By non-physicist: weight at 2 ends of 45°

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In summary, the conversation discusses the forces on a stack of planks that are attached with wedges and nails. There is a question about the distribution of weight on the different surfaces at different heights. It is suggested to take the nails out of the equation and only consider the wedges, but it is then pointed out that the system would not be stable without the nails. The conversation then shifts to discussing a trailer and the distribution of weight on its wheels and hitch. It is determined that the load is not uniform and the wheels will take the greater part of the weight.
  • #1
DynV
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It's hard to do the Relevant equations & The attempt at a solution as I know very little in physics. So I'll stick with the 1st part of template.

Homework Statement



Attached is a drawing of the basics of my issue. It's a stack of planks at 45° with 45° wedges, making the former snug the surfaces ; then nails are ran at 90° through the planks aligned so they arrive at the wedges right angle. What force would be applied on the surfaces at different heights (different hues of orange)? In other word, if you'd cut the nails at the wedges right angle, then put a scale between the wedges and the surface, what kind of numbers would the scales show?

Again, I'm very green at physics but to be the bottom end would have more weigh then the top one. Please simplify as much as possible and don't bother too much for the detail, don't be afraid to use ranges. As long as the general idea is explain, I'd be happy with that. So to me, the bottom would get 2/3-3/4 of the weight, toward the upper range.

Thank you kindly

PS: Hmm! Perhaps I was wrong to nail the planks like that. See attached smaller image. Perhaps the nail should enter upright in the planks, so at 45°, then go through the wedge along one side, and perhaps around where areas from both side would be about the same ; . That make more sense. But that's likely just a structural issue, one not lasting longer than the other, and won't affect the main point: forces on the surfaces (hues of orange).

PPS: Perhaps the nails shouldn't get out of the wedges at all. I've highlighted the former to emphasize that.
 

Attachments

  • 45_degrees-nailed_plank_stack.png
    45_degrees-nailed_plank_stack.png
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  • upright_nail_in_wedge.png
    upright_nail_in_wedge.png
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  • #2
It's not straightforward. One nail would hold things. Typically, whichever nail you put in first will be taking the weight, and unless you do something to prevent it (applying some other force during nailing), the second nail will take much less at first. However, as soon as there's an increase in load, straining the first nail a little, the second nail will start to take up more share.
There's no inherent reason why the higher or lower nail would be subject to more load by virtue of its position.
When it comes to figuring out what the load is, it will of course be a shear load.
 
  • #3
Please refer to the PPS stating that the nails should be taken out of the equation and only consider the wedges. That the small triangles should simply set on the surfaces, not attached to them.

What you're stating make sense ; and I suspected something about my premise was wrong so I tried to simplified even more (PPS).
 
  • #4
DynV said:
Please refer to the PPS stating that the nails should be taken out of the equation and only consider the wedges. That the small triangles should simply set on the surfaces, not attached to them.

If you take the nails out, why won't the planks slip down?
 
  • #5
The wedges would make the assembly flat. Wouldn't that, considering only the green (including the different hues) in the example, stand without being attached (ie: glued)?

So let's forget about the grey in the sample, and assume the oranges are attached together and so are the greens. Then try to figure out the individual weight of the top & bottom portion of the green contraption.
 
  • #6
DynV said:
The wedges would make the assembly flat. Wouldn't that, considering only the green (including the different hues) in the example, stand without being attached (ie: glued)?
No. Ignoring friction, the forces from the wedges onto the planks are normal to the surface of the planks. The force of gravity on the planks is straight down, and if the planks are touching the floor there may be a vertically upwards force from there (again, ignoring friction). So when we look at the horizontal components of the forces there's an imbalance, and the system cannot be stable.
 
  • #7
haruspex said:
No. Ignoring friction, the forces from the wedges onto the planks are normal to the surface of the planks. The force of gravity on the planks is straight down, and if the planks are touching the floor there may be a vertically upwards force from there (again, ignoring friction). So when we look at the horizontal components of the forces there's an imbalance, and the system cannot be stable.

The planks shouldn't touch the floor, only the wedges should, unless I'm not getting it right. Also, maybe I misunderstood you, but there would a nail between the wedges and the planks, albeit not pass the wedges (into the floor).
 
  • #8
DynV said:
The planks shouldn't touch the floor, only the wedges should, unless I'm not getting it right. Also, maybe I misunderstood you, but there would a nail between the wedges and the planks, albeit not pass the wedges (into the floor).
Ok, that changes things. So you're effectively reshaping the planks to have horizontal ends. In that case, the loads are the same at both ends. You can see this by taking moments about the centre of the plank. The vertically upward force at each end is the same horizontal distance from the centre, so if they were different there would be a net turning moment.
 
  • #9
Ok that was for the 1st step.

Now what about http://www.burley.com/home/bur/page_10297_316/travoysupregsup_yellow.html, is the load on the trailer wheels about the same as on the trailer hitch (on the seat-post in the page configuration) ?

Thanks
 
  • #10
DynV said:
Now what about http://www.burley.com/home/bur/page_10297_316/travoysupregsup_yellow.html, is the load on the trailer wheels about the same as on the trailer hitch (on the seat-post in the page configuration) ?
No, this is different. The load is not uniform. Its mass centre will be not far from the trailer wheels, so they will take the greater part of the load.
I guess it has a universal joint near the saddle post. Note that you'll need to take a wider line around corners when cycling with this attached.
 
  • #11
Yes a lot will be put on the small wheels but some will have to be on the large ones as it's leaning forward. Now I'm wondering about how much will be passed to the bike. Because I have bike racks & saddles and can manage with 100% on the bike but if over 2/3, possible 3/4, is on the trailer, I'd think about spending on that.

So if I pack a 50 lbs load, on the 9 lbs trailer, packing the heavier stuff at the bottom, but still needing to have some weight up, about how much would end up on the bike anyway?
 
  • #12
You need to make some estimate of what the distribution would be. Judging from the picture, you'd expect 75%-90% of the weight to be on the trailer wheels.
 
  • #13
I'd put a laptop & accessories in suitcase, a couple media-sleeves (cd), 1-2 book, tightly packed (regular) clothes, 1-2 water bottle, 1-2 meal, etc. So as much as I'd try to put the weight near the bottom, near the wheels, I couldn't. Does that give you a better idea of the weight distribution, allowing you a better estimate?
 

FAQ: By non-physicist: weight at 2 ends of 45°

What does "weight at 2 ends of 45°" mean?

This phrase refers to a specific scenario in which two objects are connected by a string or rod and are positioned at 45° angles. The weight of each object is then measured at the two ends of the string or rod.

What is the purpose of measuring weight at 2 ends of 45°?

This type of measurement is often used in physics experiments to determine the weight distribution of an object. It can also help to calculate the tension and forces acting on the string or rod.

How is the weight at 2 ends of 45° calculated?

The weight at each end is calculated by using the formula W = mg, where W is weight in Newtons, m is mass in kilograms, and g is the acceleration due to gravity (9.8 m/s²). The weight at each end can then be compared to determine weight distribution and tension.

Can this measurement be used for any type of object?

Yes, this measurement can be used for any type of object as long as it is connected by a string or rod and positioned at 45° angles. It is commonly used for simple systems in physics experiments, but can also be applied to more complex systems with multiple objects and angles.

What are some limitations of measuring weight at 2 ends of 45°?

This measurement may not accurately reflect weight distribution in more complex systems with multiple objects and angles. It also does not take into account other forces acting on the objects, such as air resistance or friction. Additionally, it may not be useful for determining weight distribution in non-static systems.

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