15. State the principle of the lever, and prove it when P and W act on opposite sides of the fulcrum. A weight of 5 lbs. is hung at one end of a uniform bar, which is balanced over a knife edge at a point 14 inches from the end at which the weight hangs. The bar weighs 30 lbs. ; find its length. (S. and A. Exam. 1894.) Ans. 32 inches. 16. State the principle of the lever. A uniform straight bar, 14 inches long, weighs 4 lbs. ; it is used as a lever, and an 8 lb. weight is suspended at one end. Find the position of the fulcrum when there is equilibrium. (S. and A. Exam. 1895.) Ans. 2 inches from 8 lb. weight. LECTURE IV. CONTENTS.-Practical Applications of the Lever-The Steelyard, or Roman Balance-Graduation of the Steelyard-The Lever Safety ValveExample I.-Lever Machine for Testing Tensile Strength of Materials -Straight Levers acted on by Inclined Forces-Bent Levers - The Bell Crank Lever-Bent Lever Balance-Duplex Bent Lever, or Lumberer's Tongs-Turkus, or Pincers-Examples II. and III.-Toggle Joints-Questions. In this Lecture we shall give a number of examples of the application of the lever. The Steelyard, or Roman Balance, is a straight lever with unequal arms, having a movable or sliding weight on the longer arm. It is very much used by butchers for weighing the carcasses of cattle and sheep, and in such cases it generally has two fulcra and two scales of division corresponding to them, the one set being, say, for hundredweights and the other for pounds. Graduation of the Steelyard.—The practical method of graduating the steelyard is to put unit weight (say 1 lb.) into the scale pan, SP (or attach it to the hook on the shorter arm if there should be no such pan), and mark the position where the sliding weight, SW, has to be placed in order to cause equilibrum. Mark this position I on the scale. Then put in two units (say 2 lbs.) into SP, and adjust SW as before, marking its new position as 2 on the scale; and so on until SW is at the end of the longer arm. In this form of steelyard, if the differences of the weights W, corresponding to successive distances. I to 2, to 3, &c., be the same, the graduations will be equal to each other. This may be proved in the following manner :-First of all, it is clear that the instrument can be so constructed that the centre of gravity of the beam and scale pan may occupy one or other of three different positions. The centre of gravity may coincide with F, or it may be on the longer arm, or it may be on the shorter arm. Suppose the centre of gravity to coincide with F, the fulcrum. Let the scale pan be loaded to the extent of W units, and suppose that the sliding weight of P units has to be placed at A in order to keep the beam horizontal. Increase W by one unit, and to restore equilibrium, let P be placed at A. Then, for equilibrium we must have Subtracting corresponding members of equations (1) and (2) P(FA1 - FA) = (W+1 ~ W) × FB, we get Increase W by n units, and let P occupy the position A,. Then, for equilibrium, we must have PX FA, (W + n) x FB = (4)· As before, subtract the corresponding members of (1) and (4). when we get PxAA,= = n× FB, FB AA, = n xp ' =nx AA by equation (3) Thus we see that the graduations are all equal for equal increments of W. The student will readily observe that the zero of the scale is at F, and by putting W = 1 in equation (1) we can fix the position of the first number on the scale Next, suppose the centre of gravity to lie in the longer arm at G. Let w = LONGER ARM AT G. weight of beam and scale pan, and suppose P at A and wat G to balance W units at B. Then, for equilibrium, we have PX FA+ w × FG = W × FB (5) As before, increase W by one unit, and let P be shifted to A, in order to restore equilibrium, then we must have PX FA,+wx FG = (W + 1) × FB Subtracting (5) from (6) we get Or, P x AA, = FB, (6) (7) Now increase W by n units, and let P occupy the position A,, then P× FA,+ w× FG = (W+n) × FB Subtracting (5) from (8) we get |