Topple a table


by kopplaw
Tags: table, topple
kopplaw
kopplaw is offline
#1
Sep25-04, 11:18 PM
P: 3
I am trying to determine how much force is necessary to make a round table topple over, or what information I need to make such a determination. The table is 30" in diameter and 1 1/4" thick, made from medium density fiberboard (MDF). The MDF has a density of 39 lbs/foot3. The table has three legs, equidistant from each other, placed at the edge of the table. The legs are made of aluminum, but I do not know their weight or exact size. The legs are 22 1/4" in height and are secured directly to the bottom of the table.

Is there a way of roughly calculating how much weight would be required to topple the table if the weight were placed precisely between two of the legs at the edge of the table?
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vsage
#2
Sep25-04, 11:49 PM
P: n/a
Interesting question. I haven't actually thought about something like this before but here's a go at it:

Since the legs are made of aluminum you might as well just discard them in the calculation. Look at the table from the side in such a way that you only see two legs (one leg is obstructed from view I see this happening on the left side in my mind). You notice the edge of the table (again on the left side in my mind) extends (1-sqrt(3)/2)*15" farther than the leg. Also since the table is so thin just disregard the table thickness and condense it to a line from the side. I hope I gave enough information that you can reduce it to a simple torque problem. That should give a rough estimate
Tide
Tide is offline
#3
Sep25-04, 11:50 PM
Sci Advisor
HW Helper
P: 3,149
Try calculating the torque about a convenient point e.g. on the floor at the midpoint joining one pair of legs at their bases. Imagine pulling from the edge of the table perpendicularly to the line on the floor. There are now two forces contributing to the torque - one from the applied force and the other from gravity acting effectively through the center of mass of the table.

There are two stages. When the first threshold is crossed the odd leg will no longer exert any force on the floor so we no longer need to consider it. As more force is applied the table will rise but will fall back to its starting position if we release the force. If we pull hard enough so that the torque pulls the table up to a point where its center of mass is directly above the line on the floor then we will have reached a state of equilibrium where the table could fall either way if the applied force is released. This is the force you're looking for since any additional force will cause the table to topple - away from its starting position.


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