What is the Maximum Load a 1.5 m Copper Wire with 1.1 mm Radius Can Suspend?

[mikefitz]

What is the maximum load that could be suspended from a copper wire of length 1.5 m and radius 1.1 mm without breaking the wire? Copper has an elastic limit of 3 × 108 Pa and a tensile strength of 4.2 × 10^8 Pa

F/A<Elastic Limit

= m(9.81) / 3.8013 = 3x10^8Pa
9.81m= 114 098 123
m= 116 248 535 N ?

Seems like an aweful lot of load, what did i do wrong on this one?

 

[mikefitz]

any ideas? I worked it out one more time with the same result...

 

[kyle8921]

As a scientist, it is important to approach problems with precision and attention to detail. In this case, it seems that there may have been a calculation error in determining the maximum load that could be suspended from the copper wire. Here is the correct calculation:

The maximum load that could be suspended from the copper wire can be determined using the formula:

F = σ × A

Where:
F is the maximum load
σ is the tensile strength of copper (4.2 × 10^8 Pa)
A is the cross-sectional area of the wire (πr^2 where r is the radius of the wire)

So, for the given wire with a length of 1.5 m and a radius of 1.1 mm, the cross-sectional area would be π(0.0011)^2 = 3.8013 × 10^-6 m^2

Plugging these values into the formula, we get:

F = (4.2 × 10^8 Pa) × (3.8013 × 10^-6 m^2)
= 1597.86 N

Therefore, the maximum load that could be suspended from this copper wire without breaking it would be 1597.86 N. It is important to note that this is assuming the wire is perfectly straight and there are no other factors (such as bending or twisting) that could affect its strength.

In terms of your calculation, it seems that you may have used the wrong value for the elastic limit of copper. The given value of 3 × 10^8 Pa is the yield strength of copper, not the elastic limit. The elastic limit is the maximum stress a material can withstand without experiencing permanent deformation. In this case, it is not relevant to the calculation of the maximum load that can be suspended from the wire.

I hope this clarification helps. Remember to always double check your calculations and use the correct values to ensure accurate results.