Question 2.8: The flat plate shown in Fig. (a) is acted on by the three co...

Part 1

The magnitudes (Fd) and senses of the couples, all of which lie in the xy-plane, are listed below.

• Couple at H: 10500N · mm clockwise.

• Couple acting on GE: (150)(90)=13500 N · mm counterclockwise.

• Couple acting on DB: (60)(150)=9000 N · mm clockwise.

Because all three couples lie in the same plane, they can be added algebraically, their sum being the resultant couple C^{R}. Choosing the counterclockwise sense as positive, we get

The negative sign shows that the sense of C^{R} is clockwise. Therefore, the corresponding couple-vector C^{R} is, according to the right-hand rule, in the negative z-direction. It follows that

Note that more dimensions are given in Fig. (a) than are needed for the solution. The only relevant dimensions are the distances between the 60-N forces (150 mm) and the 150-N forces (90 mm).

Part 2

Two forces that are equivalent to the three couples shown in Fig. (a) must, of course, form a couple. The problem states that one of the forces acts along the line OP and the other acts at point A. Because the two forces that form a couple must have parallel lines of action, the line of action of the force at point A must also be parallel to OP. FromFig. (b), we see that the perpendicular distance d between the lines of action of the two forces is d=240\sin 30^{\circ }=120mm.Having already determined that the magnitude of the resultant couple is 6000 N·mm, the magnitudes of the forces that form the couple are given by C^{R}/d=6000/120=50N. The sense of each force must be consistent with the clockwise sense of C_{R}. The final result is shown in Fig. (b).

Part 3

Here we are to determine the smallest two forces acting at points O and A that are equivalent to the three couples shown in Fig. (a). Therefore, the two forces to be determined must form a couple that is equivalent to the resultant couple (6000 N · mm, clockwise).The magnitude of a couple (Fd) equals the product of the magnitude of the forces that form the couple (F) and the perpendicular distance (d) between the forces. For a couple of given magnitude, the smallest forces will be obtained when the perpendicular distance d is as large as possible. From Fig. (b) it can be seen that for forces acting at points O and A, the largest d will correspond to θ =90^{\circ}, giving d =240 mm. Therefore, the magnitudes of the smallest forces are given by C^{R}/d =6000/240=25 N. These results are shown in Fig. (c), where again note should be taken of the senses of the forces.