Set Up: Find Weight from Magnitude of Friction on a Chain(Vectors)

[crybloodwing]

The tension, T at each end of a chain has magnitude 25N. What is the weight of the chain?

I am having trouble setting this up. In the diagram below, everything except the red is what is shown in the book. The red is what I think the possible vectors are.

However, if we are talking about weight, I would think the vector would be going completely South. I originally thought the Tension(25 N) originally went along the black line that makes the angle to the ground, however it says "N' so I am pretty sure that means it is going completely North.

Can anyone help me set up this problem and the diagram? I don't need help solving it for now, just setting it up.

 

[sinx]

The tension, T at each end of a chain has magnitude 25N. What is the weight of the chain?

I am having trouble setting this up. In the diagram below, everything except the red is what is shown in the book. The red is what I think the possible vectors are.

However, if we are talking about weight, I would think the vector would be going completely South. I originally thought the Tension(25 N) originally went along the black line that makes the angle to the ground, however it says "N' so I am pretty sure that means it is going completely North.

Can anyone help me set up this problem and the diagram? I don't need help solving it for now, just setting it up.

View attachment 10030

draw a free body diagram.
the tension is along the length of the chain.
it is true that the wt pulls down, but you are analyzing the tension. That tension is a culmination of all the weights of the links in the chain.

this is kind of a trick question, but it makes you think.
i.e. it makes you answer the question; how does a chain pull its supporing blocks to the inside when the only force on the chain is its weight?

 

[Deleted member 4993]

The tension, T at each end of a chain has magnitude 25N. What is the weight of the chain?

I am having trouble setting this up. In the diagram below, everything except the red is what is shown in the book. The red is what I think the possible vectors are.

However, if we are talking about weight, I would think the vector would be going completely South. I originally thought the Tension(25 N) originally went along the black line that makes the angle to the ground, however it says "N' so I am pretty sure that means it is going completely North.

Can anyone help me set up this problem and the diagram? I don't need help solving it for now, just setting it up.

View attachment 10030

As I interpret it, the drawing is incorrect.

The tension (25 N) should be along the 37° line

Then using equilibrium in y direction(vertical) - you would get:

2 * 25 * sin(37) = W

 

[Dr.Peterson]

The tension, T at each end of a chain has magnitude 25N. What is the weight of the chain?

I am having trouble setting this up. In the diagram below, everything except the red is what is shown in the book. The red is what I think the possible vectors are.

However, if we are talking about weight, I would think the vector would be going completely South. I originally thought the Tension(25 N) originally went along the black line that makes the angle to the ground, however it says "N' so I am pretty sure that means it is going completely North.

Can anyone help me set up this problem and the diagram? I don't need help solving it for now, just setting it up.

Are you thinking the N means north (up as opposed to south for down)? I think the N just means newtons, the unit of the tension. There is no north or south here.

You were right at first about the tension being in the angled direction.

 

[sinx]

The tension, T at each end of a chain has magnitude 25N. What is the weight of the chain?

I am having trouble setting this up. In the diagram below, everything except the red is what is shown in the book. The red is what I think the possible vectors are.

However, if we are talking about weight, I would think the vector would be going completely South. I originally thought the Tension(25 N) originally went along the black line that makes the angle to the ground, however it says "N' so I am pretty sure that means it is going completely North.

Can anyone help me set up this problem and the diagram? I don't need help solving it for now, just setting it up.

View attachment 10030

the tension in the chain is 25N.
[it is true that its vertical component is 25sin37, and its horizontal is 25cos37.]
however, the only direction the chain can transmit force is along its length.

when the 37 degrees=90 degrees, all the tension is vertical (and its tension is still 25N).
at that time, all the wt of the chain is supported vertically.

i.e. the tension in the chain equals the wt (counting both ends).

 

[Dr.Peterson]

the tension in the chain is 25N.
[it is true that its vertical component is 25sin37, and its horizontal is 25cos37.]
however, the only direction the chain can transmit force is along its length.

when the 37 degrees=90 degrees, all the tension is vertical (and its tension is still 25N).
at that time, all the wt of the chain is supported vertically.

i.e. the tension in the chain equals the wt (counting both ends).

But 37 ≠ 90, so this is false for the problem as it stands, right? (In fact, the chain couldn't hang with its ends exactly vertical. That isn't a possible shape for it. Look up "catenary".)

 

[sinx]

But 37 ≠ 90, so this is false for the problem as it stands, right? (In fact, the chain couldn't hang with its ends exactly vertical. That isn't a possible shape for it. Look up "catenary".)

before answering your question Dr. Peterson;
I will attempt to answer the original poster's question.
the wt of the chain is the sum of the vertical forces at each end; i.e. 2*25sin37⁰
to set it up, you sum the forces vertical, and they equal zero, since the chain is not moving.
since there are only two forces acting vertically opposing the wt, the wt equals this sum.

Or you sum the moments about a pt, choosing one end or the other is easiest. These also equal zero, as the chain is not rotating. i.e. sumM=F₁(L)-wt(1/2L)=0; or/ wt=2F1; Where F₁ is the vertical force at one end. This also gives the answer, in one equation.
[In both cases the center of mass is the center of the chain, and the wt acts through this point.]

for Dr. Peterson, you could suspend the chain from two vertically hanging spring scales, placed very close together.
But you are right that this is not the case we have here. I overthought the problem.
I looked up catenary. The information there was not necessary to solve this problem. However it was enlightening. Thank you.

 

[Dr.Peterson]

for Dr. Peterson, you could suspend the chain from two vertically hanging spring scales, placed very close together.
But you are right that this is not the case we have here. I overthought the problem.
I looked up catenary. The information there was not necessary to solve this problem. However it was enlightening. Thank you.

My point about the catenary is that there are no points at which it is vertical. Unless the two ends are at exactly the same point (or vertically above one another, if we don't keep them at the same height), so that the chain forms a straight line, the forces at the ends have to be angled.