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## 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 $A$.

## 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.