Calculating Height of Resting Ball Using Surface Area and Volume Equations

[ojsimon]

See the diagram attached. I am trying to find the height marked with a red arrow to height where the ball on top is resting. Each ball has a radius of 1cm... How would you do this.

I thought it may be something to do with the surface area or volume, do worked with the SA and Volume eauations of a pyramid, but had too many variables, although i may have missed something.

Does anyone know how to find this?

Thanks

 

[mrbohn1]

The diagrams seem to imply that the top ball is resting slightly skewed on the other 4. Assuming this is not the case however, this can be solved with simple trigonometry and pythagoras.

Joining the centres of the 3 circles in the second diagram creates an equilateral triangle, with all edges 2cm and all angles 60^o. Now draw a right-angled triangle connecting the centre of the top ball straight down to the ground, and then to meet the continuation of the line between the centres of two of the circles.

It remains to find the length of the line connecting the centre of the top circle to the ground (then just subtract 1 for the desired length). This can be done by considering the smaller congruent right-angled triangle formed by similarly drawing a line from the centre of one of the bottom circles to the ground, and joining it to the same continuation of the line between the centres of the triangles.

After a few very quick (so possibly wrong!) calculations I get that the answer is [tex]\sqrt{3}[/tex]cm.

 

[ojsimon]

Yeah the diagrams not meant to be skewed, just the only way i could draw it.

I understand your trig solution, but wouldn't this only acount for the 2 dimensions, surely as it is 3d it could not work like this?

 

[mrbohn1]

As far as I can see the solution still holds for 3 dimensions. It might help you to think of the bottom balls in the side on view as of being two balls diagonal to each other in the top view...

 

[mrbohn1]

Actually I take that back...tt would only work if the balls were touching, which I don't think they would be. It would be helpful to have a 3-d model!

You could do the same prrof using that the distance betweem the centres of the two bottom balls is [tex]\sqrt{8}[/tex] instead of 2 though.

 

[ojsimon]

Yeah, i don't have ping pong balls, unfortunately, there must be an easy way to work it though, what do you think of my surface area/volume formula idea?

 

[ojsimon]

I also thought it may be possible to solve using vectors, but i have no idea where to start with this..

 

[mrbohn1]

You can adapt my proof very easily: just note that the triangle between the centres of the triangles is not equilateral, but has 2 45^o angles, two sides of 2cm and one side of [tex]\sqrt{8}[/tex]cm.

Of course, I'm not sure what class this problem came out of...it may be that you are expected to use more complex methods.

 

[ojsimon]

No this was just a problem that a friend gave me but won't give me the solution till i solve it. Thanks for your help