Moment independent of direction of force?

In summary, the conversation discusses the concept of torque and the importance of taking into account the direction of the lever arm and the force when calculating it. The incorrect equation for torque is also mentioned, as well as the use of the right hand rule to determine the direction of the torque. The difference between distance and displacement is clarified and it is noted that displacement is a vector, not a scalar.
  • #1
dE_logics
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There's a PDF attached to this; have I concluded things the right way?
 

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  • #2
First, you are not applying the force radially, there is no radial component in your diagram if you take the pivot/fulcrum to be your origin. The torque is dependent upon the direction of the lever arm and the direction of the force, so you can't neglect sign or direction (the latter is what you have neglected). The reason why the torque is in the same direction for both is because you are applying the force in the same direction with relation to the lever arm.

The main problem is that you are using an incorrect equation for the torque, the torque is a vector that is the lever arm cross the force.
 
  • #3
The torque is dependent upon the direction of the lever arm and the direction of the force, so you can't neglect sign or direction (the latter is what you have neglected).

Yeah I was thinking about that, but it's just a distance...I mean, its just the magnitude of displacement...it's like taking a measured section of a scale...that doesn't have a direction.

The reason why the torque is in the same direction for both is because you are applying the force in the same direction with relation to the lever arm.

Humm...so actually it is relative. Its for me that the direction is changing...as a result I'm taking it as the modulus.

But actually its not changing for the point around which I'm trying to calculate the torque...so for that point, the direction of force will remain a constant.

Am I right?
 
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  • #4
dE_logics said:
Yeah I was thinking about that, but it's just a distance...I mean, its just a displacement...it's like taking a measured section of a scale...that doesn't have a direction.

What is just a distance? There are two interesting vectors in your picture. The first is the vector from the fulcrum/pivot to the point where the force is applied to the beam. (If we conveniently chose the fulcrum to be the origin of our coordinate system, then that vector would just be the position vector of the point on the beam that's getting pushed. But regardless of the coordinate system we choose, it's still a vector.)

The second vector is the force itself. Force is a vector, having both a magnitude and a direction.

The torque, whether clockwise or counter-clockwise, and its magnitude, depends on both of these vectors.

In space (3 dimensions), we represent torques as pseudovectors, and we take the cross product of the two vectors I mentioned. In the plane, it's simpler because torques can only be clockwise or counterclockwise. It's a bit like having a vector in only one dimension: a magnitude and a sign.

It works like this in the plane... Stand on the fulcrum and look in the direction of the vector towards where the force is applied. Now consider the direction of the force vector. If it's pointing to your left, then your torque is counterclockwise. If it's pointing to the right, then your torque is clockwise. The point is that you need both vectors to get the magnitude and orientation of the torque. (By the way, this is called the "right hand rule" restricted to the plane.)
 
  • #5
I'm still having problems with that first vector...the distance.

Ok, you took a coordinate system, that way, only the 'point' will be called negative, but not the distance from the fulcrum...that is still a distance...not a displacement. Ok...suppose we change the coordinate system to one end of the bar being rotated, i.e that end will be called as 0...then every point will be negative or positive...same is case when the torque is computed taking a coordinate system which's 0 at a point out of the bar.

Thanks for clearing that second force problem Cantab Morgan + Born2bwire. But after clearing the second problem, the first problem will persist even more. If we do take the distances as vectors, then considering we take the force as a vector too...things cancel out...it will mean the same thing as taking the modulus.

The distance also has to be taken WRT the rotating point to fix this problem i.e if we are somehow taking distance as a vector. That way there will be just one sign of distance and force throughout the rotation.
 
  • #6
We are not taking distance, we are taking the vector describing the lever arm. Distance is only a scalar.
 
  • #7
But why is the distance vector?...how can it be a vector?
 
  • #8
The torque resulting from some force F applied at some point p is defined as [itex]{\boldsymbol N} = {\boldsymbol r}\times{\boldsymbol F}[/tex] where r is the displacement vector (not distance) from the fulcrum to the point p.

In short, it's not distance (which is a scalar) that is involved here, it's a displacement vector.
 
  • #9
Ok...thanks!
 

1. What is meant by "moment independent of direction of force?"

"Moment independent of direction of force" refers to the concept that the rotational effect, or moment, of a force applied to an object is the same regardless of the direction in which the force is applied. This means that the magnitude of the moment is not affected by the direction of the force, only its distance from the pivot point.

2. How does the direction of force affect the moment?

The direction of force does not affect the moment, only the distance from the pivot point does. This is because the moment is a product of the magnitude of the force and the distance between the force and the pivot point, not the direction of the force.

3. What is the significance of "moment independent of direction of force" in physics?

The concept of moment independent of direction of force is important in physics because it allows us to calculate the rotational effect of a force on an object without having to consider the direction in which the force is applied. This simplifies calculations and makes it easier to analyze and predict the behavior of objects under the influence of various forces.

4. Are there any exceptions to the concept of "moment independent of direction of force"?

Yes, there are some exceptions to this concept. One example is when the force is applied at the pivot point itself, in which case the moment would be zero regardless of the magnitude of the force. Additionally, if the object is not rigid and can deform, the direction of the force may affect the moment.

5. How is the moment calculated for a force that is not perpendicular to the object?

The moment can still be calculated for a force that is not perpendicular to the object by using the perpendicular distance from the pivot point to the line of action of the force. This distance is known as the lever arm, and it is used in the moment equation: M = F x d, where M is the moment, F is the magnitude of the force, and d is the perpendicular distance from the pivot point to the line of action of the force.

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