Science a GoGo's Home Page Forums General Discussion General Science Discussion Forum Methane Clathrates in Gulf and Worldwide

well we could put a weight scale behind each spring.
then put the static 100kg mass on it , and each scale would read 25kg.

at least thats what should happen.

but thats because of gravity , in space zero g the mass wouldnt
show as having a weight.

What difference does size make? You haven't specified a size for your device, and you say it'll work, therefore you think it'll work for any size.

Or do you mean materials might not be strong enough to support higher forces? Well a tree can survive a car hitting it, so we could even make them out of wood.

You know the basic idea you're using is correct.

The only difference between your use of it and the correct way is you're using the _magnitudes_ of the forces. The correct way uses the complete forces - magnitude and direction.

Do the same maths with vector forces instead of scalars and it comes out right.

I never got a chance , you imediately decided that
it wouldnt work , and from the information that I
posted , you said it wouldnt work so I think its you
that thinks it is a 1 size fits all.

of course its not , but telling you that wont do
much good either will it.

I think you have that backwards , I've always used magnatudes and directions , you always just say that all the force is directed backwards.

a force cant travel through nothing , it must act on an object.
and the outward forces act on objects not nothings as your
backward forces act.

then the outward forces act on the pipe and then the springs expand tossing the mass back into the accelerator.

I would kind of enjoy seeing how you would do that.

why dont you do it , and include all forces.

just use each force and the direction of each force.

we can add the resistance to the forces after
you have that done.

and its in space so you dont need to include any g forces or forces from mass due to g.

that is why we must start with the 4000N force and its direction because its in space in zero g and we cannot use static forces because there are none in zero g that would apply in this case.

yes they are sloping and the force is directed through the slope.
inside the slope the direction of the force is downwards and lateral.

I picked this picture for that purpose.



I think if someone stood on this table and jumped upwards
when they landed on the table the outward forces applied to the legs
would bend the legs causing them to be sent outwards
from the center and the legs would break.

much in the same way that the 4 springs would get
the brunt of the 4000N force.

OK I'll do it myself. Here's my FBD.

Can we just have 2 legs instead of 4? It can just have sliders to support it from collapsing into/out of the screen.



The variables F1, F2 and F3 are just the magnitudes of the forces, and they could be zero if need be. The arrows show their directions on each member.

Not sure how you want me to calculate the forces. I could treat it statically then it won't be deflecting, just statically supporting the 4000N. It might get a bit tricky to do dynamically - the 100kg mass would have to be included and it may not be a constant 4000N.

Notice that it's possible to choose values for F1, F2 and F3 so that all forces balance on all members, allowing it to support a static load if need be. If F3=0 then it can't support any static load because there's no way to choose F1 and F2 so the blocks have zero net force on them.

I think we really should sort out how it'll behave statically before trying to do it dynamically. Even tho it's not on earth, any real thing can, in principle have a static load applied to it. Maybe from gravity or maybe from a little propeller attached to it, or a big magnet, or whatever.

Sure there'd be outward forces. I'm not disputing that.

But the downward forces wouldn't be in any way reduced. If the floor was just weak enough to break when someone jumped on a straight-legged table, then it'd equally break when they jumped on this sloping-legged table.

so your saying that F1 stays constant with the force applied at the point where the 4000N is applied.

since the F1 force pushing on the block is
pushing downwards at a 45 degree angle shouldnt this force continue through the block , which will cause the F3 force
to be placed in the F3 direction , and shouldnt there also be a F3 force pushing down on the opposing side of the block?

the 2 F3 forces on the block causing a torque that
would balance

then the two F3 forces should balance.



in the above , the F1 force is directed through
the block causing torque t to be applied around
the center of mass of the block , this torque is translational to the downward force F3 which
results in the upward force F3 , the two forces total 0.

also you left out the opposing F2 force arrows , I drew
them the same size as we are not including the magnatude
of the forces only there direction.

so in the above the F3 forces can be striken , and unless we are
going to include frictional forces only the
F1 and F2 forces remain , heres where the magnatude of the
forces come in to decide if the block will move outward
I say it will move , how bout you?

Paul, recommendation.

It has been decades since I studied statics or dynamics, but I will tell you this - you cannot solve this kind of problem consistently and correctly without free body diagrams. It is usually the first or second step to solving mechanics problems.

However, you cannot correctly set up free body diagrams without a clear understanding of vectors. I have a lot of difficulty following what you guys are arguing about, but I think kallog's FBD was correct the way it was. The red F2 is the opposing force to the component of F1 in the x direction. F3 is the opposing force for the component of F1 in the Y direction. The black F2 and F3 are not extra forces - they are the components of F1 in the x and y direction which means they can replace F1.

I admire that you're trying to wrestle this on your own, but you should focus on a few vector videos before trying to use free body diagrams.

No. I just showed a 4000N input for clarity. F1 stays constant as long as the input stays at 4000N. Obviously if you increase the input force then F1 increases also. But then the diagram doesn't apply because it only says 4000N.

The equal F1 forces everywhere also assumes the diagonal linkages are massless, or it's a static analysis. I hope we can treat them as massless, if not the whole thing.


It does. For a static analysis, the vector sum of F2 and F3 is a force antiparallel to F1 and with the same magnitude as F1.



There's nothing there to push the block down, so no such force can exist.


