Tension Question for Ship Container and Wind Gust

[physicsengineering72]

Homework Statement

A crane is lifting a 40' shipping container with a mass of 4,000 kg until the bottom of the container is 100' in the air. The remaining dimensions of the container are 8' wide and 10' tall. The load is exposed to wind gusts of 25 mph, broadside, on the largest face of the container. If there are two
taglines attached to the load as shown below (not to scale), what maximum possible tension is
expected in the taglines during a gust? Assume sea level dry air density and the worst-case loading
scenario. Ignore the very thin rigging.

See the attached image for a better description of the question.

Relevant Equations

F_wind = 1/2 pv^(2)A

I tried a method solving for the force of the wind first which I found to be 2842.95 N. From there I found the torque force using torque = F_wind x d, where I let d = 90ft. This leads to a torque of 77987.8044 Nm. From there I used Tension = torque/moment arm, with a moment arm of 20ft (half of the length of the container since there are 2 taglines). This gives a max tension answer of T = 12997.97 N.

I'm not sure that this is the correct way to approach the problem. Any help is appreciated. Thanks!

 

[haruspex]

Where do you consider the drag to act? What axis are you taking for the torque?
(It is not clear to me that torque is relevant here.)

 

[Lnewqban]

Welcome, @physicsengineering72 !

Could you show us your calculations and your reason for those 90 feet?

 

[physicsengineering72]

Hi @Lnewqban and @haruspex , thanks for the replies! I've attached my work. Let me know if you can see it.

I wasn't too sure about the torque, but the torque would be taken around the axis of the crane. In other words it would torque around a horizontal plane, perpendicular to the direction of the wind, which hits the largest face of the container. I'm not confident that torque is the way to go though; that's what I'm pretty unsure of.

Also, I wasn't sure about the drag coefficient. The question does not provide a drag coefficient, so I wasn't sure if it was needed. If it is, I wasn't sure what value to assume.

I'm realizing now that the 90ft may be wrong, if I misread the question. I was thinking that was distance from the lifting point (axis of torque) to the center of pressure, but I think I misread the question potentially.

 

[haruspex]

the torque would be taken around the axis of the crane. In other words it would torque around a horizontal plane, perpendicular to the direction of the wind,

I assume you mean around a horizontal line perpendicular to the wind, but that only gives a direction; it does not say exactly where it is.
Looks to me that you are overlooking that the support cables may also exert a torque about your axis. The cables on the windward side need not be under the same tension as those on the downwind side. They will adjust to prevent the container from rotating about its axis.

Where do you need to put the torque axis to avoid such a tension difference from mattering?

(Also, I do not understand what “height of the bottom of the container minus height of the height of the container" represents. You should mark d on the diagram.)

 

[Lnewqban]

I'm not confident that torque is the way to go though; that's what I'm pretty unsure of.

Me neither.
I believe that the provided data gives no options for torques.

I would treat the container as a pendulum, which is exposed to a horizontal wind-induced force (properly calculated by you).

Any deviation of that pendulum from a plumb attitude is prevented by the two taglines (which angle respect to a vertical is possible to calculate).

Please, see a scaled diagram of the situation.
Can you see the blue man on the ground who is pulling from one of the taglines?

Of course, I have assumed the length of wire rope from which the container is hanging, which has not been provided.

 

[haruspex]

I believe that the provided data gives no options for torques.

That was my first reaction, but see the question I ask in post #5.

 

[physicsengineering72]

I assume you mean around a horizontal line perpendicular to the wind, but that only gives a direction; it does not say exactly where it is.
Looks to me that you are overlooking that the support cables may also exert a torque about your axis. The cables on the windward side need not be under the same tension as those on the downwind side. They will adjust to prevent the container from rotating about its axis.

Where do you need to put the torque axis to avoid such a tension difference from mattering?

(Also, I do not understand what “height of the bottom of the container minus height of the height of the container" represents. You should mark d on the diagram.)

Regarding the "torque", that is what I meant. Your questions about the torque approach make me think that maybe it is not the best approach to solve the problem. It seems like you are hinting at this?

There wouldn't be enough information from the prompt to reasonably go down a torque pathway.

@Lnewqban @haruspex It seems that the best way would to take an approach similar to the drawing in post #6. The prompt gives enough details to solve for an angle I believe. Though, the only thing that makes me hesitant is assuming the "length of the wire rope from which the container is hanging".

The problem is vary vague, and it's hard to tell how to approach the problem. I'm confident in the wind force calculation, and I am almost positive it will be a big part of the calculation, I'm just unsure of the complete approach.

I'm also unsure whether the weight force needs to be factored in to the calculation. Or does the presence of the crane, and the problem set up let us factor it out?

Thanks for the help thus far. It's helpful to "talk" through it, and the questions are helpful to narrow down approaches.

 

[haruspex]

Regarding the "torque", that is what I meant.

What is what you meant? What did I write in post #5 that you are confirming here?

Please try to answer my question: Where do you need to put the torque axis to avoid a tension difference in the support cables from mattering? There is an answer.

 

[physicsengineering72]

What is what you meant? What did I write in post #5 that you are confirming here?

Please try to answer my question: Where do you need to put the torque axis to avoid a tension difference in the support cables from mattering? There is an answer.

I had been referring to the fact that I thought the torque axis would be placed at the crane, but like you point out, I need to consider that the "support cables may also exert a torque about your axis".

For this reason, I would say that the torque axis would need to be placed at the center of mass of the container (which would be the center of the container assuming even weight distribution). The torque axis would be horizontal and perpendicular to the wind. Thus, we would have direction and location of the torque axis.

 

[haruspex]

I would say that the torque axis would need to be placed at the center of mass of the container

Not the best.
To avoid having to consider the support cable tensions you want an axis about which they
exert no torque. Where would that be?

Edit…
but I don’t want to waste your time on a dead end. My view is that the right approach does involve torque, but that you have not been given the relevant data for that.
The fallback is to assume that the crane is much higher than the container, effectively at infinity. In that view, forget torques and just look at horizontal force balance.

 

[physicsengineering72]

Not the best.
To avoid having to consider the support cable tensions you want an axis about which they
exert no torque. Where would that be?

Edit…
but I don’t want to waste your time on a dead end. My view is that the right approach does involve torque, but that you have not been given the relevant data for that.
The fallback is to assume that the crane is much higher than the container, effectively at infinity. In that view, forget torques and just look at horizontal force balance.

Yeah I think I was just struggling with the torque concept since it seemed to be a dead end unless major assumptions were made as you point out. It's a poorly framed question in my opinion.

I will go ahead and use a horizontal force balance approach. In your opinion would you consider using the angle of the tagline with respect to the vertical, as @Lnewqban might've suggested?

Thanks for the help!

 

[haruspex]

I will go ahead and use a horizontal force balance approach. In your opinion would you consider using the angle of the tagline with respect to the vertical, as @Lnewqban might've suggested?

Yes.