# Hanging weight generates PSI?

jeff davis
TL;DR Summary
would a rope with a weight hanging on it have a cross sectional force?
Hello;
I have a quick question that has me thinking.
Please note that i don't much like working with imperial units because there is one too many polysemous words.

Say i had a 1 lb weight hanging on a 1in diameter rope. (rope weight neglected)

1. Could you say that at any perpendicular cross section of the rope have (approx) .785 PSI on it? The number is irrelevant i am just curious if that cross section force could be considered a "PSI".

2. pulling on a rope with psi or with lbs or with lbf?

3. would any perpendicular cross section be of the same psi if it is considered psi?

3. At what point is a force considered in its area unit opposed to just saying a pound is on there?

4. Why are we not weighed as PSI or similar instead of just pound?

I feel like at least once a year my mind rattles around the mass vs weight trap; and i blame it all on the use of the imperial system! And also my forgetful mind just a bit.

Staff Emeritus

Pressure is nothing more than force divided by area. You can see that from the units psi, pounds force per square inch.

So for a rope, first you have to neglect that the rope is made of strands, threads, and fibers. Consider it just a uniform substance with a certain cross sectional area A. Then you pull on the rope with force F. F/A does indeed carry the units of pressure and you can call it a pressure if you want.

Remember that pressure can be plus or minus. When you pull on the ends of the rope the pressure is minus, when you push on the rope to squash the middle, the pressure is plus. It is easier to think of that with a solid rod rather than a bendy rope, but the principle is the same.

I like to think of a suction cup. Because pressure under the cup is different than pressure above, the pressure difference P times the area A of the cup has units of force F, and F is how hard I have to pull to pull the cup loose. Actually, it's a bit more difficult than that because the pressure changes when I tug on the cup.

The pressure difference between the top and bottom sides of airplane wings times the area of the wings, must provide enough force to hold the airplane up.

Staff Emeritus
so a mass is not hanging from a rope giving me a weight?
If you hang a mass on a rope in zero gravity, it does not stretch the rope at all. Force and mass are related by the strength of the gravitational field. But your original question was not about gravity, so mass versus weight need not come into the question.

jeff davis
If you hang a mass on a rope in zero gravity, it does not stretch the rope at all. Force and mass are related by the strength of the gravitational field. But your original question was not about gravity, so mass versus weight need not come into the question.
ok i thought a lb took gravity into account

Mentor
Summary:: would a rope with a weight hanging on it have a cross sectional force?

Hello;
I have a quick question that has me thinking.
Please note that i don't much like working with imperial units because there is one too many polysemous words.

Say i had a 1 lb weight hanging on a 1in diameter rope. (rope weight neglected)

1. Could you say that at any perpendicular cross section of the rope have (approx) .785 PSI on it? The number is irrelevant i am just curious if that cross section force could be considered a "PSI".
Yes, and it has a name: stress
https://en.m.wikipedia.org/wiki/Stress_(mechanics)
3. At what point is a force considered in its area unit opposed to just saying a pound is on there?

4. Why are we not weighed as PSI or similar instead of just pound?
I'm not currently suspended by a rope, so I don't see a relevant reason/way to apply that. It's a concept you apply when you need it and when you don't, you don't - like any other concept.

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Mentor
Summary:: would a rope with a weight hanging on it have a cross sectional force?

Hello;
I have a quick question that has me thinking.
Please note that i don't much like working with imperial units because there is one too many polysemous words.

Say i had a 1 lb weight hanging on a 1in diameter rope. (rope weight neglected)

1. Could you say that at any perpendicular cross section of the rope have (approx) .785 PSI on it? The number is irrelevant i am just curious if that cross section force could be considered a "PSI".
Yes, the tensile stress in the rope would be 0.785 psi.
2. pulling on a rope with psi or with lbs or with lbf?
1 lbm weighs 1 lbf. So the force is 1 lbf, and the stress is 0.785 psi.
3. would any perpendicular cross section be of the same psi if it is considered psi?
any cross section perpendicular to the axis would have the same stress in psi.
3. At what point is a force considered in its area unit opposed to just saying a pound is on there?
Stress is more fundamental to determining when the rope will break.
4. Why are we not weighed as PSI or similar instead of just pound?
Both are important and relate to different considerations.

jeff davis
jeff davis
Thank you Very Much.

