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Calculate the moment of force F about the axis AC, given the coordinates of A, B, C and the force vector F.

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Answer

{4.8i+6.4j} lb⋅ft\{4.8\mathbf{i} + 6.4\mathbf{j}\}\text{ lb}\cdot\text{ft}

Step-by-step solution

6 steps

To determine the moment of the force F\mathbf{F} about the axis extending between points A and C, we will follow these steps:

  1. Identify the coordinates of points A, B, and C.
  2. Define the force vector F\mathbf{F}.
  3. Calculate the position vector from a point on the axis (A) to the point where the force is applied (B).
  4. Calculate the moment of the force about point A, MA=rAB×F\mathbf{M}_A = \mathbf{r}_{AB} \times \mathbf{F}.
  5. Determine the unit vector along the axis AC.
  6. Project the moment vector MA\mathbf{M}_A onto the axis AC to find the moment about the axis as a Cartesian vector.

  1. Identify the coordinates of points A, B, and C.

    From the diagram, we can determine the coordinates:
    Point A is at the origin: A=(0,0,0)A = (0, 0, 0) ft.
    Point C is located by moving 4 ft along the y-axis and 3 ft along the x-axis: C=(3,4,0)C = (3, 4, 0) ft.
    Point B is located by moving 4 ft along the y-axis, 3 ft along the x-axis, and 2 ft down along the z-axis: B=(3,4,−2)B = (3, 4, -2) ft.

  2. Define the force vector F\mathbf{F}.

    The force vector is given as:

    F={4i+12j−3k} lb\mathbf{F} = \{4\mathbf{i} + 12\mathbf{j} - 3\mathbf{k}\}\text{ lb}

  3. Calculate the position vector from A to B.

    rAB=B−A=(3−0)i+(4−0)j+(−2−0)k={3i+4j−2k} ft\mathbf{r}_{AB} = B - A = (3-0)\mathbf{i} + (4-0)\mathbf{j} + (-2-0)\mathbf{k} = \{3\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\}\text{ ft}

  4. Calculate the moment of the force about point A.

    MA=rAB×F\mathbf{M}_A = \mathbf{r}_{AB} \times \mathbf{F} MA=∣ijk34−2412−3∣\mathbf{M}_A = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & -2 \\ 4 & 12 & -3 \end{vmatrix} MA=i((4)(−3)−(−2)(12))−j((3)(−3)−(−2)(4))+k((3)(12)−(4)(4))\mathbf{M}_A = \mathbf{i}((4)(-3) - (-2)(12)) - \mathbf{j}((3)(-3) - (-2)(4)) + \mathbf{k}((3)(12) - (4)(4)) MA=i(−12+24)−j(−9+8)+k(36−16)\mathbf{M}_A = \mathbf{i}(-12 + 24) - \mathbf{j}(-9 + 8) + \mathbf{k}(36 - 16) MA={12i+j+20k} lb⋅ft\mathbf{M}_A = \{12\mathbf{i} + \mathbf{j} + 20\mathbf{k}\}\text{ lb}\cdot\text{ft}

  5. Determine the unit vector along the axis AC.

    First, find the vector rAC\mathbf{r}_{AC}:

    rAC=C−A=(3−0)i+(4−0)j+(0−0)k={3i+4j} ft\mathbf{r}_{AC} = C - A = (3-0)\mathbf{i} + (4-0)\mathbf{j} + (0-0)\mathbf{k} = \{3\mathbf{i} + 4\mathbf{j}\}\text{ ft}

    Next, find the magnitude of rAC\mathbf{r}_{AC}:

    ∣rAC∣=32+42+02=9+16=25=5 ft|\mathbf{r}_{AC}| = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ ft}

    Now, find the unit vector uAC\mathbf{u}_{AC}:

    uAC=rAC∣rAC∣=3i+4j5={0.6i+0.8j}\mathbf{u}_{AC} = \frac{\mathbf{r}_{AC}}{|\mathbf{r}_{AC}|} = \frac{3\mathbf{i} + 4\mathbf{j}}{5} = \{0.6\mathbf{i} + 0.8\mathbf{j}\}

  6. Project the moment vector MA\mathbf{M}_A onto the axis AC.

    The moment about the axis AC, expressed as a Cartesian vector, is given by MAC=(uAC⋅MA)uAC\mathbf{M}_{AC} = (\mathbf{u}_{AC} \cdot \mathbf{M}_A) \mathbf{u}_{AC}.
    First, calculate the scalar component of the moment along the axis:

    MAC=uAC⋅MA=(0.6i+0.8j+0k)⋅(12i+j+20k)M_{AC} = \mathbf{u}_{AC} \cdot \mathbf{M}_A = (0.6\mathbf{i} + 0.8\mathbf{j} + 0\mathbf{k}) \cdot (12\mathbf{i} + \mathbf{j} + 20\mathbf{k}) MAC=(0.6)(12)+(0.8)(1)+(0)(20)M_{AC} = (0.6)(12) + (0.8)(1) + (0)(20) MAC=7.2+0.8+0=8 lb⋅ftM_{AC} = 7.2 + 0.8 + 0 = 8\text{ lb}\cdot\text{ft}

    Now, multiply this scalar component by the unit vector uAC\mathbf{u}_{AC} to get the Cartesian vector:

    MAC=MACuAC=8(0.6i+0.8j)\mathbf{M}_{AC} = M_{AC} \mathbf{u}_{AC} = 8(0.6\mathbf{i} + 0.8\mathbf{j}) MAC={4.8i+6.4j} lb⋅ft\mathbf{M}_{AC} = \{4.8\mathbf{i} + 6.4\mathbf{j}\}\text{ lb}\cdot\text{ft}

    The final answer is {4.8i+6.4j} lb⋅ft\{4.8\mathbf{i} + 6.4\mathbf{j}\}\text{ lb}\cdot\text{ft}.

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