Creating hanging mobiles using algebra

Before Thanksgiving break, Algebra1 students created hanging mobiles out of plastic autumn garlands, dowels, and string. Students provided the imagination and creativity for their floral arrangements, and mathematics provided the algorithm for balancing their creations in a hanging mobile.  

The algorithm can be observed at children's playgrounds (at least pre-2000) with the simple seesaw.  Two evenly weighted children, seated at either end can move and down easily because balance is achieved.  However, when one child is much heavier than the other, the seesaw stays tipped to the heavier child's side.  There was no balance for the seesaw to go up and down easily.  The solution is for the heavier child to move closer to the middle of the seesaw.  The explanation can be explained with algebra. In order for the seesaw to balance, the weight of one child multiplied by their distance from the center of the seesaw has to equal the weight of the second child multiplied by their distance from the center. Thus, the greater the weight difference between one child and the other, the less distance from the "fulcrum" the child needed to sit.  

This same seesaw principle can be applied to hanging mobiles: the heavier the floral arrangement on one side of the dowel, the closer that arrangement needs to sit from the center (thus where the string holding the dowel is tied).  Below are some photos of the project and its results.

Algebra as a decision-making tool

Suppose you are in the market for a new car, and the particular one you like is offered in

both gas-only and hybrid models. You are wondering for the hybrid at what specific

number on the odometer does the savings in gasoline purchased offset the additional

amount in the purchase price? I enjoy teaching algebra because this question can be

answered with the knowledge of writing and solving systems of algebraic equations. 

Algebra allows one to simplistically model the cost of driving based on the purchase

price (P), the rate of the cost of driving per mile (m), and the variable of miles driven (x).

(Cost = P + mx) Some online research gives you the values for the purchase price,

gas price, and fuel economy (mpg). Graph your two equations, and where they

intersect is the exact mileage point at which operating the two types of car is equal. 

Any miles driven beyond this point and the higher-priced hybrid starts saving you money.

Students in my Algebra 1 class conducted this research and solved the systemsof equations written for operating the two models in order to answer the original

question for themselves. Some of them were surprised by the answer.