Overarch 1

5848 /www/nrich/html/content/id/5848/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/4769">Warm-up</a></li> <li><a href="http://nrich.maths.org/4767">Try this next</a></li> <li><a href="http://nrich.maths.org/299">Think higher</a></li> <li><a href="http://en.wikipedia.org/wiki/Moment_%28physics%29">Read: mathematics</a></li> <li><a href="http://plus.maths.org/content/unspinning-boomerang">Read: science</a></li> <li><a href="http://plus.maths.org/content/spaghetti-breakthrough">Explore further</a></li> </ul> <div> </div> Note: you may first like to play the <a href="http://nrich.maths.org/public/viewer.php?obj_id=5863&part=" style="font-style: italic;">Tower Rescue</a> game which will give you a feeling for the mechanics of building the structures described in this problem A uniform square tile of side 20cm is placed on the ground and another identical tile placed on top so that it overhangs as far as possible without toppling over, as in the following diagram: <mdo:image alt="" height="121" src="one%20balancing.JPG" width="396"></mdo:image>: Clearly the top tile can be placed as far as half of the way along the base tile without toppling over, so that the overhang will be of length 10 cm. These two tiles are then joined in this configuration and placed on top of a third tile so that the whole construction just balances. What will the size of the overhang be in this case? <mdo:image alt="" height="107" src="two%20balancing.JPG" width="382"></mdo:image> In a similar way, find the maximum overhang when three tiles are balanced in this way on top of a fourth tile. For a more challenging extension of this please see <a href="http://nrich.maths.org/public/viewer.php?obj_id=299&part=">Overarch 2</a> . Investigation: Try making a tower of CDs or other similar objects in this fashion. How will the fact you are using real world objects affect your answers? Think carefully about the physical effects at play and how you could model these mathematically. <a href="http://nrich.maths.org/public/viewer.php?obj_id=6817&part=">Click here for a poster of this problem</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Luke from St Patricks suggested that 1. The answer is 5 cm. The answer is 2.5 cm. For every tile you add on the overhang will have to be halved to have equal balance with both of the other tiles. Yuliang Djanogly City Academy identified that the best overhang in the three tile case is 6.6cm over the base tile. The explanation for the sort of reasoning for this was given by Jason, by using centre of mass considerations Whatever amount of tiles there are, when they are placed on another tile, half the mass must be on the tile for them to just balance. So, for the first two, the top tile may sit 10cm on and 10cm off the bottom one. This gives a pivot point for the mass at the halfway point, ie. 15cm. ergo these two may be placed 15cm on and 15cm off the bottom one. Now the pivotal point of the mass of these three tiles is about 16.6cm. An algebraic extension was given by Lindsay from Collyer's Sixth Form College So we've established that the maximum overhang for two tiles is 10cm. Now we can consider three tiles: The centre of mass of the first two tiles must sit on the edge of the third tiles to achieve maximum overhang. Using the idea of centre of mass we can produce the following equation, where $x$ is the new centre of mass $$10m + 20m = 2m\times x$$ . Rearranging this gives $$ \frac{10m + 20m}{2m} = x = 15cm$$ The two tiles can therefore be placed 5cm from the end of the third, giving an additional overhang of 5cm and an overall overhang of 10 + 5 = 15cm. This process can be repeated to add a fourth tile: Centre of mass of three tiles $$ \frac{10m + 20\times 2m}{3m} = \frac{50}{3} = 16 \frac{2}{3} $$ So they can be placed $20 - 16\frac{2}{3} = 3\frac{1}{3}$cm from the end of the fourth tile, creating an overall overhang of $10 + 5 + 3\frac{1}{3} = 18\frac{1}{3}cm$. There seems to be a pattern in these results, which becomes clearer once we tabulate them. This can be generalised for n tiles in the following way: $$ 10m + 20\times(n-1)m = nmx, \mbox{ where } x \mbox{ is the centre of mass} $$ This rearranges to $\frac{10+20(n-1)}{n} = x$. Because the edge of the bottom tile must be directly below the common centre of mass of the stack of n tiles above, x represents the total overhang of a stack of n tiles. We can see this in a table <table border="1"> <tbody> <tr> <td>Number of tiles being added</td> <td>Additional overhang (cm)</td> <td>Total overhang (cm)</td> <td></td> </tr> <tr> <td>1</td> <td>10</td> <td>10</td> <td>10/1</td> </tr> <tr> <td>2</td> <td>5</td> <td>10 +5 = 15</td> <td>10/1+10/2</td> </tr> <tr> <td>3</td> <td>$3\frac{1}{3}$</td> <td>$10+5+3\frac{1}{3}$</td> <td>10/1+10/2+10/3</td> </tr> </tbody> </table> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Before attempting this problem is it interesting to ask students to make a scale sketch of their best intuitive guess at the answer. Once the answer is found the students can compare their intuition with reality. Were they roughly correct? Does their answer give them any clues as to the form of a larger compound structure? Ask students to explain in words why their structures only just balance, as producing a clear verbal explanation will reinforce the understanding of the underlying mechanical principles. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> When we have 3 tiles stacked in equilibrium the overall weight of the tiles is (obviously) 3. To balance on the fourth tile the centre of mass of the 3-tile combination must (obviously) be exactly on the edge of the fourth tile. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">First tile is one half sticking out, second tile is one third sticking out, third tile is one quarter sticking out and so on. </mdoxml> 2 3 0 0 0 0 1 Overarch 1 This short question asks if you can work out the most precarious way to balance four tiles. Admin Short problems Mechanics Centre of mass Applications engineering Admin Short problems