1. Fig. 11 represents a body weighing 800 pounds sus pended from a ring which is supported by two ropes as shown. Compute the pulls on the ropes.
Ans. Pull in the horizontal rope = 400 pounds. Pull in the inclined rope = 894 pounds.
2. Suppose that in Fig. 12 the rope supporting the body on the plane is so fastened that it is horizontal. Determine the pull on the rope and the pressure on the plane if the inclination of the plane to the horizontal is 30 degrees and the body weighs 120 pounds.
Ans. 5 Pull68.7 pounds.
Pressure = 138 pounds.
3. A sphere weighing 400 pounds rests in a V-shaped trough, the sides of which are inclined at 60 degrees with the horizontal. Compute the pressures on the sphere.
Ans. 400 pounds.
' 17. Algebraic Conditions of Equilibrium. Imagine each one of the forces of a concurrent system in equilibrium replaced by its components along two lines at right angles to each other, horizontal and vertical for example, through the point of concur rence. Evidently the system of components would also be in equilibrium. Now since the components act along one of two lines (horizontal or vertical), all the components along each line must balance among themselves for if either set of components were not balanced, the body would be moved along that line. Hence we state that the conditions of equilibrium of a system of concurrent forces are that the resultants of the two sets of com ponents of the forces along any two lines at right angles to each other must equal zero.
If the components acting in the same direction along either of the two lines be given the plus sign and those acting in the other direction, the negative sign, then it follows from the foregoing that the condition of equilibrium for a concurrent system is that the algebraic sums of the components of the forces along each f two lines at right angles to each other must equal zero.
Examples. 1. It is required to determine the pull on the rope and the pressure on the plane in Example 1, Art. 16 (Fig. 12), it being given that the inclination of the plane to the horizontal is 30 degrees.
Let us denote the pull of the rope by and the pressure of the plane by F2. The angles which these forces make the horizon tal are 30' and 60° respectively; hence the horizontal component of = X cos 30' = 0.8660
and " " " " F2 = F2 X cos 60° = 0.5000 F2; also " " " " the weight = 0.
Tho angles which and F2 make with the vertical are 60° and 30° respectively, hence the vertical component of = X cos 60° = 0.5000 and the vertical component of F2 = F2 X cos 30° = 0.8660 F2; also the vertical component of the weight 120. Since the three forces are in equilibrium, the horizontal and the vertical components are balanced, and hence 0.866 F, = 0.5 and 0.5 + 0.866 F, = 120.
From these two equations and may be determined; thus from the first, 0.866 F2 _ = 1.732 Substituting this value of F2 in the second equation we have 0.5 + 0.866 X 1.732 = 120, or 2 F, = 120; 120 hence, = = 60 pounds, and F2 = 1.732 X 60 = 103.92 pounds.
2. It is required to determine the pulls in the ropes of Fig. 13 by the algebraic method, it being given that the angles which the left- and right-hand ropes make with the ceiling are 30 and 70 degrees respectively.
Let us denote the pulls in the right- and left-hand ropes by and F2 respectively. Then the horizontal component of = X cos 70° = 0.342 F1, the horizontal component of F2 = F2 X cos 30° = 0.866 F2, the horizontal component of the weight = 0, the vertical component of = X CO3 20° = 0.9397 the vertical component of F2 = F2 X cos 60° = 0.500 F27 and the vertical component of the weight = 100.
Now since these three forces are in equilibrium, the horizontal and the vertical components balance; hence 0.342 = 0.866 F, and 0.9397 + 0.5 F, = 100.
These equations may be solved for the unknown forces; thus from the first, 0.866 - 2.532 F,.
0.342 Substituting this value of in the second equation, we get 0.9397 X 2.532 F, 0.5 F, = 100,or, 2.88 100 ; 100 hence = 2.88 = 34.72 pounds, and 2.532 X 34.72 = 87.91 pounds.
1. Solve Ex. 1, Art. 16 algebraically. (First determine the angle which the inclined rope makes with the horizontal; you should find it to be 63° 26'.) 2. Solve Ex. 2, Art. 16 algebraically.
3. Solve Ex. 3, Art. 16 algebraically.