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. RULE 1. the fulcrum, * is to the distance from the fulcrum to the end of the lever where the power is applied ; EXPLANATION. is to the longer; 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 4900 CASE 3. [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 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. is to the diameter 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. |