An important example of vector multiplication is the concept of moment of a force with respect to a point in space. The moment Mo is a measure of the capacity of force F acting on point P to cause rotation about point O. It is defined as follows:
where rp/o is the position vector connecting point O to P (Fig. 3.6b). The magnitude of Mo is the product of the magnitude of the force and the perpendicular distance of the point O from the line of action of the force. Note that in the evaluation of the moment, the position vector from O to P can be replaced with any other that connects point O with a point on the line of action of force F (Fig. 3.6b). Note also that, if the distance between point O and the line of action of force F is zero, then this force creates no moment with respect to point O.
If two forces are equal in magnitude and opposite in direction, their sum is equal to zero. Nevertheless, these two forces exert a moment with respect to any point in space so long as they do not share the same line of action (Fig. 3.6c). Such a pair of forces form what is called a force couple. The moment created by the force couple about a point in the plane of the couple can be shown to be equal to the product of the perpendicular distance d between the line of action of the opposing forces and the mag nitude of one of the forces comprising the couple. The moment of a force couple about any point in its plane is the same.
Example 3.5. Abduction of Arms. An athlete whose arms are 66 cm long stands with his hands at the thighs holding 10-kg dumbbells. The athlete contracts his front, middle, and rear deltoids and pulls the weights up directly to the side (Fig. 3.7). He raises his arms to the full-flexed shoulder position with the weights above the elbow joint and higher than the shoulder level. Then he slowly lowers the weight to the starting position, and repeats the exercise. Compute the moment generated by the weight of the dumbbell at the shoulder when the arm makes 0°, 45°, and 90° with the vertical axis.
Solution: Let us draw a coordinate system E whose origin coincides with the center of the shoulder joint. The position of the dumbbell with respect to the origin is then given by the following equation:
where L denotes the length of the arm and 6 is the angle the arm makes with the vertical axis as shown in Fig. 3.7. The force exerted by the dumbbell on the athlete equals to the weight of the dumbbell if the exercise is done slowly. The moment this force generates with respect to the center of the shoulder is
M = (L sin 6 ex - L cos 6 e2) X (-M g e2) = -M g L sin 6 e3 (3.21)
Thus, the magnitude of M, ||M|| = 0, 45.8 N-m, and 64.7 N-m for 6 = 0°, 45°, and 90°, respectively.
Note that we could have computed the moment M without going through the vector product. From the definition of vector product, we know that the magnitude of r X F must be equal to the magnitude of F times the distance d from point O to the line of action of F. As an exercise, identify d for each of the cases considered.
Example 3.6. Bent-Knee Abdominal Crunch. Lie on back on floor, resting lower legs across bench with arms behind base of neck. Slowly curl head and upper torso up off floor in one even-paced movement. Slowly lower until almost touching head and torso and repeat (Fig. 3.8). Compute the moment created by the weight of the upper body on the pelvic joint at the beginning of the crunch where the torso is only slightly off the ground. For the athlete shown in the figure, the distance L between the pelvic joint and the center of mass of the upper body is 34 cm. The weight of the upper body is 25 kg.
Solution: We can find the answer to this question without utilizing vector mathematics. The magnitude of moment M must be equal to the moment e
Figure 3.8. Schematic diagram of bent-knee abdominal crunches.
e e arm (0.34 m) times the force (25 kg X 9.81 m/s2). Its direction, from the right-hand rule, is clockwise, or into the paper. Thus M = —83.4 e3 (N-m).
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