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Moment About a Point: 3D - Statics

  1. A wrench applies a force F⃗=20j^\vec{F} = 20\hat{j} N at point P(0, 0.3, 0) m on a bolt. If the moment about the bolt axis (z-axis) must not exceed 4 N⋅m in magnitude, what is the maximum distance the point of application can be moved in the x-direction while maintaining the same force direction?
    Answer: 0.2 m The moment about the z-axis is given by Mz=r⃗×F⃗⋅k^M_z = \vec{r} \times \vec{F} \cdot \hat{k}. If the force is moved to point (x, 0.3, 0), then r⃗=xi^+0.3j^\vec{r} = x\hat{i} + 0.3\hat{j} and F⃗=20j^\vec{F} = 20\hat{j}. The cross product is r⃗×F⃗=(xi^+0.3j^)×(20j^)=x(20)(i^×j^)=20xk^\vec{r} \times \vec{F} = (x\hat{i} + 0.3\hat{j}) \times (20\hat{j}) = x(20)(\hat{i} \times \hat{j}) = 20x\hat{k}. The z-component is Mz=20xM_z = 20x. For ∣Mz∣≤4|M_z| \leq 4, we need ∣20x∣≤4|20x| \leq 4, so ∣x∣≤0.2|x| \leq 0.2. The maximum distance in the x-direction is 0.2 m. Choice B incorrectly uses the force magnitude instead of the moment arm. Choice C assumes the constraint applies to the total moment magnitude. Choice D neglects the force magnitude in the calculation.
  2. A force F⃗=20i^+30j^−15k^\vec{F} = 20\hat{i} + 30\hat{j} - 15\hat{k} N is applied at point P(4, -2, 3) m. What is the magnitude of the moment of this force about point O(1, 2, -1) m?
    Answer: 125.6125.6 N⋅m When you encounter a moment calculation problem in 3D space, you're dealing with the cross product of a position vector and a force vector. The moment M⃗=r⃗×F⃗\vec{M} = \vec{r} \times \vec{F}, where r⃗\vec{r} is the position vector from the reference point to the point of force application. First, find the position vector from O to P: r⃗=(4−1)i^+(−2−2)j^+(3−(−1))k^=3i^−4j^+4k^\vec{r} = (4-1)\hat{i} + (-2-2)\hat{j} + (3-(-1))\hat{k} = 3\hat{i} - 4\hat{j} + 4\hat{k} m. Now calculate the cross product M⃗=r⃗×F⃗\vec{M} = \vec{r} \times \vec{F}: M⃗=∣i^j^k^3−442030−15∣\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 4 \\ 20 & 30 & -15 \end{vmatrix} =i^[(−4)(−15)−(4)(30)]−j^[(3)(−15)−(4)(20)]+k^[(3)(30)−(−4)(20)]= \hat{i}[(-4)(-15) - (4)(30)] - \hat{j}[(3)(-15) - (4)(20)] + \hat{k}[(3)(30) - (-4)(20)] =i^[60−120]−j^[−45−80]+k^[90+80]= \hat{i}[60 - 120] - \hat{j}[-45 - 80] + \hat{k}[90 + 80] =−60i^+125j^+170k^= -60\hat{i} + 125\hat{j} + 170\hat{k} N⋅m The magnitude is: ∣M⃗∣=(−60)2+(125)2+(170)2=3600+15625+28900=48125=125.6|\vec{M}| = \sqrt{(-60)^2 + (125)^2 + (170)^2} = \sqrt{3600 + 15625 + 28900} = \sqrt{48125} = 125.6 N⋅m. Answer (B) is correct. Answer (A) likely comes from incorrectly calculating just one component of the cross product. Answer (C) might result from sign errors in the cross product calculation. Answer (D) could come from arithmetic mistakes in the final magnitude calculation or incorrectly setting up the position vector. Always double-check your cross product components and remember that the position vector goes from the reference point TO the force application point.

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