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Here's an elegant - and useful - type of physics problem:

A heavy weight is supported in the following ways: It hangs suspended vertically from a rope which dangles at point A from a strut coming out and upward from the side of a wall. The strut (and thus the weight) is also supported by a cable coming down at an angle from the wall and which is also attached at point A. At point A, the angle of the strut below the horizontal is 45 degrees and the angle of the cable above the horizontal is 30 degrees. (point A, where three entities meet -- robe, cable, strut -- is some distance out from the wall. Visualize a store sign or movie marquee suspended over a sidewalk by a projecting strut and a cable.)

If the maximum tension (pull) the cable can withstand is 1000 lbs. and the maximum compression (outward push) the strut can withstand is 2000 lbs., what is the maximum weight that can be supported by this frequently found structure? (Assume the rope the weight dangles from is strong enough.)

Hint: This is an introductory physics "equilibrium problem"(from mechanics). This means force components must balance along the x-axis and also along the y-axis. I don't know how to do a diagram in a post, so I tried to explain the problem visualizably in the first two paragraphs.

Note: Show your work. Explain each step along with the mathematical or physical reason why it is valid as if you were writing a proof in geometry.

Edited by Philip Coates

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"Show Your Work" -- one can scarcely imagine a more appropriate thread title to mark Phil's return from self-imposed exile.

Ghs

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"Show Your Work" -- one can scarcely imagine a more appropriate thread title to mark Phil's return from self-imposed exile.

Ghs

Yes! For example, it's both civil and benevolent!

JR

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"Show Your Work" -- one can scarcely imagine a more appropriate thread title to mark Phil's return from self-imposed exile.

Ghs

Yep...

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Oh jeez I did about eight thousand variants of this problem in my Statics & Strengths of Materials course. As soon as I saw the words "strut", "heavy weight", and "cable", I knew exactly what was coming.

"Show Your Work" -- one can scarcely imagine a more appropriate thread title to mark Phil's return from self-imposed exile.

Ghs

Yep...

If we can't dazzle them with brilliance...

Mike

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Here's an elegant - and useful - type of physics problem:

A heavy weight is supported in the following ways: It hangs suspended vertically from a rope which dangles at point A from a strut coming out and upward from the side of a wall. The strut (and thus the weight) is also supported by a cable coming down at an angle from the wall and which is also attached at point A. At point A, the angle of the strut below the horizontal is 45 degrees and the angle of the cable above the horizontal is 30 degrees. (point A, where three entities meet -- robe, cable, strut -- is some distance out from the wall. Visualize a store sign or movie marquee suspended over a sidewalk by a projecting strut and a cable.)

If the maximum tension (pull) the cable can withstand is 1000 lbs. and the maximum compression (outward push) the strut can withstand is 2000 lbs., what is the maximum weight that can be supported by this frequently found structure? (Assume the rope the weight dangles from is strong enough.)

Hint: This is an introductory physics "equilibrium problem"(from mechanics). This means force components must balance along the x-axis and also along the y-axis. I don't know how to do a diagram in a post, so I tried to explain the problem visualizably in the first two paragraphs.

Note: Show your work. Explain each step along with the mathematical or physical reason why it is valid as if you were writing a proof in geometry.

--Brant

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"Show Your Work" -- one can scarcely imagine a more appropriate thread title to mark Phil's return from self-imposed exile.

Ghs

Yes! For example, it's both civil and benevolent!

JR

BOTTOM LINE:

I will never post on your site again. I would prefer not to have anything further to do with you or to communicate with you again. Quite frankly, your weakness and double standards disgust me.

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Ah yes...

Your lips are saying NO NO, but there's yes yes in your eyes...

1924 Roy B. Carson Orchestra!

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I will never post on your site again. I would prefer not to have anything further to do with you or to communicate with you again. Quite frankly, your weakness and double standards disgust me.

Makes me think of the saying in Kölsch (Cologne dialect): "Wat kümmert mich ming Jeschwätz von jestern?" ('I couldn't care less about my gibberish from yesterday').

Edited by Xray

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I hope Jonathan soon chimes in so we get this gauntlet running behind us.

--Brant

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> Statics....As soon as I saw the words "strut", "heavy weight", and "cable", I knew exactly what was coming. [Mike H]

This one is actually a bit more difficult than the 'run of the mill' of those problems.

> I did about eight thousand variants of this problem

I'm curious if anyone who remembers their Physics 101 is actually able to do this one and explain the steps: There's one tricky part.

Edited by Philip Coates

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> Statics....As soon as I saw the words "strut", "heavy weight", and "cable", I knew exactly what was coming. [Mike H]

This one is actually a bit more difficult than the 'run of the mill' of those problems.

