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Chapter 2

Q. 2.8

The flat plate shown in  Fig. (a) is acted on by the three couples. Replace the three couples with (1) a couple-vector; (2) two forces, one acting along the line OP and the other acting at point A; and (3) the smallest pair of forces, with one force acting at point O and the other at point A.

The flat plate shown in Fig. (a) is acted on by the three couples. Replace the three couples with (1) a couple-vector; (2) two forces, one acting along the line OP and the other acting at point A; and (3) the smallest pair of forces, with one force acting at point O and the other at point A.

Step-by-Step

Verified Solution

Part 1
The magnitudes (Fd) and senses of the couples, all of which lie in the xy-plane, are listed below.
• Couple at H:350\:lb\cdot in. clockwise.
• Couple acting on GE:(150)(3)=450\:lb\cdot in. counterclockwise.
• Couple acting on DB:(60)(5)=300\:lb\cdot in. 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

\underset{+}{\large\curvearrowleft}\:\:C^{R}=−350+450−300=−200\:lb\cdot in.

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 C^R=−200k\:lb\cdot in.

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-lb forces (5\:in.) and the 150-lb forces (3\:in.).

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. From Fig. (b), we see that the perpendicular distance d between the lines of action of the two forces is d=8\sin30^\circ=4\:in. Having already determined that the magnitude of the resultant couple is 200\:lb\cdot in., the magnitudes of the forces that form the couple are given by C^R/d=200/4=50\:lb. 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 (200\:lb\cdot in., 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=8\:in. Therefore, the magnitudes of the smallest forces are given by C^R/d=200/8=25\:lb. These results are shown in Fig. (c), where again note should be taken of the senses of the forces.

The flat plate shown in Fig. (a) is acted on by the three couples. Replace the three couples with (1) a couple-vector; (2) two forces, one acting along the line OP and the other acting at point A; and (3) the smallest pair of forces, with one force acting at point O and the other at point A.
The flat plate shown in Fig. (a) is acted on by the three couples. Replace the three couples with (1) a couple-vector; (2) two forces, one acting along the line OP and the other acting at point A; and (3) the smallest pair of forces, with one force acting at point O and the other at point A.