Calculating Table Torque to Tip Over: 36 kg Weight

[Jacob87411]

A round table weighing 36 kg is supported by three equally placed supports. How much weight is required to tip the table over if the weight is placed equally inbetween two supports?

The answer is 36, I am curious if it is just a coincedence that its 36 or if it is always equal to the weight. Say if the table weighed 48 KG would it take 48 to tip it, and how would i start proving this mathematically

 

[Yapper]

I don't understand this problem if the weight is placed inbetween two of the supports why would it tip over? since the weight is on the pivot point. Maybe I don't understand the configuration of the legs and where the weight is placed. But if that's all the information the problem gives you then I wouldn't be surprised if it is always equal to the weight

 

[wais]

It is not just a coincidence that the weight required to tip the table over is equal to the weight of the table itself. This is due to the principle of moments, which states that for an object to be in equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments. In this case, the weight of the table is acting as a clockwise moment and the weight placed between the two supports is acting as a counterclockwise moment.

To prove this mathematically, we can set up an equation using the principle of moments. Let's assume that the weight of the table is W and the weight placed between the two supports is x. We can then set up the equation:

W x distance from support + x x distance from support = W x distance from support

Since the table is supported by three equally placed supports, the distance from each support to the center of the table will be the same. Therefore, we can simplify the equation to:

W x distance + x x distance = W x distance

Now, we can solve for x by dividing both sides by the distance:

W + x = W

Subtracting W from both sides, we get:

x = 0

This means that for the table to remain in equilibrium, the weight placed between the two supports must be zero, which is not possible. Therefore, the weight required to tip the table over will always be equal to the weight of the table itself.

To answer your second question, if the table weighed 48 kg, it would indeed take 48 kg to tip it over if the weight is placed equally between two supports. This can be proven using the same equation and following the same steps as above. This principle can also be applied to any number of supports, as long as they are equally placed and the weight is distributed evenly between them.

In conclusion, the weight required to tip a table over when the weight is placed equally between two supports will always be equal to the weight of the table itself. This can be proven using the principle of moments and is not just a coincidence.