Crushing Multiple Cans: 8n Force Needed?

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Dan32
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This is possibly a very silly question, I am not experienced in this field.

I am creating a Can Crusher for a project.

If it takes 8n to crush a can. Will it still take 8n to crush multiple cans joined together end by end?
Also, if multiple cans were being crushed on a platform, against another platform. Let's say, 4 cans on the platform side by side, all vertical. Will it take 8n to crush all 4 cans side by side?

Thanks,
Dan.
 

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Have you tried doing a free-body diagram of the forces on the cans in both configurations?
 
Dan32 said:
This is possibly a very silly question, I am not experienced in this field.

I am creating a Can Crusher for a project.

If it takes 8n to crush a can. Will it still take 8n to crush multiple cans joined together end by end?
Also, if multiple cans were being crushed on a platform, against another platform. Let's say, 4 cans on the platform side by side, all vertical. Will it take 8n to crush all 4 cans side by side?

Thanks,
Dan.

Tim's hint is good for your question. In addition, though, on a project optimization issue, are you just pushing down on the cans to crush them, or are you doing something else first to lower the force (significantly) needed to crush the can?
 
CanCrusher.png


Is this correct? So no matter how many rows or columns long, It will crush them all?

berkeman said:
are you just pushing down on the cans to crush them, or are you doing something else first to lower the force (significantly) needed to crush the can?

Originally, I designed it to crush the cans at an angle, to encourage the cans to buckle easier. However, It is much more efficient and easier to crush multiple cans at once.

Thanks for the help
 
Surely each column has the 8N crush strength, so multiple columns will require multiple times 8N to crush reliably. Obviously there could be uneven stacking, so one column is carrying the load for a bit, but once the load is shared, 8N will not do the job.