*Where is the balancing point: what is the center of gravity?*

Then into unreachable cup, stuck to the walL, straight back chair lift: human center of gravity

- straight back chair
- tape measure
- cup
- pieces of cardboard cut in regular and irregular shapes
- head pins
- stone or masses
- thread
- scissors
- nail
- pencil
- ruler
- meter stick
- string
- masses

- The unreachable cup
- Who can lift the chair?

- Place a cup out in the middle of the floor. Stand 30 cm away and pick it up without bending your knees.
- Move to a wall. Place your heels against the wall. Have someone set the cup 30 cm in front of you. Pick up the cup without bending your knees.

What happened? Why?

This one requires a demonstration by the instructor.

- Measure two foot lengths from the wall by measuring off toe to heel, toe to heel. This works better with zoris or shoes on.
- Standing two foot lengths away, face the wall. Bend over and with a straight back, place your head on the wall.
- Have someone else hand you a straight back chair.
- Lift the chair, keeping your head on the wall.
- With the chair lifted, stand up.

Who could lift the chair? Who could not? Why could some lift the chair and some could not?

The above activities involve center of gravity. In this activity the center of gravity of an irregular piece of cardboard will be found as an illustration of the center of gravity as the balancing point.

- Cut an irregular shape from cardboard.
- Make a hole with a nail near any edge of the cardboard.
- Secure the cardboard to corkboard using a pin
- Hang a light weight or rock from the pin.
- Mark the plumb line on the cardboard.
- Repeat for a hole roughly 90° around the edge.
- The intersection of the lines is the center of gravity, the balance point.
- Balance the cardboard on a finger at the balance point.

A **cantilever** is a beam that has one end extending into open space without support. With no weights attached, a meter stick balances in the middle.

The center of gravity for a meter stick with no weights attached is in the middle of the meter stick.

Attach a weight to one end and the fulcrum string has to move to rebalance the meter stick. The left end of the meter stick is termed a cantilever. The force of the weight is keeping the cantilever up. The new center of gravity, center of balance, is at the new location for the fulcrum string.

For the meter stick to balance, the mass of the meter stick to the left of the string must balance the mass to the right of the meter stick. Data was taken for a meter stick that had a mass of 193 g. Masses were used for weights, hence the force is called "gram-force." The table shows the length of the cantilever Lc and the mass m needed to balance that length of cantilever.

Data table

Lc (cm) | m (g) |
---|---|

50 | 0 |

59.5 | 20 |

62 | 40 |

64.5 | 60 |

66.5 | 80 |

68.7 | 100 |

70.3 | 120 |

72 | 140 |

73.5 | 160 |

74.5 | 180 |

75.6 | 200 |

76.5 | 220 |

77.8 | 240 |

A graph of the above data produces the following chart.

This system is not a direct variation nor a linear equation. The curve above is a section of a rational function.

For a 100 centimeter meter stick, Lc in centimeters, and m in grams hung from the 99 cm point, the governing equation is found by setting the torques equal to each other. ρ is the "linear density" of the meter stick in grams per centimeter of length.

$m=\frac{\rho (5000-100\mathrm{Lc})}{(\mathrm{Lc}-99)}$

The equation contains information about system. As Lc approaches 99 from zero, the denominator becomes smaller and smaller. This represents the fulcrum string moving towards the mass. The decreasing denominator means that an increasing mass can be supported. As the denominator decreases towards zero, the mass that is balanced by the cantilever increases towards infinity. Of course in the real world the string would break when the total weight exceeded the strength of the string.

At the zero in the denominator the mass would be infinite. This is a vertical asymptote in the function.

Another interesting feature is that below 50 cm the mass goes negative. In other words a lifting force must be applied on the right arm if the cantilever is less than half the meter stick. This also makes sense: with the cantilever less than 50 cm, the right arm is longer and thus produces more force downwards. A lifting force, effectively a negative weight, would have to be applied. Note that the equation is calculating force using gram-force.

For the particular meter sticks in use in class, ρ = 1.93. Substituting x for Lc and y for m yields an equation more familiar to an algebra student.

$y=\frac{(9645.5-192.91x)}{(x-99)}$