Problem 44. Two forces, each equal 10 lb., act in a vertical plane so as to form a positive couple. The distance between the forces is 2 ft.; another couple whose moment is equal to 20 in.-lb. acts in a horizontal plane and is negative. Required the resultant couple, its plane, and direction of rotation.

Problem 45. A couple whose moment is 10 ft.-lb. acts in the xy-plane; another couple whose moment is -30 in.-lb. acts in the xz-plane, and another couple whose moment is - 25 ft. in-lb. acts in the yz-plane. Required the amount, direction, and location of the resultant couple that will hold these couples in equilibrium. (x, y, and z-axes are at right angles with each other in this case.)

a case. Let the forces be P1, P2, P3, P4, etc., as shown in

Fig. 51, and let them have the directions shown. For the sake of analysis, introduce at the origin two equal and opposite forces P1, parallel to P1, two equal and oppo

site forces P2, parallel to P2, and so on for each force. The introduction of these equal and opposite forces at the origin cannot change the state of motion of the rigid body.

1

The original force P1 taken with one of the forces P introduced at the origin, forms a couple whose moment is Pid. The same is also true of the forces P, P

P4

etc., giving respectively moments Pad2, Pgdg, P4d4, etc. In addition to these couples there is a system of concurring forces at the origin P1, P2, P3, P4, etc. The resultant of this system is, as has been shown (Art. 16), R = √(Σx)2 + (Σy)2.

The moments Pid1, P2d2, Pd, etc., being all in one plane, may be added algebraically (see Art. 32), giving the moment of the resultant couple as ΣPd.

The system of non-concurrent forces in a plane may be reduced, then, to a single force R at the origin (arbitrarily selected) and a single couple whose plane is the plane of the forces.

=

For equilibrium, R=0 and ΣPd = 0, or Σx 0, Zy = 0, and XPd=0; that is, for equilibrium, the sum of the components of the forces along each of the two axes is zero and the sum of the moments with respect to any point in the plane is zero.

Considering as a special case the case where the forces are concurring, it is seen that Pd is always zero (see Art. 16). The case of a system of parallel forces in a plane may also be considered as a special case of the above. (Art. 20 and Art. 31).

Problem 46. The following forces act upon a rigid body: a

force of 100 lb. whose line of action makes an angle of 45° with the horizontal, and whose distance from an arbitrarily selected origin is

distance from the origin is 6 ft. Find the resultant force and the resultant couple.

Problem 47. It is required to find the stress in the members AB, BC, CD, and CE of the bridge truss shown in Fig. 52.

NOTE. The member AB is the member between A and B, the member CD is the member between C and D, etc. This is a type of Warren bridge truss. All pieces (members) are pin-connected so that only two forces act on each member. The members are, therefore, under simple tension or compression; that is, in each member the forces act along the piece. Usually, in such cases there are no loads on the upper pins.

SOLUTION OF PROBLEM. The reactions of the supports are found by considering all the external forces acting on the truss. Taking moments about the left support, we get the reaction at the right support, equal 4500 lb. Summing the vertical forces or taking moments about the right-hand support, the reaction at the left-hand support is found to be 3500 lb.

T

Cutting the truss along xy and putting in the forces exerted by the left-hand portion, consider the right-hand portion (see Fig. 53). The forces C and T act along the pieces, forming a system of concurring forces. For equilibrium, then, Στ O and y = 0, giving two equations, sufficient to determine the unknowns C and T. The forces in the members CD and CE may now be considered known.

=

4500 LBS.

FIG. 53

This

2T

Cutting the truss along the line ZW and putting in the forces exerted by the remaining portion of the truss, we have the portion represented in Fig. 54. gives a system of concurring forces of which C and 2 T are known, so that from the equations x 0 and Σy = 0 the remaining forces d and e may be found.

Problem 48. Find the stress

in each of the members AB and

d.

FIG. 54

BC of the simple roof truss shown in Fig. 55. The truss is pin-connected, and all the members are under simple tension or compression ex

piece. The simplest method of solution for such a case is to take the whole member in question and consider all of the forces acting.

Problem 49. In the crane shown in Fig. 57 (a) find the forces acting on the pins and the tension in the tie AC. The method of cutting cannot be used in this case since the vertical and horizontal