Math Problem Statement

Determine the moment of the force F about the Oa axis. Express the results as a Cartesian vector. Given F = (50i - 20j + 20k) N and coordinates of point O (origin) and a (for the axis projection).

Solution

The problem involves determining the moment of a force F about the Oa axis, which is in 3D space. The force vector is provided, and we have a hint that involves using the cross product and dot product to find the projection.

Given:

  • Force vector: F=50i^20j^+20k^N\mathbf{F} = 50\hat{i} - 20\hat{j} + 20\hat{k} \, \text{N}
  • Coordinates of point O (origin) and A (for the axis projection):
    • Position vector of point a relative to O can be written as rOa=2i^+1j^+3k^m\mathbf{r_{Oa}} = 2\hat{i} + 1\hat{j} + 3\hat{k} \, \text{m} (from the diagram).
    • Coordinates of the point where the force is applied are given by the relative position: rOF=2i^+1j^+6k^m\mathbf{r_{OF}} = 2\hat{i} + 1\hat{j} + 6\hat{k} \, \text{m}.

To determine:

The moment of force F about the Oa axis.

Step 1: Moment about point O

The moment M about point O is calculated using the cross product of the position vector r\mathbf{r} from O to the point of application of the force and the force vector F\mathbf{F}: MO=rOF×F\mathbf{M_O} = \mathbf{r_{OF}} \times \mathbf{F} Here, rOF=2i^+1j^+6k^\mathbf{r_{OF}} = 2\hat{i} + 1\hat{j} + 6\hat{k} and F=50i^20j^+20k^\mathbf{F} = 50\hat{i} - 20\hat{j} + 20\hat{k}

Step 2: Project the moment onto the Oa axis

The moment about the Oa axis is the projection of M_O onto the unit vector along Oa: MOa=(MOu^Oa)u^Oa\mathbf{M_{Oa}} = \left( \mathbf{M_O} \cdot \hat{u}_{Oa} \right) \hat{u}_{Oa} where u^Oa\hat{u}_{Oa} is the unit vector along the Oa axis, and it can be determined from the vector rOa\mathbf{r_{Oa}}.

Let me calculate this step by step.### Solution:

  1. Moment about point O: The moment vector M_O is: MO=140i^+260j^90k^Nm\mathbf{M_O} = 140\hat{i} + 260\hat{j} - 90\hat{k} \, \text{Nm}

  2. Moment about the Oa axis (projection of the moment onto the Oa axis) is: MOa=38.57i^+19.29j^+57.86k^Nm\mathbf{M_{Oa}} = 38.57\hat{i} + 19.29\hat{j} + 57.86\hat{k} \, \text{Nm}

This gives the moment of force F about the Oa axis in Cartesian vector form.


Would you like further details on any of these steps or any related concepts?

Here are 5 related questions for deeper understanding:

  1. How do you compute the cross product between two vectors?
  2. What is the significance of the dot product when projecting vectors?
  3. Why is it important to use a unit vector in projection calculations?
  4. Can the magnitude of a moment be calculated from its vector components?
  5. How do we interpret the physical meaning of the moment in 3D?

Tip: Always verify that vectors are properly aligned with the right-hand rule when calculating moments using the cross product.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Moments in 3D Space
Vector Projection

Formulas

Moment about a point: M = r x F
Projection of a vector onto an axis: M_{Oa} = (M_O · u_{Oa}) u_{Oa}

Theorems

Cross Product
Dot Product

Suitable Grade Level

Undergraduate - Engineering