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Moment About an Axis: 3D - Statics

  1. A force system consists of F⃗=18i^+24j^−12k^\vec{F} = 18\hat{i} + 24\hat{j} - 12\hat{k} N applied at point A(4, 2, 3) m. What is the perpendicular distance from the y-axis to the line of action of the force?
    Answer: 4.8 m When you encounter problems asking for the perpendicular distance from a coordinate axis to a line of action, you're working with the concept of moment arms and the geometric relationship between forces and reference axes. To find the perpendicular distance from the y-axis to the force's line of action, you need to determine which components of the position vector and force vector matter. Since the y-axis runs along the j-direction, the perpendicular distance depends only on the x and z coordinates and force components. The perpendicular distance formula is: d=∣r⃗×F⃗∣y−component∣F⃗∣xz−planed = \frac{|\vec{r} \times \vec{F}|_{y-component}}{|\vec{F}|_{xz-plane}} From point A(4, 2, 3), the relevant position components are x = 4 m and z = 3 m. From F⃗=18i^+24j^−12k^\vec{F} = 18\hat{i} + 24\hat{j} - 12\hat{k}, the relevant force components are Fx=18F_x = 18 N and Fz=−12F_z = -12 N. The perpendicular distance is: d=∣x⋅Fz−z⋅Fx∣Fx2+Fz2=∣4(−12)−3(18)∣182+(−12)2=∣−48−54∣324+144=102468=10221.6=4.8d = \frac{|x \cdot F_z - z \cdot F_x|}{\sqrt{F_x^2 + F_z^2}} = \frac{|4(-12) - 3(18)|}{\sqrt{18^2 + (-12)^2}} = \frac{|-48 - 54|}{\sqrt{324 + 144}} = \frac{102}{\sqrt{468}} = \frac{102}{21.6} = 4.8 m Answer C (4.8 m) is correct. Answer A (2.4 m) likely comes from using only half the numerator. Answer B (3.6 m) might result from incorrectly using just the z-coordinate. Answer D (6.0 m) could stem from using the full 3D force magnitude instead of just the xz-plane components. Remember: for perpendicular distances to coordinate axes, only consider the components perpendicular to that axis in your calculations.
  2. A uniform rod of length 6 m has forces applied at its ends: F1⃗=20i^+15j^+25k^\vec{F_1} = 20\hat{i} + 15\hat{j} + 25\hat{k} N at one end and F2⃗=−10i^+30j^−15k^\vec{F_2} = -10\hat{i} + 30\hat{j} - 15\hat{k} N at the other end. If the rod is oriented along the line from origin to point (3, 4, 0) m, what is the moment about an axis parallel to the rod and passing through the midpoint of the rod?
    Answer: 0 N⋅m When analyzing moments about an axis in three-dimensional statics, you need to understand that only force components perpendicular to the axis of rotation can create moments. Forces parallel to the axis pass through it and produce zero moment. The rod extends from the origin to point (3, 4, 0) m, so its direction vector is u⃗=3i^+4j^32+42=0.6i^+0.8j^\vec{u} = \frac{3\hat{i} + 4\hat{j}}{\sqrt{3^2 + 4^2}} = 0.6\hat{i} + 0.8\hat{j}. The midpoint is at (1.5, 2, 0) m. To find the moment about an axis parallel to the rod, you must first determine what components of each force are perpendicular to the rod's direction. However, there's a more direct approach: since both forces act at points along the rod's axis, and we're calculating moments about an axis parallel to (and containing points on) this same rod, both forces effectively act along lines that either intersect or are parallel to our moment axis. For F1⃗\vec{F_1} at (3, 4, 0) and F2⃗\vec{F_2} at (0, 0, 0), the position vectors from the midpoint are (1.5, 2, 0) and (-1.5, -2, 0) respectively. When you calculate r1⃗×F1⃗+r2⃗×F2⃗\vec{r_1} \times \vec{F_1} + \vec{r_2} \times \vec{F_2} and project onto the rod's direction, the components cancel due to the symmetric positioning and the specific force distributions. The answer is A) 0 N⋅m. Options B, C, and D represent common calculation errors where students incorrectly include parallel force components or make sign errors in the cross products. Study tip: For moments about axes, always identify which force components are truly perpendicular to your axis of rotation—parallel components contribute nothing to rotational effects.

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