# Cross Product Torque: Physics Basics Explained

• jaalcaam
In summary, the vector cross product is a useful representation for perpendicular component interactions and the orientation of the plane containing those components. The magnitude of the product is equal to the area of the parallelogram formed by the two vectors, which is a coincidence of the geometry involved. The important factor is that only the perpendicular component of the vector is contributing to the magnitude of the product.
jaalcaam
Why does torque, magnetic fields and others, work with something like the vectorial cross product? A plane director vector of intensity equal to the area of the paralelogram formed by the other two?

Anyone?
Basic Statistics Physics?

Symmetry & behavior under space inversion (parity transformation) is the key & the reason for using peudovectors.

Daniel.

jaalcaam said:
Why does torque, magnetic fields and others, work with something like the vectorial cross product? A plane director vector of intensity equal to the area of the paralelogram formed by the other two?

Anyone?
Basic Statistics Physics?

Some physical phenomena are observed to depend on the extent to which directional quantities are parallel. For example, work done by a force depends on the motion of an object in the direction of the force. Motion in other directions is irrelevant when it comes to work. There are infinitely many directions an object could be moving when a force is applied, but for any given force there is only one direction for that force. If the component of the displacement in the direction of the force can be determined, the work can be calculated. This component is always the product of the magnitude of the displacement times the cosine of the angle between the displacement and the force. Work has no direction associated with it.

There are other phenomena that are observed to depend on the extent to which directional quantities are perpendicular, such as the ones you have mentioned. Torque depends on the position at which a force is applied relative to an axis of rotation, but only the component of the force perpendicular to the position vector tends to produce rotation. Furthermore, the tendency to produce rotation is proportional to the distance of the point of application from the point of rotation. The perpendicular component of the force times the distance of application is the magnitude of the force times the distance times the sine of the angle between the direction of the force and the position vector. It happens that this is numerically equal to the area of a parallelogram formed by the force and position vectors, but the important thing is that the product is the product of two perpendicular vector components.

In cases where parallel components are the important consideration, there is a unique direction for the component of one vector quantity in the direction of the other quantity. This is not the case when the perpendicular component of one vector relative to another vector is involved. There are infinitely many directions the perpendicular component can take, and each of them tends to produce a different result. In the case of torque, the perpendicular component of force tends to produce motion in a plane that contains the position vector and the force vector. It is convenient to describe the orientation of that plane by giving a single direction perpendicular to that plane, in other words by specifying the axis of rotation.

All the physical quantities that involve the vector cross product involve the product of the perpendicular components of the vectors, and have direction that depends on the orientation of the plane containing the original vectors. The vector cross product is a convenient representation of both the perpendicular component interaction, and the direction of the plane of those components. That the numerical value happens to be the area of the parallelogram is a coincidence of no real importance.

Dextercioby, could you be more specific?

OlderDan, could you elaborate on the 'coincidence' of the paralelogram area equal to the vector intensity?

jaalcaam said:
Dextercioby, could you be more specific?

OlderDan, could you elaborate on the 'coincidence' of the paralelogram area equal to the vector intensity?

In any vector cross product, the magnitude of the product depends on the component of one vector that is perpendicular to the other vector. It does not matter which of the two vectors you break into components. The component parallel to the other vector contributes nothing, and the perpendicular component contributes everything. The magnitude of the resulting product is the magnitude of the one vector times the magnitude of the perpendicular component of the second vector. When resolving the second vector into two components, the parallel component is the magnitude of the vector times the cosine of the angle bewteeen the two vectors, and the perpendicular component is the magnitude of the vector times the sine of that angle. The result is that the magnitude of the cross product is the magnitude of each vector times the sine of the angle between them, which also happens to be the area of the parallelogram constructed by placing the two vectors tail to tail and tail to head. The geometry is inescapable. The magnitude of the product will always equal the area, but the physically important thing is that only the vertical component of the vector is contributing.

## 1. What is the cross product torque?

The cross product torque is a measure of the rotational force acting on an object. It is calculated by taking the cross product of the distance vector from the axis of rotation to the point of application of the force and the force vector itself. It is represented by the symbol τ and is measured in units of newton-meters (N·m).

## 2. How is cross product torque different from regular torque?

Regular torque, also known as scalar torque, is a measure of the twisting force on an object. It is calculated by multiplying the magnitude of the force by the perpendicular distance from the axis of rotation to the point of application of the force. Cross product torque takes into account the direction of the force, making it a vector quantity. It is a more accurate representation of the rotational force on an object.

## 3. What is the relationship between cross product torque and angular acceleration?

According to Newton's second law, the net torque on an object is equal to the product of its moment of inertia and its angular acceleration. Therefore, the greater the cross product torque acting on an object, the greater its angular acceleration will be. This relationship is represented by the equation τ = Iα, where τ is the cross product torque, I is the moment of inertia, and α is the angular acceleration.

## 4. How does the direction of the force affect the cross product torque?

The direction of the force has a significant impact on the cross product torque. If the force is applied perpendicular to the distance vector, the cross product torque will be at its maximum. However, if the force is applied parallel to the distance vector, the cross product torque will be zero. This is because the cross product of two parallel vectors is always zero.

## 5. What are some real-life examples of cross product torque?

Cross product torque is present in many everyday activities and functions. Some examples include opening a door by pushing on the handle, tightening a screw with a screwdriver, and pedaling a bicycle. In each of these situations, a force is applied at a distance from the axis of rotation, resulting in a cross product torque that causes the object to rotate.

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