Tension Direction: Hello, What Is Correct?

[Max T]

Hello , which of the following has the correct T direction?

 

[Doc Al]

Depending upon the object being analyzed, I'd say they are all correct.

The first diagram shows the tension acting on the box and the tension acting on the hand. Both correct.
The second diagram shows the tension acting on the box. Looks OK to me.

Realize that in common usage, "tension" refers to the force that the rope exerts on whatever it is pulling. That can be confusing at times.

 

[Stephen Tashi]

Realize that in common usage, "tension" refers to the force that the rope exerts on whatever it is pulling. That can be confusing at times.

To emphasize that point, we can say that "tension" is not a force. "Tension" is property of a rope that can be used to deduce forces at many different places. So there are different force vectors that are "due to tension".

The direction of a force due to tension depends on whether we are considering the force exerted by the rope on something or the force of something pulling against the rope (i.e. what we are considering to be the "free body" in a free body diagram). So "tension" is not a vector. It is the forces due to tension that are vectors.

Of course, it's common for people talking about a physics problem to pick a particular force vector and call it "the tension" instead of saying it is "a force due to tension". Common speech in inexact.

 

[Chestermiller]

There is a reason that tension in a rope is so confusing. It is because, in reality, the entity we call tension is not a scalar quantity, nor is it a vector quantity. It is a second order tensor, related directly to the so-called stress tensor in the material. Here's how it all plays out mathematically:

Suppose you have a rope under tension laid out from left to right, and let $\vec{i}$ represent the unit vector pointing from left to right along the rope. Then the second order tension tensor for the rope is given by $\vec{T}=T\vec{i}\vec{i}$, where T is the magnitude of the tension tensor (a scalar). Suppose you have a cross section at a given location along the rope and you want to find the tension force (vector) that the portion of the rope to the right of the cross section is exerting on the portion of the rope to the left of the cross section. To do this, you take the dot product of the tension tensor $\vec{T}$ with of a unit vector drawn from the left of the cross section to the right of the cross section:
$$\vec{T}\centerdot \vec{i}=T\vec{i}\vec{i}\centerdot \vec{i}=T\vec{i}(\vec{i}\centerdot \vec{i})=+T\vec{i}$$
Next, suppose you want to find the tension force (vector) that the portion of the rope to the left of the cross section is exerting on the portion of the rope to the right of the cross section. To do this, you take the dot product of the tension tensor $\vec{T}$ with of a unit vector drawn from the right of the cross section to the left of the cross section:
$$\vec{T}\centerdot (-\vec{i})=T\vec{i}\vec{i}\centerdot (-\vec{i})=-T\vec{i}(\vec{i}\centerdot \vec{i})=-T\vec{i}$$

Doing the math in this way guarantees that you always get the correct sign for the tension force (vector).

Another entity which is also causes confusion for these same reasons is pressure. The pressure tensor is the "isotropic part" of the stress tensor, and can be represented mathematically by:

$$\vec{P}=P(\vec{i}_x\vec{i}_x+\vec{i}_y\vec{i}_y+\vec{i}_z\vec{i}_z)$$ where P is the scalar magnitude of the pressure tensor. See what happens if you dot this tensor with a unit vector (perpendicular to an area element) in an arbitrary direction. This is how the present mathematical formalism automatically satisfies the condition that the pressure at a given location in a fluid acts equally in all directions.

Chet

 

[Subductionzon]

Worse yet in reality a rope does not follow a straight line. It follows a catenary and then the math gets even more fun.

 

[Chestermiller]

Worse yet in reality a rope does not follow a straight line. It follows a catenary and then the math gets even more fun.

What I said in my previous post applies locally along a rope, even if it is in the shape of a catenaery. In that case, the vector $\vec{i}$ is the local unit normal vector to the rope cross section.

Chet