I want to know all of the maths concerning this scissor mechanism!
- Thread starter Maglatron
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this might sound stupid but if the forces are in equilibrium (downwards force 2.7N and upward force from the edge of the cam 2.7N) how/why will the scissor/cam mechanism move the weight upwards at the pivot? can someone explain please, thanks
so I had a look around and found that I do need an out of balance force
the 3.05N as I said before will accelerate the block upward at 1.26m/sec^2 so to rise 1.26m It would take a second!
now all I need to know is how to relate the wheel of 10.8kgm^2 inerta to the cam and then to the weight through the scissor
so I had a look around and found that I do need an out of balance force
the 3.05N as I said before will accelerate the block upward at 1.26m/sec^2 so to rise 1.26m It would take a second!
now all I need to know is how to relate the wheel of 10.8kgm^2 inerta to the cam and then to the weight through the scissor
Last edited:
It won't. As I said originally, "If the platform is stationary and there are no horizontal forces acting on the mechanism, then the vertical upward force at each of those pivots must equal the weight of the (loaded) platform.".
The applied upwards force must exceed the downward load if you want the load to move up.
Hello,
Wich DNS is set in Brave?
You can change the DNS to OpenDNS using the following steps:
Goto Settings , then Brave Shield & privacy , then Use secure DNS.
Tick the Choose other provider and press the triange and choose OpenDNS.
Bertus
Wich DNS is set in Brave?
You can change the DNS to OpenDNS using the following steps:
Goto Settings , then Brave Shield & privacy , then Use secure DNS.
Tick the Choose other provider and press the triange and choose OpenDNS.
Bertus
so the torque into the cam is positive but the same torque is negative from the flywheel and negative torque/inertia of the flywheel gives decceleration I need to find the torque required to give a force of 3.05N at the top of the snail/drop of cam and this torque will be efficient enough because it's the max radius of the cam! so the flywheel is turning 3.3873422rad/sec and its inertia is 10.8kgm^2 how do you work out, from the desired torque at the cam to produce a force of 3.05N radially (upwards) I need a force to relay back to the flywheel to work out it's acceleration or rather decceleration the faster I can do this action the less velocity the flywheel will lose! thanks could even incorporate gears and have the radius of the gear to be 0.01 on the flywheel and add a compound to make the cam turn faster
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Hello,
Here is a page about the error you get and possible solutions:
www.howtogeek.com
Bertus
Here is a page about the error you get and possible solutions:
How to Fix the NET::ERR_CERT_AUTHORITY_INVALID Error
It's possible your computer may be causing it!
www.howtogeek.com
Bertus
Hello,
Here is an other page about clearing the cache in firefox:
www.howtogeek.com
Bertus
Here is an other page about clearing the cache in firefox:
How to Clear the Cache in Firefox
Keep Firefox running neat by clearing the cache once in a while.
www.howtogeek.com
Bertus
tried it, believe it or not I've been getting this error on certain sites since I've had my connexion kinda just gotten used to it although you given me a few more options to explore I think I've spent enough time trying to fix it and geting on with it regardless but again thank you for the time and effort. If you have any info on the cam normal to force of 3.05Newtons (what torque I'd need) I'd be well happy
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