Math Problem Statement

A force F of magnitude 840 N acts at a point as shown. Determine (a) the moment of the force about point B; (b) the direction associated with the unit vector λ along the axis of the moment; and (c) the perpendicular distance d from point B to the line of action of the force.

Solution

The problem involves a 3D force vector F\mathbf{F} acting on a block, and the task is to determine several related quantities.

Let's break down each part of the problem:

Given Data:

  • The magnitude of force F=840N\mathbf{F} = 840 \, \text{N}
  • Coordinates:
    • Point AA: (275 mm, 0, 400 mm)
    • Point BB: (150 mm, 200 mm, 0 mm)
    • Point OO: (250 mm, 175 mm, 0 mm)
  • You need to find:
    1. The moment of the force about point BB.
    2. The direction associated with the unit vector along the axis of the moment.
    3. The perpendicular distance from point BB to the line of action of the force.

Approach:

  • (a) Moment of the force about point BB: The moment of a force F\mathbf{F} about a point BB is given by: MB=rB/A×F\mathbf{M}_B = \mathbf{r}_{B/A} \times \mathbf{F} Where rB/A\mathbf{r}_{B/A} is the position vector from point BB to point AA, and F\mathbf{F} is the force vector.

    To find rB/A\mathbf{r}_{B/A}, we subtract the coordinates of BB from those of AA.

  • (b) Direction associated with the unit vector λ\mathbf{\lambda}: The unit vector along the axis of the moment can be obtained by normalizing the moment vector MB\mathbf{M}_B: λ=MBMB\mathbf{\lambda} = \frac{\mathbf{M}_B}{|\mathbf{M}_B|}

  • (c) Perpendicular distance from BB to the line of action of the force: The perpendicular distance dd is related to the magnitude of the moment and the force by: d=MBFd = \frac{|\mathbf{M}_B|}{|\mathbf{F}|}

Now, I'll calculate the required quantities.### Results:

  1. Moment of the force about point BB: MB=[19.21N\cdotpm,76.83N\cdotpm,32.41N\cdotpm]\mathbf{M}_B = \left[ -19.21 \, \text{N·m}, -76.83 \, \text{N·m}, -32.41 \, \text{N·m} \right]

  2. Unit vector along the axis of the moment: λ=[0.224,0.898,0.379]\mathbf{\lambda} = \left[ -0.224, -0.898, -0.379 \right]

  3. Perpendicular distance from point BB to the line of action of the force: d=0.102m(or 101.87 mm)d = 0.102 \, \text{m} \, \text{(or 101.87 mm)}

Would you like more detailed explanations or have any other questions?

Related Questions:

  1. How is the position vector between two points calculated?
  2. What does a cross product represent in the context of moments?
  3. Why is the direction of the moment important in physical applications?
  4. How can the unit vector be interpreted geometrically?
  5. What are practical examples where perpendicular distance is relevant in force systems?

Tip: Always ensure the units are consistent when performing vector operations, especially in problems involving distances, forces, and moments.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Moments
Cross Product
Unit Vectors
Perpendicular Distance

Formulas

Moment of force: M_B = r_B/A × F
Unit vector: λ = M_B / |M_B|
Perpendicular distance: d = |M_B| / |F|

Theorems

Moment of a Force
Cross Product in 3D Space
Properties of Unit Vectors

Suitable Grade Level

College Level - Engineering Mechanics