Energy Required To Knock Something Over

[sjd0004]

Homework Statement

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

Homework Equations

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

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.

 

[gneill]

Homework Statement

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

Homework Equations

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

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.

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?

 

[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?

 

[gneill]

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?

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.

 

[sjd0004]

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

 

[gneill]

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

Yes.

 

[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?

 

[gneill]

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?

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.