> I did about eight thousand variants of this problem

I'm curious if anyone who remembers their Physics 101 is actually able to do this one and explain the steps: There's one tricky part.

It seems Phil has put this problem up on multiple sites, loosing track of where he is. Mike Hardy posts on Atlantis II--or did. I'm simply assuming that's where these quotes are coming from.

Starting to get pissed off at Phil, again, and his tiresome, purblind posting techniques.

--Brant

edit: it's not from Atlantis II

Edited by Brant Gaede

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> Statics....As soon as I saw the words "strut", "heavy weight", and "cable", I knew exactly what was coming. [Mike H]

This one is actually a bit more difficult than the 'run of the mill' of those problems.

> I did about eight thousand variants of this problem

I'm curious if anyone who remembers their Physics 101 is actually able to do this one and explain the steps: There's one tricky part.

Okay. I'll bite.

If we define C to be the force on the cable, S to be the force on the strut, and W to be the weight of the heavy weight, we proceed with a force balance to get:

x-component: Ccos(30) = Scos(45)

y-component: Csin(30) + Ssin(45) = W

The 'tricky' part: the x-component of the balance gives us an equation for two variables (S and C) that we probably think we already know (from their given maximum values).

Plugging in the maximum values, the x-component shows that S = 1225 lbs, instead of its maximum of 2000 lbs. (the logic here is that 2000*cos(45) is greater than 1000*cos(30), which means that, for the equality to be satisfied, C must increase (impossible. already at maximum) or S must become the value that gives the equality).

Then plugging in the value of C = 1000 lbs, S = 1225 lbs, we get the maximum weight W = 1366 lbs.

Mike

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Ah yes...

Your lips are saying NO NO, but there's yes yes in your eyes...

1924 Roy B. Carson Orchestra!

I've heard that line in films that involves rape. That, to the rapist justifies the act somehow...

Edited by David Lee

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> Statics....As soon as I saw the words "strut", "heavy weight", and "cable", I knew exactly what was coming. [Mike H]

This one is actually a bit more difficult than the 'run of the mill' of those problems.

> I did about eight thousand variants of this problem

I'm curious if anyone who remembers their Physics 101 is actually able to do this one and explain the steps: There's one tricky part.

Okay. I'll bite.

If we define C to be the force on the cable, S to be the force on the strut, and W to be the weight of the heavy weight, we proceed with a force balance to get:

x-component: Ccos(30) = Scos(45)

y-component: Csin(30) + Ssin(45) = W

The 'tricky' part: the x-component of the balance gives us an equation for two variables (S and C) that we probably think we already know (from their given maximum values).

Plugging in the maximum values, the x-component shows that S = 1225 lbs, instead of its maximum of 2000 lbs. (the logic here is that 2000*cos(45) is greater than 1000*cos(30), which means that, for the equality to be satisfied, C must increase (impossible. already at maximum) or S must become the value that gives the equality).

Then plugging in the value of C = 1000 lbs, S = 1225 lbs, we get the maximum weight W = 1366 lbs.

Mike

Wow, is this guy looking for the answer for his son's physics homework? Have other people do the "heavy lifting" without paying? he he he.

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Mike H. is exactly correct! (I got 1369, but that's probably just a rounding issue).

And the tricky part [at least to me! - I am not a physicist or an engineer and I messed it up the first time I tried it] was indeed NOT to mindlessly plug in maximum values and try to solve, since one max value (2000 for the strut) would make the cable go up beyond its allowable maximum. Solving for the tension in the cable as a function of the compression in the strut shows you this. Tension in cable always must equal .82 times compression in the strut.

Edited by Philip Coates

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Subject: Impossibly large compression and tension in everyday structures

As Mike said in post 13:

" If we define C to be the force on the cable, S to be the force on the strut, and W to be the weight of the heavy weight, we proceed with a force balance to get:

x-component: Ccos(30) = Scos(45)

y-component: Csin(30) + Ssin(45) = W "

Here's the 'thought experiment' problem I came up with regarding extremely sharp angles in some 'statics' situations. Suppose we change the above slightly:

Let the strut come out at a horizontal angle (angle is now zero) and create a *very sharp angle* (say 5 degrees) above horizontal for the cable. And let's say we have a fairly heavy weight suspended from the structure, like a ton (2000 lbs.) Then we have:

x-component: Ccos(5) = Scos(0) = S

y-component: Csin(5) + Ssin(0) = W

Solving:

(1) x-component equation: .996C = S.

