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INTRODUCTION TO ARITHMETIC.

PART XII.

CONTAINING,

MECHANICAL POWERS, EXERCISING QUESTIONS, A SECRET FOR TEACHERS, AND AN APPENDIX ON BOOK-KEEPING.

MECHANICAL POWERS.

OF THE LEVER.

To find what weight may be raised by any given power.

RULE 1.

As the distance between the weight to be raised and

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is to the distance from the fulcrum to the end of the

lever where the power is applied;

So is the power applied,

to the weight it will raise.

EXPLANATION.

As the shorter distance,

is to the longer;

So is the lesser weight,

to the greater.

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We will suppose A to be the fulcrum of a lever 12 feet long: O, the weight to be raised, and P, the power applied.

* Fulcrum, is the point of suspension on which the lever bears. On steelyards, it is the short crossbar turning in the upper hook when weighing

LESSON 1.
CASE 1.

If 140 pounds rest on the end of a lever 12 feet long, how many pounds will the said 140 balance at the other end, allowing the fulcrum to be placed 18 inches from the weight to be raised as at A? Answer, 980 pounds.

OPERATION.

12 feet the length of the lever.

1.5 18 inches from the fulcrum to the weight.

=

10.5 feet from the fulcrum to the end of the lever where the power is applied.

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980)1470.0(1.5 ft. Ans. 980

4900

4900

CASE 3.

If we have an object of 980 [raised. pounds to raise with a lever 12 feet long, and the fulcrum is placed 18 inches from the foot of the lever, so that we have 10.5 feet for the long preponderating part, how many pounds must be applied to the long end to balance the said 980lb.? Ans. 140.

CASE 2. If 140lb. be placed on the end of a lever 10.5 feet from the fulcrum, and at the same time balance 980lb. at the

OPERATION.

other end, what is the length greater. lesser. greater.

of the lever from the fulcrum

to the weight so balanced?

Ans. 18 inches.

OPERATION.

greater. lesser. greater.

10.5 : 1.5 :: 980 to a 4th 980 [number

10.5)14700(140lb. to be

120

135

980; 140: 10.5 to a fourth

105

[applied.

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140 [number

RULE 2.

Divide the distance from the fulcrum to the power applied by the distance from the fulcrum to the weight raised; then multiply that quotient by the power applied, the product will be the weight raised.

EXAMPLE

10.5 ft. distance from the fulcrum to the power applied. 1.5)10.5(7 the quotient.

105

140 the power applied.

7 the quotient for a multiplier.

980 the weight raised or balanced.

OF THE WHEEL AND AXLE.

How to find a proportion in raising a greater weight suspended from the Axle than that applied to the wheel.

RULES.

1. As the diameter of the axle,

is to the diameter of the wheel;

So is the power applied to the rim of the wheel; to the weight raised by the axle.

2. As the lesser diameter,

is to the greater diameter;

So is the lesser weight applied on the wheel,

to the greater weight suspended from the axle.

And to the contrary :

3. As the greater diameter,

is to the lesser;

So is the greater power suspended from the axle, to the lesser applied to the wheel.

CASE 1.

Suppose the diameter of the axle be 4 inches, the diameter of the wheel 48 inches, and 2 pounds are applied to the rim; how many pounds suspended from the axle will balance those 2 pounds?

Answer 24.

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