Physics 40s scaling and proportions .

[mp3guy]

physics 40s scaling and proportions ...please help

the physics of a wire supporting a sphere...

A wire of .2cm in diameter is just strong enough to support a steel sphere. What must the diameter of the wire be if the following changes are made to the sphere?


A. new sphere has 4x the original volume
B. new sphere has 4x the original diameter


for part A.
-determine the relationship between the volume and the strength
-determine the relationship between the strength and the linear dimension
-determine the actual diameter

for part B.
-determine the relationship between the diameter and the strength
-determine the actual linear dimension


any help would be very much appreciated. I'm not really sure how you would relate changes in the wire to changes in the sphere...

thanks, mp3guy

 

[Doc Al]

I would assume that the wire being "just strong enough" means that the material is at maximum stress (force per cross-sectional area). Thus, to support more weight, you will need a thicker wire to maintain the same F/A.

Your job is to figure out how the weight of the sphere changes in each case, and then figure out how the diameter of the wire must change to support it.

 

[mp3guy]

what complicates things is they don't give you a lot of info...besides increasing the sphere's diameter and volume by 4x in separate instances, you have no numbers to go by...what formulas would i use to solve this problem, or how do i find out the volume and linear dimension of the wire?

 

[Doc Al]

You are given all the information you need. For example: If the sphere's volume gets multiplied by 4, what happens to its weight? If its diameter gets multiplied by 4, what happens to its volume?

How does the volume of a sphere depend on its radius (or diameter)? (How does the volume of any shape depend on its linear dimensions?)

 

[HallsofIvy]

Generally speaking, the strength of a wire is proportional to the area of a cross section of the wire which is itself proportional to the square of the diameter of the wire.

The volume (and therefore the weight) of a sphere is proportional to the cube of the diameter.

If the diameter of a sphere is multiplies by 4 (but the density remains the same) its weight is multiplied by the cube root of 4 so the wire must have diameter square root of that: sixth root of 4.

 

[Doc Al]

If the diameter of a sphere is multiplies by 4 (but the density remains the same) its weight is multiplied by the cube root of 4 so the wire must have diameter square root of that: sixth root of 4.

You might want to rephrase that answer, Halls!