### Brain Teasers

# The Flag Pole

Oh no! Our flag has fallen down from the flag pole (stupid cheap string)! Quick, we need to figure out how much string we need to get it back up there...

The problem is, I don't remember its exact length!

All I can remember is that the pole was built to be exactly 14 times the height of the flag.

The flag's height is equal to the ground to the pulley system.

The string wraps evenly around the pole 4 times.

The string only goes up from the pulley system to the bottom of the flag (the returning string is built into the pole).

The pulley system is attached 1 meter off the ground.

Oh yes, and the diameter of the pole is 10cm.

So quickly! We don't have much time until everyone sees, how much string do we need?

Conversions:

100cm = 1m

2.54cm = 1 inch

The problem is, I don't remember its exact length!

All I can remember is that the pole was built to be exactly 14 times the height of the flag.

The flag's height is equal to the ground to the pulley system.

The string wraps evenly around the pole 4 times.

The string only goes up from the pulley system to the bottom of the flag (the returning string is built into the pole).

The pulley system is attached 1 meter off the ground.

Oh yes, and the diameter of the pole is 10cm.

So quickly! We don't have much time until everyone sees, how much string do we need?

Conversions:

100cm = 1m

2.54cm = 1 inch

### Hint

The string covers 12 meters of distance in 4 rotations.Use the Pythagorean theorem.

### Answer

If the pulley system is attached 1 meter off the ground and the flag's height is equal to the ground to the pulley system, then the flag must be 1 meter high. If the string only goes up from the pulley system to the bottom of the flag and the flag pole is 14 meters in height, then the string must cover 12 meters in total (14-1-1).If the diameter of the pole is 10cm and the height covered is 12 meters (1200cm), then, unwrapped in to a rectangle, it would be pi(10) (since the circumference of a circle is pi{d}) by 1200 cm rectangle.

If the string wraps around in 4 even rotations, that means the rectangle divided on the y plane into 4 equal sections would represent one rotation per section. Finding the hypotenuse of each section (a^2 + b^2 = c^2) would then yield the distance covered per rotation.

This may be better represented with the following diagram:

._________

|________/| -- The Unwrapped Pole --

|_______/_|

|______/__|

|_____/___|

|____/____|

|___/_____|

|__/______|

|_/_______|

|/________|

|________/|

|_______/_|

|______/__|

|_____/___|

|____/____|

|___/_____|

|__/______|

|_/_______| 1200 CM

|/________|

|________/|

|_______/_|

|______/__|

|_____/___|

|____/____|

|___/_____|

|__/______|

|_/_______|

|/________|

|________/|

|_______/_|

|______/__|

|_____/___|

|____/____|

|___/_____|

|__/______|

|_/_______|

|/________|

pi(10) CM

._________

|________/| -- ONE SECTION --

|_______/_|

|______/__|

|_____/___|

|____/____| 300 CM

|___/_____|

|__/______|

|_/_______| <---- The string is the hypotenuse of this right angled triangle

|/________|

pi(10) CM

The string's distance in one rotation would be expressed by the following equation:

a^2 + b^2 = c^2

c = [300^2 + pi(10)^2]^1/2

Therefore, in 4 rotations it would be the above * 4.

The final equation is thus expressed as:

let a represent the total height of the pole, then f represent the height of the flag, then rot represent the number of rotations, then p represent the height of the pulley, then d represent the diameter of the pole

s = rot * [ ({a-f-p}/4)^2 + d(pi)^2 ]^½

= 4 * [ ({1400-100-100}/4) ^ 2 + 10(pi)^2 ]^½

= 4 * [ 90 000 + 10(pi)^2 ]^½

= 1206.561796

Therefore, the string covers approximately 1206.561796 cm in 12 meters when there are 4 even rotations.

( Or: The string = number of rotations {#to multiply by the amount of distance covered in ONE rotation} [ ([total distance covered]/4 {#to equal the height in one rotation} ) ^ 2 + circumference ^ 2 {#to equal the width in one rotation ] ^ square root {#to calculate the distance of the string in one rotation} )

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