whose direction cosines are cos Ry, cos y =· = R Σπ R Since the moment arrows also follow the same laws as the force arrows, we may also write the moment of the resultant couple, where M=√M2+M2 + M2, M1 = P1d1 cos λ1 + P2d2 cos λ2 + etc., The direction angles of Mare λ, μ, v, and these are de- M eos λ = M M M This system of forces produces a translation of the body in the direction of Rand a rotation about a gravity axis parallel to M. If R=0 and M÷0, the body only rotates or is translated with uniform motion, and if M= 0 and R0, the body has only translation with possibly uniform rotation. For equilibrium both R=0 and M=0; that is, Σx= 0, Σy = 0, Σz=0, M ̧= 0, F Y FIG. 64 Mx X M1 = 0, and M2 =0, or expressed in words: the sum of the components of the forces along each of the three arbitrarily chosen axes is zero, and the sum of the moments with respect to each of these axes is zero. It may be further shown that the single force and resultant of the preceding article may be replaced by a single force and a couple whose plane of rotation is perpendicular to the line of action of the force. Suppose M and R both drawn at the origin and let a be the angle between them. M may be resolved into components along and perpendicular to R, G cos a along R, and G sin a perpendicular to R. G sin a may be replaced by another couple having the same moment. forces be R and + R and allow the the line of action of the resultant force. Let the R to act along The other force of the couple acts along a line parallel to the direction of the resultant force. The forces + R and R along the line of action of the resultant force neutralize each other, and we have left (a) a force, R, parallel to the original resultant, and (b) a couple, & cos a, acting in a plane perpendicular to R. That is, the system reduces to a single force and a single couple whose plane is perpendicular to the line of action of the force, or we may say, the effect of any system of forces acting on a rigid body, at any instant, is to cause an angular acceleration about the instantaneous axis of rotation and an acceleration of translation along that axis. Such a system of forces is called a screw wrench. The instantaneous axis is called the central axis. This axis passes through the center of gravity of the body if the body is free to rotate. Problem 59. A vertical shaft is acted upon by the belt T1 and T2, the crank pin pressures P, and the reactions of the supports. See Fig. 65. Write down the six equations for equilibrium. NOTE. The y-axis has been chosen parallel to the force P, and T1 and T2 are parallel to the x-axis. Σx = x2 + x" - T1 - T1 = 0, Σzz" - G - G' = 0; Mx == - Pb — y''l = 0, M1 = x'l — T ̧c — T2c - G1 = 0, M2 = Pa + T1r - T2r = 0. From these six equations six unknown quantities can be found. If G1, G2, T1, and T2 are known, the reaction of the supports and P may be found. T2 pressures α Problem 60. A crane shown in Fig. 66 has a boom 45 ft. long and a mast 30 ft. high. It is loaded with 20 tons, and the angle between the boom and mast is 45°. The two stiff legs each make angles of 30° with the mast and an angle of 90° with each other. Find the pin pressures in boom and mast, also the stress in FIG. 65 B FIG. 66 the legs when (a) the plane of the crane bisects the angle between the legs, and (b) the plane of the crane makes an angle of 30° with one of them. If the boom weighs 4000 lb., find the stress in the legs when the plane of the crane bisects the angle between them. Assume that the pulleys A and B are at the ends of the boom and mast respectively. Problem 61. Suppose the shaft of Problem 59 to be horizontal, find P and the reactions of the supports. Assume y horizontal and perpendicular to the shaft, and x vertical. CHAPTER VII MOMENT OF INERTIA P 37. Definition of Moment of Inertia. The study of many problems considered in mechanics brings to our attention the value of the integral of the form SydF, where dF represents an area and y the distance of the center of gravity of that area from an axis of reference. general definition of moment of inertia would be the prodA more uct of an elementary area, mass, or volume by the square of its distance from a designated point, line, or plane. The integral given above simply adds these products to give the moment of inertia of an entire area. In the case of a mass, the integral becomes SydM, and of volume Syd V. If the area, mass, or volume is not continuous throughout, the limits of integration must be properly taken to account for the discontinuity. moment of inertia by the letter I. We shall designate Thus we write: = Sy2dF, I= Sy2dM, I= Sy2dV, for area, mass, and volume, respectively. In finding the moment of inertia of several disconnected parts, it is often 69 |