Calculating Mass of Steel Ball for Ultimate Strength

  • Thread starter aquabug918
  • Start date
  • Tags
    Strength
In summary, the question is asking for the mass of the biggest ball that a 1mm steel wire with a length of 2 meters and an ultimate strength of 1.1 X 10^9 N/m^2 can bear. To find this, we need to calculate the area of the cross-section of the wire and then use the equation \frac{x}{A} = \frac{1.1 \times 10^9}{1m^2} to find the maximum force that the wire can resist. This will give us the maximum mass of the ball that the wire can bear.
  • #1
aquabug918
17
0
A solid steel ball is hung at the bottom of a steel wire length 2 meters and radius of 1mm. The ultimate strength of steel is 1.1 X 10^9 N/m^2. What is the mass of the biggest ball the wire can bare.

This seems like a pretty straight forward question. I am guessing the 2 meter radius doesn't matter. I am thinking that you need to find the area of a cross section of the 1mm wire. I am not sure what to do next.




2nd part ... what is the period of torsional oscillation of the system?
The shear modulus of steel = 8x10^10 N/m^2 and the interia is (2MR^2)/5.


Here I think you need to use the equation... T = 2pi * (I/c)^.5 where C is the shear modulus. I can't figure this part out. Do i need to worry about the cross sectional area here also?


Thank you very much everyone!
 
Physics news on Phys.org
  • #2
aquabug918 said:
A solid steel ball is hung at the bottom of a steel wire length 2 meters and radius of 1mm. The ultimate strength of steel is 1.1 X 10^9 N/m^2. What is the mass of the biggest ball the wire can bare.

This seems like a pretty straight forward question. I am guessing the 2 meter radius doesn't matter. I am thinking that you need to find the area of a cross section of the 1mm wire. I am not sure what to do next.

Thank you very much everyone!

Let's say you found the area of the cross-section to be A. Well, if it is [tex] \frac{1.1 \times 10^9}{1m^2}[/tex], how much is it for [tex]\frac{x}{A}[/tex]. And notice that this is not the final answer. It will give you the maximum force that that specific thickness of steel wire can resist.
 
  • #3


I would approach this problem by first identifying the relevant equations and variables. For the first part, we are looking for the mass of the largest steel ball that the wire can support, given its length, radius, and the ultimate strength of steel. We can use the equation for stress, which is force divided by area, to calculate the maximum force that the wire can withstand before breaking. Then, we can use the equation for weight, which is mass multiplied by acceleration due to gravity, to find the mass of the ball that would produce this maximum force when hung from the wire.

For the second part, we are asked to find the period of torsional oscillation of the system. This would involve using the equation for the period of a simple harmonic oscillator, which is dependent on the moment of inertia and the torsional constant. The moment of inertia can be calculated using the given formula, and the torsional constant can be found by using the equation for shear modulus and the cross-sectional area of the wire.

In summary, to solve both parts of this problem, we would need to apply relevant equations and use the given information to calculate the necessary variables. It is also important to pay attention to the units and make sure they are consistent throughout the calculations.
 

1. How do you calculate the mass of a steel ball for ultimate strength?

To calculate the mass of a steel ball for ultimate strength, you need to know the density of steel (usually around 7.85 g/cm^3) and the diameter of the ball. Then, you can use the formula: mass = density * volume. The volume of a sphere is (4/3) * π * (diameter/2)^3. Plug in the values and you will get the mass of the steel ball.

2. Why is it important to calculate the mass of a steel ball for ultimate strength?

Calculating the mass of a steel ball is important because it helps determine the strength and durability of the material. By knowing the mass, we can assess if the steel ball is suitable for a particular application or if it needs to be reinforced to withstand greater forces.

3. What is the ultimate strength of a steel ball?

The ultimate strength of a steel ball refers to the maximum tensile stress or load that the steel ball can withstand before breaking. It is an important factor in engineering and design, as it determines the safety and reliability of structures and materials.

4. What factors can affect the mass of a steel ball for ultimate strength?

The mass of a steel ball for ultimate strength can be affected by various factors such as the diameter and density of the steel, any impurities or defects in the material, and the manufacturing process used. The environment in which the steel ball will be used, such as temperature and humidity, can also impact its mass and ultimately, its strength.

5. How can I use the calculated mass of a steel ball for ultimate strength in my research or project?

The calculated mass of a steel ball for ultimate strength can be used in various applications, such as in designing structures, creating simulations, and conducting experiments. It can also be a useful reference for selecting materials and determining the appropriate safety measures for a particular project or research.

Similar threads

  • Introductory Physics Homework Help
Replies
15
Views
312
  • Introductory Physics Homework Help
Replies
4
Views
337
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top