and the horizontal part of R1 or R, or, as it is called, the horizontal thrust when the beams form a roof, is equal to W' cot. a+cot. B which is the reaction in the horizontal beam AB, called the tiebeam. If the side AB were taken away, and its place supplied by a cord, the last expression would be the tension in the cord. If the points A and B rested on a smooth horizontal plane, we should have the pressure on the plane 77. PROP. A number (n) of beams connected by hinges form a frame-work in a vertical plane, having weights hung from the hinges: to find the position of equilibrium, and to shew that the horizontal reaction at each hinge is the same. Let ABCDEF be the frame-work, of which the extremities A and F are fixed points. Let AH and FH, a horizontal and vertical line, be given, as well as the lengths R b1, b2, bз, &c. of the beams. The beams being given, A we may suppose the weight to be applied at the ends, and W1, W2, W3, &c. to be the whole weights acting at the hinges, B, C, D, &c. respectively. Let a1, B1 be the angles which the beams AB, BC make with the vertical line at B; a2, B2, 3, B3, &c. the corresponding angles at C, D, &c. 'n Let R1, R2, R3, &c. . . . R be the reactions in the beams. commencing at A. We shall have to find n reactions, and 2(n-1) angles. From the geometrical relations we have - AH-b1sin. α-b2 sin. α-b3 sin.az-&c. . . .-b, sin. ẞn-1=0 (a) FH-b1 cos. a1-b2 cos. a2-b3 cos. a2-&c. ... - bn × Resolving horizontally and vertically at each hinge, we have, in equilibrium, at B, R1sin. — R2sin.ẞ1=0 R1cos. a1+R2 cos. B1- W1=0 (1) We have also ẞ1+α2 =π, B2+αz=π, &c.... ẞn-2+an-1=π, or n-2 equations. These equations being 3n-2 in number suffice to find the n tensions and 2n-2 angles. The expressions would have been identical with those for the funicular polygon if we had taken F in the horizontal line through A; and they lead to the same consequences. The first of each pair of equations (1), (2), &c. shews that the horizontal component of the reactions at each hinge is the same; and by the same process as in Article 75 we shew that it is the same at every hinge, and equal to CHAPTER X. ON FRICTION. WE have hitherto considered the surfaces on which bodies. pressed to be perfectly smooth, so that they offered no resistance to motion parallel to themselves, their only reaction being perpendicular. When rough surfaces are in contact, the motion, or tendency to motion, parallel to the surfaces, is affected by the roughness, and we call the effect friction. Experiments have been made to determine the laws of friction, which we may subdivide into rubbing friction, when one body rubs on the other, and rolling friction, when one rough surface rolls upon another: the former only will be considered here, under the term friction or statical friction. m F 78. If a body rest, as at A, upon a rough plane, BC, it is found that a force, within certain limits, may act upon it parallel to the plane without motion ensuing, as would be the case if the plane were smooth. The greatest force which can be so applied, B W A without the body moving, measures the friction. If W be a weight acting by a cord passing over a pulley, as in the figure, on the body A, when motion is about to take place, F being the opposing force of friction which balances W, we have F=W. It is found that if we put various weights, as m, upon the body A, then W or F is proportional to the weight of A and m, or is proportional to the pressure perpendicular to the surfaces in contact. It is also found that it is independent of the magnitude of the surfaces in contact, the friction being the same when the pressure is the same, whether the surface of the body be increased or diminished; except in extreme cases, where the pressure is exceedingly great compared with the surfaces in contact. So that if R be the pressure on the plane which is equal and opposite to its normal reaction, we have F=μ R where μ is a constant, depending on the nature of the surfaces in contact, and called the coefficient of friction. We may determine the coefficient of friction by placing the body on a plane of which we can increase the inclination to the horizon until the body begins to slide down. R a W R 79. PROP. To shew that the coefficient of friction between two given substances is the tangent of the inclination of the plane formed of one of the substances, when the body formed of the other is about to slide down it. The body a in the above figure is in equilibrium from the normal reaction of the plane R, the friction μR acting up the plane, and its weight acting vertically. Resolving parallel and perpendicular to the plane, we have Friction being considered as an inert force resisting the tend ency to motion, will act up or down the plane as the body is on the point of moving down or up respectively. Resolving parallel and perpendicular to the plane, we have P cos. e±μR-W sin. a = 0 P sin. e+R—W cos. α = 0 Multiply (2) by μ and subtract and add, we have P(cos. eμ sin. e) — W (sin. a μ cos. a) = 0 (1) (2) where the upper sign is to be taken when friction acts up, and the lower when friction acts down the plane. No motion will occur whilst the relation of P to W lies between the two values. 81. PROP. To find the limits of the ratio of P to W in the screw when friction acts assisting the power or the weight. Proceeding as in Article 54, let ABC be the inclined plane formed by the unwrap- p ping of one revolution of the thread; the angle BAC = a. Let W be the whole weight sustained by the screw; w be the part of it supported at a; Q the whole force acting at the circumference of the cylinder, whose radius ED=r, FD= a the lever at which the power P acts, and w and let q be the part of Q which supports w at a. The forces which are in equilibrium at a are the weight w, the reaction R, the friction acting up or down the plane μR, and the horizontal pressure q. |