35 dynes. Determine the total force and also its vertical component.

6. Four forces of 24, 10, 16, 16 dynes act on a particle, the angle between the first and second being 30°, between the second and third 90°, and between the third and fourth 120°. Calculate the magnitude of their resultant.

7. Three forces, proportional to 1, 2, 3 respectively, act on a point; the angle between the first and second is 60°; the angle between the second and third is 30°. Find the angle which the resultant makes with the first.

8. On a smooth plane, rising 2 in 5, a weight of 10 lbs. is kept from sliding by a force in the direction of the plane. Determine the pressure on the plane.

9. Three cords are tied together at a point. One of these is pulled in a northerly direction with a force of 6 pounds, and another in an easterly direction with a force of 8 pounds. With what force must the third cord be pulled in order to keep the whole at rest?

10. A weight of 10 tons is hanging by a chain 20 feet long. Find how much the tension in the chain is increased by the weight being pulled out by a horizontal force to a distance of 12 feet from the vertical through the point of support.

11. A weight of 4 pounds is suspended by a string, and it is also acted on by a horizontal force. If, in the position of equilibrium, the tension of the string is 5 pounds, what is the horizontal force?

12. A mass of 10 lbs. is supported by strings of lengths 3 and 4 feet respectively, attached to two points in the ceiling 5 feet apart. What is the tension of each string?

13. Answer the above question in the case of equal strings, of such a length that the mass is only an inch from the ceiling.

14. A body rests on a horizontal plane, whose coefficient of friction is 1/2; at what inclination must a force equal to the weight of the body be applied so that it may be just on the point of moving the body?

15. Suppose the resistance of the air to the motion of a hailstone to be equal to one tenth of the weight, when the speed is 16 feet per second, and the increase to be as the square of the speed. What is the greatest speed the hailstone can acquire by falling?

16. A weight of 112 lbs. rests on a rough plane, inclined at 15° to the horizon, and the coefficient of friction is 75. Calculate the limiting forces which, applied horizontally, must be exceeded to give the weight upward, downward, and horizontal movement respectively on the plane.

SECTION XXIX.-DEFLECTING FORCE.

ART. 147.-Deflecting Force. The force required to produce a given curvature of the path of a body moving with a given speed is proportional to the mass of the body and to the necessary acceleration. The body has a tendency to move uniformly in a straight line, and this tendency, looked upon as a force urging it outwards from the curve, is called centrifugal force. According to Art. 121 the acceleration is given by the equivalence

1 L per T per T = (L per T)2 by (radian per L arc), hence the force is given by

1 FM by (L per T)2 by (radian per L arc).

Now, the units upon which F per M depends can be transformed into any form in which they are combined into units, provided the quality of each unit is preserved; thus

1F=M by (L arc per T)2 per L radius,

=M by (L arc per T) by (radian per T),

= M by (radian per T)" by L radius.

=

=

As 2 radian revolution, by substituting in the last, we get 42 FM by (revolution per T)2 by L radius.

In the case of the British system we have

1 poundal = lb. by (ft. arc per sec.)2 per ft. radius.

In the C.G.S. system

1 dyne = gm. by (cm. arc per sec.)2 per cm. radius.

ART. 148.--Simple Harmonic Motion. The acceleration on a particle describing a simple harmonic motion is given by the equivalence

1 L per T per T = (radian per T)2 by L displacement, therefore the force required is given by

1 F per M = (radian per T)2 by L displacement. In the motion of a simple pendulum, instead of L displacement we have L length of pendulum; hence

or

1 F per M = (radian per T)2 by L length,

4 F per M (revolution per T)" by L length.

=

EXAMPLES.

Ex. 1. If a stone weighing 2 lbs. be attached to a yard of string, which can just support 20 lbs., at what rate must it be whirled round horizontally so as to break the string?

42 poundal = lb. by (revolution per sec.) by ft. radius, 2 lbs. by 3 ft. radius,

Ex. 2. A toy-car, whose mass is lb., runs at the rate of 5 miles an hour on a level circular railway 20 feet in circumference ; calculate the horizontal pressure on the rails.

1 poundal = lb. by (ft. per sec.)2 per ft. radius,

Ex. 3. A railway train moves smoothly at the rate of 30 miles an hour over a curve of 500 yards radius; find the angle at which a plumb-line in one of the carriages will be inclined to the vertical.

1 poundal per lb. = (ft. per sec.)2 per ft. radius, per sec.)2 per ft. rad.,

i.e.,

500 × 3 × 32.2

0.4 ft. outwards per ft. downwards.

Hence the tangent of the angle is 04, and the angle about 6°.

Ex. 4. Find the maximum force in a case of simple harmonic motion in which the moving mass is a gramme, the range on each side of the middle position 10 centimetres, and the period second.

=

42 dyne gm. by (rev. per sec.)2 by cm. displacement,
1 gm. by 3002 (rev. per sec.)2 by 10 cm.,

Ex. 5. Find how many vibrations a simple pendulum, 4 inches long, would make in a minute at Glasgow.

42 poundals per pound = (rev. per sec.)2 by ft. length,

Ex. 6. The time of a complete vibration at Paris of a pendulum 6,400 centimetres long is 16 seconds; show that the value of g is 986 approximately.

1

162

42 dyne per gm. = (rev. per sec.)2 by cm. length,

(rev. per sec.)2 by 6,400 cm. length,

1. Calculate, in pound's weight, the tension of a string 4 feet long which has a mass of 10 pounds attached to it, describing a horizontal circle once in half a second.