(2) y-component equation: .087C + 0 = W.

Which means for a 2000 pound weight, the strut has to be able to withstand a compression of nearly 23,000 pounds--more than 11 tons. And the tension force the cable has to be able to withstand is nearly the same.

Not possible for normal materials!!! (How can this be? Or is there a mistake in the force equations--I've checked and double checked and they seem correct.) YIKES!!

{I apologize in advance if I'm making some sort of careless mistake. But it seems as if the force equations above break down and don't reflect reality for very sharp angles.}

Edited by Philip Coates

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Not possible for normal materials!!! (How can this be? Or is there a mistake in the force equations--I've checked and double checked and they seem correct.) YIKES!!

{I apologize in advance if I'm making some sort of careless mistake. But it seems as if the force equations above break down and don't reflect reality for very sharp angles.}

No mistake in the force equations. Your force balance for point A is correct.

I've tried to figure out a few things to write about the quoted material above, but I'm a little confused about your thought that the force equations break down. Please elaborate.

Mike

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What I'm saying is that 23,000 lbs (of tension or compression) to support 2000 lbs makes no sense and is not what happens in the real world.

And I could make the numbers much worse:

If instead of a five degree angle, it were five-hundredths or five-millionths of a degree, the cable and struts would be thousands of times greater to support the same weight.

In other words, the angle theta could be made indefinitely small so as to make sine theta indefinitely small.

For example make it so that w = 2 lbs. and the second equation instead of sine of theta times C = .087C = w = 2 lbs, make theta so small [the angle in this case would be .006 degrees, since arc sine .0001 is .006] that the equation becomes sine theta times C = .0001C = 2.

Then one is concluding that 2,000 lbs of strut compression and cable tension are required to support a two pound weight. Ridiculous!

Edited by Philip Coates

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What I'm saying is that 23,000 lbs (of tension or compression) to support 2000 lbs makes no sense and is not what happens in the real world.

And I could make the numbers much worse:

If instead of a five degree angle, it were five-hundredths or five-millionths of a degree, the cable and struts would be thousands of times greater to support the same weight.

In other words, the angle theta could be made indefinitely small so as to make sine theta indefinitely small.

For example make it so that w = 2 lbs. and the second equation instead of sine of theta times C = .087C = w = 2 lbs, make theta so small [the angle in this case would be .006 degrees, since arc sine .0001 is .006] that the equation becomes sine theta times C = .0001C = 2.

Then one is concluding that 2,000 lbs of strut compression and cable tension are required to support a two pound weight. Ridiculous!

Well when you've got a force at a five degree angle, only nine percent of the force is in the y-direction. And a purely horizontal force can't do anything to counter a vertical force.

Here's a thought experiment, or if you really want to be convinced, give it a try for real:

Take a light weight and attach two strings to it. One outwards to the right and the other to the left. Now suspend the weight by holding the ends of the strings. If the strings are vertical, your hands only have to supply a force equal to the weight of the object. Now place your hands at approx. forty-five degree angles from the horizontal (or vertical I guess!). It's quite a bit harder to hold the weight up, because only half of your supplied force is actually acting vertically. So you literally need twice the force to suspend the object.

Now move your hands such that the strings get closer and closer to horizontal. Because you're essentially dealing with one quarter of the unit circle when moving your hands, you will feel a "parabolic" change in the force required to suspend the object. It's not really parabolic, though, because it will asymptote to infinity as you get closer to the horizontal. It is literally impossible to suspend the object if there is no angle with the horizontal.

(For extremely light weights we can get the string to be apparently straight, but our eyes can only detect angles above a certain threshold. If you want to try this for real, use a weight of a pound or more.)

Mike

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Wow! I can't stand the tension!

I can't wait to see how this is going to turn out!

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Mike, your post very briefly and elegantly explains the principle: "When you've got a force at a five degree angle, only nine percent of the force is in the y-direction. And a purely horizontal force can't do anything to counter a vertical force" and I really like the idea of a thought or actual experiment you can do at home.

(You would have been a better physics teacher than the ones I had in high school and college! We seldom had a chance to actually -see- the physics in action.)

I didn't have string or a heavy enough weight i could tie them to. However, I came up with and performed a basically equivalent experiment this morning: I took several volumes of Encyclopedia Britannica, loaded them in my backpack and tried lifting them in a complete half-circle out to the side --> almost straight down, moving up to and past horizontal, and finally straight up over my head. My shoulder felt the greatly increased strain the closer my arm was to horizontal. {Weight trainers do similar experiments all the time at the gym when they lift two dumbbells with both arms out to the side.}

However, I have a problem with your idea that a purely horizontal structure can do -nothing- to counteract a vertical weight and with your statement that "the force required to suspend the object...will asymptote to infinity as you get closer to the horizontal. It is literally impossible to suspend the object if there is no angle with the horizontal."

