Statics Problem: The table shown weighs 30 lb and has a diameter of 4 ft.?

1. Draw a line joining any two of the legs.

You can put as much load as you want anywhere along this line without tipping the table, but as you move the load out from this point the table will begin to be unstable.

2. If you calculate the distance from the center of this line to the center of the table you will get 12",

(12" = 24" x sin 30deg.)

3. The moment about the line due to the weight of the table which acts at the center of the table = 12" x 30# = 360 inch pounds.

4. Therefore as long as the moment due to the 100 pound load about the line is less than 360 inch pounds the table will not tip, but as soon as the moment due to the 100 pound load exceeds 360 inch pounds the able will start to rotate or tip.

So as soon as the 100 pound load exceeds 3.6" from the line the table will start to tip over.

That distance is 3.6" + 12" = 15.6 from the center of the table.

The maximum value of "a" is 1/10 of the radius of table

the center of gravity of table is on the center but we can also imagain it as 3 equale loads on the theree vertics then on C 10Ibs on D 1000Ibs then a*100 = r*10 where r is the radius

tehn a = 0.1r ##

Unless we know the weight of the leg that will be lifted with the force, we can't calculate the answer correctly.

After we know the weight of the legs, we can subtract that from the weight of the tabletop to calculate the moment arms that will cause tipping.

Sorry that we can't give you a better answer with the incomplete problem statement.