Vector products -Finding the direction of angular momentum

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Homework Help Overview

The discussion revolves around understanding the vector product in the context of torque, specifically when dealing with vectors that lie in different planes, such as the zx and xy planes. Participants are exploring how to determine the direction of torque given a force and a position vector in a three-dimensional space.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning how to apply the right hand rule when vectors are in different planes and whether the vectors can be considered to lie in a common plane. There are discussions about the relationship between the position vector and the force vector, particularly when they are perpendicular.

Discussion Status

Some participants have offered insights into the nature of the vectors and the application of the right hand rule, suggesting that the vectors can be treated as lying in a common plane despite initial concerns. There is ongoing exploration of how to visualize and calculate the torque using the cross product of the vectors.

Contextual Notes

Participants are working within the constraints of a homework problem that specifies the magnitude and direction of the force and the position vector, while also grappling with the implications of these vectors being in different planes.

ctpengage
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How do you resolve a vector products when the two vectors lie in different planes, i.e. how do you find the vector product of a vector that lies in the zx plane and a vector that lies in the xy plane. Using the right hand rule for vectors is redundant as the two vectors lie in different planes.

Can anyone help resolve this problem. Thanks
 
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The specific problem I have is finding out what the torque is, about the origin O, due to a force, of magnitude 2.0N on a Point A, given by the position vector r in the xz plane where r is equal to 3.0m and the angle it makes is 30 degrees. The force is directed along the y axis. What is the direction of the torque? I can't use the right hand rule because the vector products are in a different plane.
 
Welcome to PF!

ctpengage said:
The specific problem I have is finding out what the torque is, about the origin O, due to a force, of magnitude 2.0N on a Point A, given by the position vector r in the xz plane where r is equal to 3.0m and the angle it makes is 30 degrees. The force is directed along the y axis. What is the direction of the torque? I can't use the right hand rule because the vector products are in a different plane.

Hi ctpengage! Welcome to PF! :smile:

Two vectors (from the same point) are always in a common plane. :smile:​

In this case, the common plane is sloping at 30º to the "horizontal", and goes through the whole y-axis. :wink:
 


tiny-tim said:
Hi ctpengage! Welcome to PF! :smile:

Two vectors (from the same point) are always in a common plane. :smile:​

In this case, the common plane is sloping at 30º to the "horizontal", and goes through the whole y-axis. :wink:

How do you use the right hand rule in that case?
 
ctpengage said:
How do you use the right hand rule in that case?

Hi ctpengage! :smile:

I'm wondering what version of the right hand rule you've been taught. :confused:

The one I know is where you point your right thumb along one vector, your right index finger along the other vector, and then the cross-product is where your middle finger is pointing.

If you use that, there should be no difficulty. :wink:
 
tiny-tim said:
Hi ctpengage! :smile:

I'm wondering what version of the right hand rule you've been taught. :confused:

The one I know is where you point your right thumb along one vector, your right index finger along the other vector, and then the cross-product is where your middle finger is pointing.

If you use that, there should be no difficulty. :wink:

How do I use it to solve the problem given above? In that the force vector is directed perpendicular to the position vector? As said two vectors from a point lie in a single plane. But how do you figure it out with respects to the x, y and z axes? Hopefully you'll understand what I mean.
 
ctpengage said:
How do I use it to solve the problem given above? In that the force vector is directed perpendicular to the position vector?

It's supposed to be!

The easiest case is where they are perpendicular!
As said two vectors from a point lie in a single plane. But how do you figure it out with respects to the x, y and z axes? Hopefully you'll understand what I mean.

Nooo … :confused:

If you know a coordinate method, you just use that.

If you just know the ordinary vector method, then just remember that the cross-product must be perpendicular to that sloping plane I mentioned earlier. :smile:

(which automatically makes it perpendicular to the original two vectors :wink:)
 
ctpengage said:
The specific problem I have is finding out what the torque is, about the origin O, due to a force, of magnitude 2.0N on a Point A, given by the position vector r in the xz plane where r is equal to 3.0m and the angle it makes is 30 degrees. The force is directed along the y axis. What is the direction of the torque? I can't use the right hand rule because the vector products are in a different plane.
The force is along the y-axis and its magnitude is 2.0N. Okay, write it as <0, 2.0, 0>.

The position vector (assuming you mean "that angle it makes with the x-axis is 30 degrees) is <3.0cos(30), 0, 3.0sin(30)> = <[itex]\3\sqrt{3}/2, 0, 3/2>. The torque is the cross product of those two vectors.[/itex]
 

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