This question arose from some questions i had regarding pull testing a product. A tube connected to a plastic molded hub to be more specific. Someone proposed that the pull force should be 1lb minimum. This had me second guessing myself on a few fundamentals that i had previously considered "clear" in my mind.

At this point would it be documented as "1 lb minimum" or "1 lbf minimum" or some PSI relation.

I had assumed "1 LbF minimum" and when i went to reference myself the thought process arose that the idea that whatever force needed to break it should be gravity sensitive but not necessarily just Earth's gravity sensitive.
More like "X" to break it... but ultimately "X - gravity" would be the force required.

So then does this become a mass issue? I ask this because my mind is hung up on the idea that weight is just mass * gravity. If gravity is not a constant then how can my simple brain understand it?

thanks,
Jeff

Mentor
Let's temporarily table the discussion of stresses and psi's, and focus exclusively on masses and forces.

Here's how it plays out. In English units, the units of mass and force are ##lb_m## and ##lb_f##, respectively. In terms of these units, Newton's 2nd law of motion reads not F=ma, but instead $$F=m\frac{a}{g_c}$$ where a is the acceleration in ##ft/sec^2## and ##g_c=32.2\ \frac{lb_m}{lb_f}\frac{ft}{sec^2}##. Within this framework, the weight of an object is given by: $$W=m\frac{g}{g_c}$$ where g is the local acceleration of gravity. So, at a location where ##g=32.2\ ft/sec^2##, the weight of the object in ##lb_f## is numerically equal to its mass in ##lb_m##. So ##1\ lb_f## is the weight of a body of mass ##1 \ lb_m## at a location where the acceleration of gravity is the nominal value of ##g=32.2\ ft/sec^2##.

jeff davis
so $$1lbf = Xlbm \frac{(32.2 ft/sec^2)} {32.2(ft/sec^2)(lbm/lbf)}$$
Then say we were on venus g = (29ft/sec^2)
$$1lbf = Xlbm \frac {(29ft/sec^2)} {32.2(ft/sec^2)(lbm/lbf)}$$
goes to $$1lbf = Xlbm \frac{.901}{(lbm/lbf)}$$
goes to $$\frac{1lbf(lbm/lbf)} {.901}=Xlbm$$
goes to $$1.110(lbm) = X(lbm)$$
$$1.110=X$$

Mentor
If you are saying that, at a location where g is 29, if a body weighs 1 pound (force), it’s mass is 1.11 pound mass, that is correct.

jeff davis
Thanks I understand that concept better now.

When we would refer to a "lb" needing to break something; would this pound be assuming the local acceleration due to gravity? (32.2 ft/sec^2)? Would this be the reason lbm and lbf are specified when needing to be technical?

I am also a bit curious now why Newtons law is F = M(a/gc). Why is that "gc" in there. Is it solely to relate the lbs two values?

thanks; the more i work thru it the more things become connected.
jeff

Mentor
Thanks I understand that concept better now.

When we would refer to a "lb" needing to break something; would this pound be assuming the local acceleration due to gravity? (32.2 ft/sec^2)? Would this be the reason lbm and lbf are specified when needing to be technical?
How much do you think the value of g where you are located differs from 32.2?
I am also a bit curious now why Newtons law is F = M(a/gc). Why is that "gc" in there. Is it solely to relate the lbs two values?
This is because the natural unit of mass for use in Newton's law to get F is pounds force is not the pound mass; it is the "slug", which is 32.2 ##lb_m##. A force of 1 pound force causes a 1 lb mass to accelerate at 32.2 ft/s^2, but it causes a mass of 1 slug to accelerate at 1 ft/sec^2.

jeff davis
jeff davis
How much do you think the value of g where you are located differs from 32.2?
Probably very little; but is it just a blanket assumption?

How can someone sell a "calibrated" test weight with lbs as its unit? I suppose this is a misstatement and is actually a test mass?

jeff davis
This is because the natural unit of mass for use in Newton's law to get F is pounds force is not the pound mass; it is the "slug", which is 32.2 ##lb_m##. A force of 1 pound force causes a 1 lb mass to accelerate at 32.2 ft/s^2, but it causes a mass of 1 slug to accelerate at 1 ft/sec^2.

ahh