# Moment independent of direction of force?

dE_logics
There's a PDF attached to this; have I concluded things the right way?

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Gold Member
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.

dE_logics
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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Cantab Morgan
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.)

dE_logics
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.

Gold Member
We are not taking distance, we are taking the vector describing the lever arm. Distance is only a scalar.

dE_logics
But why is the distance vector?...how can it be a vector?

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