Hi guys, is it required to apply external gravity force when using remote mass?I been confused about it, because not if not applying gravity load direction, how will solidworks detect the direction of load object.
and next question is, how to simulate 20g force on that part that has treated as remote mass.
thanks in advance.
A remote mass is just that, a mass. Without acceleration (gravity), Force=Mass x acceleration is zero.
You can tailor the direction of gravity/acceleration when you define a gravity load in the Property Manager. It allows entry in x, y, or z directions. The term "Gravity" is confusing when not referring to earth's gravitation field, so think of it as an acceleration vector.
Unfortunately it does not accept entry in units of "g" as in your 20g load/force. You must enter 20 x 9.81 m/s^2 to reference multiples above standard gravity. And yes, build a simple model to exercise these concepts because they are not intuitive. If working in English units a refresher on Lb force versus Lb mass would be helpful.
The answer to the first question is "yes".
The answer to the second question is to apply gravity in the direction of your choosing and multiply the number shown times 20. (9.8*20=196)
Ok, that's assuming that 'real' gravity is acting in the direction you chose, i.e. dropping a ball. Otherwise the remote mass method won't work, i.e. car crashing into a tree.
this comes up quite often and there should be several posts about it
when performing an impact type analysis, it is best to use linear dynamic rather than "fake it" with a gravity load
and gravity needs to be on for a mass to be affected by it, remote masses are no different. otherwise it is a remote force.
Why not just conduct a simple experiment?