division of forces

Dear kkrish:

Forces are vectors, and division of vectors is not easily done.

What do you mean? What are you trying to do?

David A. Smith

Thanks for your reply,

I want to find the ratio of effect of one vector from a reference point to another vector from the same reference point.Yes, they are vectors.How to find the ratio?Is there any method to find the ratio.It will be of great help. 1.vector A at an angle theta1 2.vector b at an angle theta2 how to find the ratio of vector A to vector B. I have a circular disk of metal.There are two points in the disk which are not uniform and have some manufacturing defect on the surface.Due to non uniformity there will be two forces acting on the disk when the disk rotates.I want to measure the two forces from a reference point on the disk.And i want to study the effect of the forces when the reference point or any of the other two points change. Thank you.

N:dlzc D:aol T:com (dlzc) wrote:

Dear kkrish:

Each force then can be broken down into components:

- from the center to the point of action: "dot product"

- perpendicular to the line above, not necessarily in the plane of the disk: "cross product"

..."effect of the" ...

Typically you will form the products I mention above, and create a "free body diagram". The forces, once the products above have been formed for each reference point, can be shown acting on the reference point. The cross product terms are torques. Be very careful with the signs. The dot product terms describe linear acceleration.

If the disk is free to rotate, then the forces must provide whatever angular acceleration is observed.

If the manufacturing defect only provides a place for friction to be creating one or both forces, then you will also have a component of one or both forces that is normal to the plane of the disk, and this component will be *included* in the component you get with the cross product. Then you will need torque information about the bearing the disk can rotate on.

If it was easy, they would not call it work. And you would not get paid for it.

David A. Smith