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A vector which is drawn parallel to a given vector through a specified point unlike free vector in space is called a localised vector. The effect of a force acting on a body depends not only on the magnitude & direction but also on its point of application & line of action.

What is localized vector and how is force a localized vector? Can anyone clarify me the book's reasoning?

Another property of the force is that the point of application of a given force acting on a rigid body may be transferred to any other point on the line of action without altering the effect of the force.

What is this all about? I couldn't understand this property infact the whole para. What does the book talk about force in this para? Plz help clarifying the meanings.

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  • $\begingroup$ this doesn't really make a lot of sense. sure, a vector is just a direction and a amount. (one way to think of it.) sure, if you have a vector acting on a specific point on an object -ie, you are pushing it, obviously that will affect it differently depending on where you push the object. are you working on a video game engine like Unity3D or Unreal? it would be tremendously easier if you explained what you are doing and told us what the book is. $\endgroup$ – Fattie Oct 11 '14 at 12:04
  • $\begingroup$ Oh!no,no I was just studying the localized vector in an old textbook of an unfamed native author...that's all! $\endgroup$ – user36790 Oct 11 '14 at 12:15
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A force is a vector acting along a line in space. It may act anywhere along the line and the end result will be the same (2nd paragraph).

The reason pure forces act on lines is that if they move away from the line there will be a torque applied also and the resulting motion will not be the same. The line of action of a force is the locus of the point where no net torque exists for this force.

Given a set of forces and torques specified at an arbitrary point, the location (and direction) of the line of action can be found. This line is unique, meaning that only one exists for each set of force and torque components.

In addition, these principles are dual to the idea of rotations about axes. Rotational motion needs to have the location of the rotation axis specified to be fully defined. Any other point away from the rotation axis will exhibit linear motion.

To fully specify the loading on a rigid body you need one of these two pieces of information:

  • A force vector $\vec{F}$ and a point which the line of action goes through $\vec{r}$. The equivalent torque at the origin is $\vec{\tau}=\vec{r} \times \vec{F}$.
  • A force vector $\vec{F}$ and the equivalent torque at the origin $\vec{\tau}$. The location of the line of action closest to the origin is $\vec{r} = \frac{\vec{F}\times \vec{\tau}}{\|\vec{F}\|^2}$.

To fully specify the motion of a rigid body you need one of these two pieces of information:

  • The rotation vector $\vec{\omega}$ and a point which the rotation axis goes through $\vec{r}$. The linear velocity at the origin is $\vec{v}=\vec{r} \times \vec{\omega}$.
  • The rotation vector $\vec{\omega}$ and the linear velocity at the origin $\vec{v}$. The location of the rotation axis closest to the origin is $\vec{r} = \frac{\vec{\omega}\times \vec{v}}{\|\vec{\omega}\|^2}$.

Quantities that require one free vector and one line vector to fully specify something are called screws (or screw vectors). All screws transform from one place to another with the same laws. Some quantities that are screws are:

  • Small motions (infinitesimal movement)
  • Rigid Body Motion (velocity)
  • Momenta (linear and angular)
  • Impulses and Contacts
  • Forces & torques
  • Spatial Acceleration (motion of entire rigid body)

Quantities that are not screws are:

  • Body Rotations (large angles are not vectors)
  • Material Accelerations (motion of each particle on a body)

Note $\times$ is the vector cross product, and $\cdot$ the vector dot product

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The localized vector is a vector which we know its magnitude and direction and its point of application , which is the point where the force acts , if the force were free vector , how you will calculate for example the moment ? that won't make any sense ok the second principle i found it in Vector mechanics book called PRINCIPLE OF TRANSMISSIBILITY. check that picture enter image description here

a clearer example which is if u pull the car by a rope from the front that will be the same effect if u pushed it from the rear given that the two forces are equal and on the same line of action , the line of action is the dotted line in the image.

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For a force, you need the following bits of knowledge:

  1. The direction and magnitude of the force.
  2. The point it acts on.

Some people call that pair of specifications a "localized vector", because it consists of both a location (#2) and a vector (#1).

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  • $\begingroup$ Can you tell me what is localized vector and what does the book want to tell? $\endgroup$ – user36790 Oct 11 '14 at 14:47
  • $\begingroup$ The "location" is a vector too. $\endgroup$ – Solomon Slow Oct 28 '15 at 17:05
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Vectors are considered free (of location) and are the same if they have the same magnitude (length) and orientation. So a vector A in one coordinate system with origin O is the same as a vector A in another coordinate system with origin O* even if O* is moving (translational and/or rotational motion) with respect to O. However, the meaning of the vector depends on the origin. If r is a vector describing the location of a particle with respect to O, the same vector r with respect to O* is not the location of the particle even though both vectors are equal. The location of the particle with respect to O* is r* = r - h where h is the vector from O to O*. similarly, r* is the same vector in O and O* but only in O* is it the position of the particle with respect to O*. If O* is rotating with respect to O, although the vector A is the same in both coordinate systems, its Cartesian components (magnitudes in x, y, z directions) differ- e.g., rx != rx*; also dA/dt is not the same in both coordinate systems. See Mechanics, Keith Symon; chapters 3, 7, and 11.

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  • $\begingroup$ Please use MathJax in your answer. It will look better. $\endgroup$ – SchrodingersCat Oct 28 '15 at 16:19
  • $\begingroup$ To extend Aniket's comment, a short primer on MathJax can be found here. $\endgroup$ – Kyle Kanos Oct 28 '15 at 16:42

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