SOLUTION: Solve: The force needed to keep a car from skidding on a curve varies jointly as the weight of the car and the square of the car's speed, and inversely as the radius of the curv

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Question 144912: Solve:
The force needed to keep a car from skidding on a curve varies jointly as the weight of the car and the square of the car's speed, and inversely as the radius of the curve. If a force of 3600 pounds is needed to keep an 1800 pound car traveling at 20 mph from skidding on a curve of radius 600 feet what force would be required to keep the same car from skidding on a curve of radius 570 feet at 50 mph? Round to the nearest pound of force?

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
The force needed to keep a car from skidding on a curve varies jointly as the weight of the car and the square of the car's speed, and inversely as the radius of the curve. If a force of 3600 pounds is needed to keep an 1800 pound car traveling at 20 mph from skidding on a curve of radius 600 feet what force would be required to keep the same car from skidding on a curve of radius 570 feet at 50 mph? Round to the nearest pound of force.

This is the general formula for all variation problems:



>>...The force needed to keep a car from skidding on a curve varies...<<

This force is the VARYING_QUANTITY.  We will use the letter F. So

F=VARYING_QUANTITY

>>...jointly as the weight of the car and the square of the car's speed,...<<

The JOINT_VARIABLES are the weight, W, and speed, S, squared, which gives S%5E2

>>...and inversely as the radius of the curve...<<

There is just one INVERSE_VARIABLE, the radius R.

So we write the equation using the general formula for all
variation problems:



eliminating the ones that don't apply:

VARYING_QUANTITY+=+k%2A%28%28PRODUCT_OF_JOINT_VARIABLES%29%2F%28INVERSE_VARIABLE%29%29

F+=+k%2A%28%28WS%5E2%29%2FR%29

>>...If a force of 3600 pounds is needed to keep an 1800 pound car traveling at 20 mph from skidding on a curve of radius 600 feet...<<

Plug those values in: F=3600,W=1800, S=20, R=600

F+=+k%2A%28%28WS%5E2%29%2FR%29

3600+=+k%2A%28%28%281800%29%2820%29%5E2%29%2F600%29

Solve for k:

3600+=+k%2A1200

3=k

So go back to this equation:

F+=+k%2A%28%28WS%5E2%29%2FR%29

and substitute only the value of k=3 

F+=+3%2A%28%28WS%5E2%29%2FR%29

or we can just make it

F+=+%283WS%5E2%29%2FR

Now our formula is complete. We are now ready to use it:

>>...what force would be required to keep the same car
from skidding on a curve of radius 570 feet at 50 mph?...<<

That just asks: What is F when W=1800, R=570, and S=50

So we plug those in:

F+=+%283WS%5E2%29%2FR

F+=+%283%281800%29%2850%29%5E2%29%2F570

F+=+23684.21053

>>...Round to the nearest pound of force...<<

we round that to the nearest pound.

F+=+23684pounds 

Edwin