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using force to tip over an object
Mahir Abrahim Jan 14, 2008 10:04 PM (in response to Andrew Thompson)Try pointing the tipping force just slightly upward to nudge Cosmos into lifting it instead of squishing it into the floor. 
using force to tip over an object
Ian Hogg Jan 14, 2008 11:58 PM (in response to Andrew Thompson)Hi Andrew,
A standard approach to this is by various companies is using a "tilt table". What they do is mount the object on a table that has a hinge at 1 end and a ram or somethim tilting it up. As long as you know the weight of the object, you can determine the angle at which the object starts to rotate off the tile table. Use trigonometry to work out the force that acts at the CG (parallel to the table) when the object starts to tip.
In COSMOSMotion, you add a hinge/revolute joint to a point on the object closest to the table hinge line and use 3D contact (or an impact force on the furtherst point) to stop it falling through the tile table. Then euse either a rotart motion on the tilt table hinge or a transaltional motionon an actuator driving the tabel to force it through a rotation as a function of time (ensure the time is suffiicently long to exlude acceleration due to rotation, or rotate at a constant velocity) and then just observe when the rotation of the object relative to the table starts.
I like this approach because it is consistent with experimental tests, across all objects, and takes out the guesswork of the force needed to tilt the object.
Hope it gives you some ideas.
cheers,
Ian
using force to tip over an object
Andrew Thompson Jan 15, 2008 3:12 PM (in response to Ian Hogg)Thanks for the tips. I set up a tilt table, driven at the hinge by a 1rpm motor.
On one particular design, I found that the sample begins to tip at 24 deg. The sample's mass is 26lbs.
My physics is many, many years rusty, so I'm not clear about how visualize the force vectors and solve for the unknown, namley the minimum weight required to tip the table.
Additional help will be greatly appreciated.
Best,
Andrew...
using force to tip over an object
Ian Hogg Jan 22, 2008 3:26 AM (in response to Andrew Thompson)Hi Andrew,
I think I may have gotten ahead of myself. Looks like my hand calcs are getting rusty also.
At an incline when the cg is over the pivot point (ie at the point of tipping), the force acting parallel to the tilt table is mass*gravity*sin(angle)
This is what I was thinking, but I think I am missing something nad need to sit down and go through it again.
For a box just sitting on the ground, assuming that y is the height of the cg above the tipping point, and x is the horizontal distance of the cg away from the pivot, and F is the horizontal force at the cg to tip the object, mg*x  F*y = 0. So F > mg*x/y.
Now sin(angle) = x/sqrt(x^2+y^2), so this is different to x/y (hence I still need to see where I'm going wrong on the calc with the tilt table).

