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300 difference. 100 Answer.

END OF PART ELEVENTH,

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

the fulcrum, * 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.
O

P

А We will suppose A to be the fulcrum of a lever 12 feet long : 0, 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.

980)1470.0(1.5 ft. Ans. lesser, greater. lesser.

980 1.5 : 10.5 ; : 140 to a 140 [fourth num

4900
[ber.

4900
420
105

CASE 3.
1.5)14.00(980 lb.weight If we have an object of 980
135

[rażsed. pounds to raise with a lever

12 feet long, and the fulcrum . 120

is placed 18 inches from the 120

foot of the lever, so that we

have 10.5 feet for the long ...0

preponderating part, how CASE 2. If 140lb. be placed on the to the long end to balance the

many pounds must be applied end of a lever 10.5 feet from said 980lb.? Ans. 140. the fulcrum, and at the same

OPERATION. time balance 980lb. at the other end, what is the length greater. lesser. greater. of the lever from the fulcrum

10,5 : 1.5 : : 980 to a 4th

980 to the weight so balanced ?

[number Ans. 18 inches.

120 OPERATION.

135 greater. lesser. greater.

10.5)14700(140lb. to be 980 : 140 :: 10.5 to a fourth

105 140 [number!

[applied 420

420 105

420 980)1470.0(1.5 ft.!

..0

[ocr errors]

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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