![]() So with that, we can find the answer easily with this: If you choose pounds (weight), feet (length), and foot-pounds (torque force) as your units, everything works out nicely. And torque in this situation manifests as a force pushing horizontally. Distance is the distance to the center of mass, which is half of the total distance. ![]() We can use weight instead of force because the only force is gravity acting upon the mass-a.k.a. (1/4 the length of the fully extended door) The center of mass is halfway between the two hinges.The center of mass is also 90° out from the top hinge (in reality it will be a little lower).The door is at a 90° angle (perfectly horizontal). ![]() The greatest forces will occur at the top of the track. So you'll need to restrict the door opening with a stop in the track somewhere to keep things within reason. If the door is open 89 degrees, then the force at R is 57/2 W. If the door is fully open, tan(90) -> infinity and the universe ends. Sum Torque = 0 -> (L/2 sin(theta) * 2 W)(clockwise+) - R 2L cos(theta)(counterclockwise, -ve)Ĭonsider the results - if the door is closed, theta is zero and tan(theta) is zero. ![]() Take the sum of the torques about P (hinge pin at top) The forces P and T disappear as the moment arm is zero. The X component of the length will be L sin( theta). Thus, when the door is closed, theta is 0, and when open all the way (impossible in real life) theta is 90Ĭonstraint: The door will form an isosceles triangle at all times when closing. Let Theta be the angle between vertical, and the upper portion, measured from the opening. Assume: The two halves of the door are of equal length and weight. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |