Homework Help: Energy Required To Knock Something Over

1. Feb 14, 2012

sjd0004

1. The problem statement, all variables and given/known data

I need to figure out how much force it would take to knock a can over.

2. Relevant equations

As of now, I have no clue where to even start with this.

3. The attempt at a solution

I was told by a friend that the way to solve it is by finding the difference between the value at equilibrium and the value when the can is just about to tip. Unfortunately, I have no idea how to do this.

2. Feb 14, 2012

Staff: Mentor

Draw a diagram showing a can standing in a stable position. Locate its center of gravity. Now draw the can in a position where it's tipped so it's just balanced and could go either way...fall over or return to its stable position. Where's the center of gravity in that case?

3. Feb 15, 2012

sjd0004

Okay, I have the diagram drawn. Would I just do the distance between the two points x the force required to get it there? I believe that the formula for work which is equal to energy would be distance * force?

Last edited: Feb 15, 2012
4. Feb 15, 2012

Staff: Mentor

You could do that, but it could involve a fair amount of effort to sort out the math (the force will be changing direction over the distance it's applied). It would be much simpler to apply conservation of energy. In particular, find the change in gravitational potential energy between the two positions of the center of mass.

5. Feb 15, 2012

sjd0004

Once the gravitational potential energy difference is found, would the difference of them be the total needed to tip the can?

6. Feb 15, 2012

Staff: Mentor

Yes.

7. Feb 15, 2012

sjd0004

Thank you. One more question though to make sure I have it down. I know that the gpe would be mass*gravity*height. So for the cans would I use the mass of the can*9.8*center of gravity height?

8. Feb 15, 2012

Staff: Mentor

Right. You use the height of the center of mass of the object to determine the overall GPE for it. While some parts of the object are higher and some are lower (and thus individually have higher or lower GPE's), the sum of the GPE over the whole object will be the same as though all the mass happened to be located at the center of mass.