How to calculate torque with cross product?

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Homework Help Overview

The problem involves calculating the torque generated by a force acting on a rod positioned in three-dimensional space. The force is defined by its vector components, and the torque is to be determined about the origin using the cross product of the force vector and the position vector from the origin to the point of application of the force.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss identifying the position vector OP, which corresponds to the vector r in the torque equation. There are suggestions to use the definition of the cross product and to consider the arrangement of the force components in relation to the axes.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the calculation of torque. There are multiple interpretations of how to apply the cross product and the torque formula, and participants are exploring these different avenues without reaching a consensus.

Contextual Notes

Participants note the importance of considering the correct signs in the calculations and the conventions for determining the direction of moments. There is an emphasis on ensuring clarity in the definitions and relationships between the vectors involved.

mohemoto
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Homework Statement


A rod has one end at the origin and one end at the point P whose coordinates are (1m, 2m, 2m). A force F = (3i+2j-1k) N acts on the rod at the point P. What is the torque about the origin due to F?


Homework Equations


torque = F x r


The Attempt at a Solution


I'm not sure what to multiply by what. Are there any suggestions?
 
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Well what is the vector OP which is the same as your vector r?
 
Use the definition of cross product. (If it's still unclear, do an internet search on "determinant of a matrix.")

[tex]\vec a \times \vec b = <br /> \left|<br /> \begin{array}{ccc}<br /> \hat \imath & \hat \jmath & \hat k \\<br /> a_x & a_y & a_z \\<br /> b_x & b_y & b_z<br /> \end{array} <br /> \right|[/tex]
 
Since the force is already given in mutually perpendicular directions x , y, and z, (i, j, and k unit vectors), then use Torque = force times perpendicular distance from line of action of force to the axis about which you are summing moments.
T_x = F_y(z) + F_z(y)
T_y = F_z(x) + F_x(z)
T_z = F_y(x) + F_x(y)

Please watch plus and minus signs. By convention, counterclockwise moments about an axis are taken as positive (x axis points right positive, y-axis points up positive, and z axis points out of plane toward you as positive).
 

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