You have a 12 inch ruler being used as a miniature seesaw. A pencil is placed under the 4 inch mark. A weight is placed at the 2 inch mark that balances the seesaw. Now the uneven seesaw is balanced. if you then placed a pencil at the 12 inch end, how many similar pencils approximately will you need to place at the other 1 inch end to get a balance again?

The number of pencils that needs to be placed at the other 1 inch end to get a balance again is approximately 3 pencils

Step-by-step explanation:

The given parameters are;

The length of the ruler used as a seesaw = 12 inch

The location where the pencil is placed = Under the 4 inch mark

The location where the weight is placed to balance the seesaw = The 2 inch mark

Therefore, the pencil at the 4 inch mark is fulcrum

The location where a pencil is placed on the ruler = The 12 inch end

The number of pencils, x, that needs to be placed at the other 1 inch end to get a balance again is given as follows;

The distance of the pencil at the 12 inch end from the fulcrum = 12 inches - 4 inches = 8 inches

The distance of the pencils placed at the 1 inch end from the fulcrum = 4 inches - 1 inch = 3 inches

Therefore at balance point, (equilibrium) the anticlockwise moment = The clockwise moment, which gives;

The mass of 1 pencil × 8 inches = x × The mass of 1 pencil × 3 inches

x = (The mass of 1 pencil × 8 inches)/(3 inches × The mass of x × 1 pencil)

x = 8/3 = 2.  ≈ 3 pencils

The number of pencils that needs to be placed at the other 1 inch end to get a balance again = x ≈ 3 pencils.

that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values

so, converting firstly the mixed fractions to improper fractions and then "adding".