Appying a force bar + Young's Modulus + Force Applied

In summary, the conversation discusses a problem involving a bar with a height of 17.2 cm, cross-sectional area of 12.32 cm2, applied force of 6130 N, and Young's modulus of 410 MPA. The equations for stress, strain, and Young's modulus are provided, and the individual is unsure of their approach and asks for reassurance. They also mention finding a value for ΔL and subtracting it from the original height to get a final value of 16.34 cm. However, they are unsure if their thought process is correct and ask for clarification on Young's modulus.
  • #1
Jtwa
9
0
1. Homework Statement
h = 17.2 cm
A = 12.32cm2
E = 410 MPA
Fapp = 6130 N
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2. Homework Equations
Stress
σ = F/A where F is force applied and A is cross sectional area of side where force is applied

Strain
ε = ΔL/L where L is the length of the bar and ΔL is the change of length of bar

Young's Modulus
E = stress/strain = FL/AΔL

3. My attempt at solution

We are solving first solving for ΔL which I got as 0.864cm and then we subtract that from 17.2cm which I got as 16.34cm

I'm not sure if my thought process is correct or how to approach this question. What is Young's modulus ? Am I approaching the problem correctly? Am I looking for the right variable? Did I get the correct answer ? Just need reassurance. Thanks in advance guys!
 
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  • #2
Jtwa said:
1. Homework Statement
h = 17.2 cm
A = 12.32cm2
E = 410 MPA
Fapp = 6130 N
2. Homework Equations
Stress
σ = F/A where F is force applied and A is cross sectional area of side where force is applied

Strain
ε = ΔL/L where L is the length of the bar and ΔL is the change of length of bar

Young's Modulus
E = stress/strain = FL/AΔL

3. My attempt at solution

We are solving first solving for ΔL which I got as 0.864cm and then we subtract that from 17.2cm which I got as 16.34cm

I'm not sure if my thought process is correct or how to approach this question. What is Young's modulus ? Am I approaching the problem correctly? Am I looking for the right variable? Did I get the correct answer ? Just need reassurance. Thanks in advance guys!
You just have a bunch of numbers where a problem statement should be. What problem are you supposed to be solving? What is the text which should go with this data?

If you don't know what Young's modulus is, look it up! You've got the whole internet at your disposal.
 
  • #3
oops. I forgot to upload a photo
 
  • #4
Jtwa said:
oops. I forgot to upload a photo
I get a different number. Please post all your working.
 

1. What is a force bar?

A force bar is a physical object that is used to apply a force to an object. It is typically a long, slender rod made of a strong and durable material, such as steel or aluminum. The force bar is used to transfer a specific amount of force to the object it is applied to.

2. What is Young's Modulus?

Young's Modulus, also known as the modulus of elasticity, is a measure of the stiffness of a material. It represents the ratio of stress to strain within a material when a force is applied. In simpler terms, it measures how much a material will deform under a given amount of stress.

3. How is Young's Modulus calculated?

Young's Modulus is calculated by dividing the stress (force applied per unit area) by the strain (change in length divided by original length) of a material. This gives a measure of how much a material will deform under a given amount of force.

4. How is a force bar used in relation to Young's Modulus?

A force bar is used to apply a specific amount of force to an object, while Young's Modulus is used to measure the resulting strain within the object. By applying a known force with the force bar and measuring the resulting strain, we can calculate the Young's Modulus of the material being tested.

5. Why is Young's Modulus important?

Young's Modulus is an important concept in material science and engineering as it allows us to predict how materials will behave under different levels of stress. This information is crucial in designing and creating structures and objects that can withstand specific forces and perform their intended functions safely and effectively.

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