Again, there's nothing there to apply a force purely in that direction. The only things touching the block on that end are the base, which, being slippery can only apply an upward force (F3), and the diagonal linkage which is only capable of transmitting force parallel to its length because it's pin-jointed (F1).

The horizontal component of that F1 can balance F2. No need for an extra F2.

Haha I might have got a bit carried away here, but I made a computer simulation. I had to adjust the spring constants to get approximately 2s turnaround time.



Here's the reaction force measured at the point shown on the diagram. It's the same as on the opposite side. All other vertical reaction forces are zero. The horizontal reaction forces at the ends of the springs depend on the angle on the diagonal members, but they came out at an average of 1500N each for the geometry shown. That kind of makes sense because they're a bit steeper than 45deg. I suppose at 45deg, there'd be 2000N on each spring.




The model assumes small displacements for everything, so it might be a bit inaccurate. However the larger you make the real thing, the closer it'll approximate the model.

thats nice kalog.

but what does it mean?

does the 3200 mean 3200N?

you used fully constrained , does that mean that you
removed the springs?

and the force is just stopped there.

by the way the slides arent just laying against a floor
or something they are in space they must ride on some type
of track that would hold them from floating around.

so there is a torque from the 45 degree force placed on
the slider.

you said you made a program to get your results , did the
program take distance into consideration?

ie... how far did the slider move outward?

I cant see how you got 3200N from 1 leg if that is what
you are saying.

or trying to say.



if your working a 2 second time frame you should use a distance that would equal 40 meters in that 2 seconds
I would like to know how much your distance was.

or at least the displacement in each direction not both
together and then say it was 0 , as you mostly do.

the distance that the 100kg mass moved in the
(-) direction.


when you did get the 2 seconds what spring constants did
you use?



the force F1 goes to zero in 1 second , did your program take that into consideration?

your diagram doesnt look like it would fit.

at 1 second your showing the highest vertical forces
as the 4000N has been reduced to zero , how do you account for that?

Yea I suppose I didn't really explain it properly.


Yes. That's the peak force, occurring in the middle of the motion.


The line segments closest to the ends are the springs. So the outside ends of the springs are fixed. That's also where the horizontal 1500N (average) reaction force occurs. The inside ends are free to move horizontally, and are connected to the blocks.


Yep, the 4 bottom elements are all prevented from moving vertically, but free to slide sideways.


There could be depending on the details of the connection. I assumed there's some slider material directly below the diagonal linkage's connection. That way no moments are applied.


It's a general purpose program I used. No, that's one of the weaknesses. It assumes the digonal linkages have the same fixed angle when finding the forces, tho is does have them moving. The springs compressed a fairly spectacular amount, but that would just be because I made the dimensions quite small. If it were all scaled up a lot, there'd be hardly any movement of anything (relative to the size).



Yes. 3200N on each leg. So 6400N total. Of course that's only the maximum and only occurs for an instant. Most of the 2s are spent at less force.

The average is 2000N/leg. Which is really quite close to the 4000N total I've been saying all along :P

your diagram should look more like a big "M" instead of
a mountain.

the largest forces your showing as resultant forces
dont have anything to result from.

the 4000N is zero at 1 second , you must have the program
wrong somewhere.



that 1500N average would also be wrong.



so the torque is present.
and there is a matching F3 force , that is striken then.



whats the name of the program?
can it be downloaded?
I would like to plug in your numbers to see what results I
can get from it using different spring constants.



you must be using only 2 legs , thus the 6400N
that couldnt be the result from a force of 4000N
applied at a angle of 45 degrees.



we'll just have to wait and see , I think your program
doesnt include the proper elements or you wouldnt get
such large reaction forces when there are no action forces
present.

Sink in the middle?? There's no joint that would allow that. Or do you mean the blocks tilt? I assumed the blocks were prevented from tilting. If they can then it's unstable because the springs will just pop them out of straightness as soon as any load is applied.



Not at all. 1 second is where the springs are the most compressed. Obviously they have quite a lot of force on them at that time.

But anyway, that simulation was a bit of a side-track. It won't prove anything because nobody's going to go through all the code checking it.

We should focus on the static analysis since we still disagree even on that.

The computer says there's no force there, It just needs to be restrained vertically to prevent instability. The FBD says there's no moment applied to the blocks, so there shouldn't be any anyway.

so the springs are at the point where you measured.
the legend of your diagram only shows the verticle reaction force.

there is no lateral force mentioned.

basically what your saying is that the springs are
compressed with 6400N , for 2 springs.

I dont know kallog it sounds suspicious to me.



I think that without the math to examine , I will have
to decline your programs results.

and the computer program can be designed to neglect
forces that would be considered elements in a force diagram.

so I dont think the results are viable.

theres almost allways a force there , wherever there may be.

Oh yea I didn't mention that. When I looked at the animation it sunk down then bounced up. The whole thing took about 2s and it was at the bottom of its deflection at 1s.


No, that's the vertical force. The springs got an average of 1500N each.

Nonetheless, it's pretty amazing that the vertical force averaged to 4000N on both legs. I didn't have to tweak anything (except the springs to get the 2s you specified), despite the arbitrary angles of the legs and arbitrary overall dimensions.


Fair enough. Lets focus on the static. Can you respond to my last message about that?