My problem with this is that the real world is full of all sorts of exactly and purely horizontal structures supporting weight: A boom coming out from a crane with a wire dangling underneath it. Or a beam supporting a weight.** The support for a dangling theater marquee coming out from a building. In fact, as I did the backpack experiment above, I moved my arm up -through- the horizontal slowly and my shoulder did not experience any impossibility, nor did my shoulder socket rupture as I moved to and past the horizontal - very, very slowly.

**photos of several such structures:

In thinking about this problem, I suspect I may have an answer: The horizontal member is never infinitely thin. It has thickness, and so the force is -never- transmitted purely horizontally even through a horizontal member. Suppose the perfectly horizontal titanium or steel beam below supports a weight:

------------------------------------------------------------------

| A |

| |

| B |

-------------------------------------------------------------------

|

|

|

WEIGHT

Then, according to my hypothesis, the compressive force is not transferred horizontally but diagonally between points A and B.

[i couldn't get my diagram to post properly on OL: Point A should be at the top left of the elongated rectangular boom or strut and B at the bottom right...and the weight is supposed to be dangling off the right edge.)

Does that make sense and solve the 'approaching infinite force' problem? (And when I lifted my shoulder slowly through the horizontal, the force transmitted by the bookbag changed across several diagonals as I rotated my arm.)

Edited by Philip Coates

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It may be a horizontal structure, but the support force it provides (through its supports i.e. the nails holding it in the wall) is vertical.

Just do a force diagram on the backpack with encyclopedias system:

Let T represent the tension in the ropes, and W be the weight of the pack. The resulting force balance is:

T = T (x-component)

W = 0 (y-component)

The only way this system may be balanced is if W = 0. You may have been able to move the backpack above the horizontal for a moment, but I guarantee you will not be able to suspend it there.

However, if we hang the backpack on a bar that is fixed in the wall, then we should consider not the backpack, but the backpack AND the bar. And when we do that, we move from a point mass to a rigid body. (It's not really important in this case, but a rigid body is different because it can rotate, so we obtain one more equilibrium condition: the sum of the torques equals zero). If we write a force balance for the backpack and bar system, and let Rx and Ry be the reaction forces of the wall, we get:

Rx = 0

Ry = W

So while the bar is horizontal, it supplies a vertical force (well, really the wall supplies the vertical force, the bar just translates it). You're thought about the non-uniform translation of shear stress is correct, but it doesn't matter here. Would the distribution of stress matter at all if there was no vertical force countering the weight? If Ry in the equation above is zero, the system cannot hold a weight and be at equilibrium, no matter how the stress varies.

The statements that "a purely horizontal force can't do anything to counter a vertical force" and "the force required to suspend the object...will asymptote to infinity as you get closer to the horizontal. It is literally impossible to suspend the object if there is no angle with the horizontal" are really the exact same thing.

And on the topic of seeing the physics in action, it's definitely very important! I had a terrific physics professor who always had about ten 'toys' out for every lecture. I had an awful physics teacher in high school, though. Maybe one or two demonstrations for a whole year. Pretty hard to learn physics without a large number of guided observations.

Mike

Edited by Mike Hansen

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> Statics....As soon as I saw the words "strut", "heavy weight", and "cable", I knew exactly what was coming. [Mike H]

This one is actually a bit more difficult than the 'run of the mill' of those problems.

> I did about eight thousand variants of this problem

I'm curious if anyone who remembers their Physics 101 is actually able to do this one and explain the steps: There's one tricky part.

It seems Phil has put this problem up on multiple sites, loosing track of where he is. Mike Hardy posts on Atlantis II--or did. I'm simply assuming that's where these quotes are coming from.

Starting to get pissed off at Phil, again, and his tiresome, purblind posting techniques.

--Brant

edit: it's not from Atlantis II

Embarassed to say I got the two Mikes mixed up. Lousy non-use of the quote function, as usual.

--Brant

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I will never post on your site again. I would prefer not to have anything further to do with you or to communicate with you again. Quite frankly, your weakness and double standards disgust me.

Makes me think of the saying in Kölsch (Cologne dialect): "Wat kümmert mich ming Jeschwätz von jestern?" ('I couldn't care less about my gibberish from yesterday').

So is Phil going to apologize to the site owners for his rants against them? How about explaining why he has changed his mind about posting here? It seems not.

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