4 Replies Latest reply on Jan 22, 2008 3:26 AM by Ian Hogg

    using force to tip over an object

    Andrew Thompson
      Got another newbie question:

      I'd like to test various table designs to see how much force is required to tip it over. I've set up a few simulations in cosmosmotion but am getting seemingly inconsistent results.

      Here's my setup:

      1. fixed part acting as the floor
      2. the table design (either multibody part or an assembly)
      3. gravity applied downward and normal to the floor plane
      4. force applied downward and normal to a specific area of the table top
      5. 3d contact between the floor and the table base (or in the case of the multipart body, the entire body). I've tried this several different ways, using material settings, friction, etc. and also by just setting the elasticity of the interaction.

      Results:

      Sometimes it works; other times the table will pass through the floor, not totally freely, but more like dropping a table through jello.

      Any ideas? I'm sure I'm overlooking stuff.

      Thanks again.

      Best,
      Andrew...
        • using force to tip over an object
          Mahir Abrahim
          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
            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 3-D 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
                Thanks for the tips. I set up a tilt table, driven at the hinge by a 1-rpm 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
                    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).