To solve this problem, we need to find the magnitude of the projected component of the force vector F along the line of the cable AB. The given force vector is:

$\mathbf{F} = 64\mathbf{i} + 11\mathbf{j} - 43\mathbf{k} \, \text{N}$

The unit vector along the cable AB is:

$\mathbf{u}_{AB} = -0.9231\mathbf{i} - 0.2308\mathbf{j} + 0.3077\mathbf{k}$

Step 1: Use the dot product

The projection of F onto the direction of the cable AB is given by the dot product of F with the unit vector u_AB, followed by taking the magnitude of the result:

$\text{Projection of } \mathbf{F} \text{ onto } \mathbf{AB} = \mathbf{F} \cdot \mathbf{u}_{AB}$

$\mathbf{F} \cdot \mathbf{u}_{AB} = (64)(-0.9231) + (11)(-0.2308) + (-43)(0.3077)$

Step 2: Compute the dot product

$\mathbf{F} \cdot \mathbf{u}_{AB} = 64(-0.9231) + 11(-0.2308) + (-43)(0.3077)$ $= -59.0784 - 2.5388 - 13.2201$ $= -74.8373 \, \text{N}$

Step 3: Interpretation

The negative sign indicates that the force vector is in the opposite direction of the unit vector, but since we are asked for the magnitude, we take the absolute value:

$|\mathbf{F} \cdot \mathbf{u}_{AB}| = 74.8373 \, \text{N}$

Thus, the magnitude of the projected component of the force along cable AB is approximately 74.84 N.

Would you like further details or have any questions about this solution?

Related questions:

1. How is the dot product useful in physics when calculating force projections?
2. Can the direction of the force change the result if the unit vector is different?
3. Why is the magnitude of a vector always non-negative?
4. What other applications involve projecting vectors in three-dimensional space?
5. How do changes in the coordinates of point A and B affect the unit vector?

Tip:

When working with vectors, always ensure that the unit vector has been correctly normalized (i.e., it has a magnitude of 1).