Answer :
The column is compressed as much as 1.8 × 10⁻³ m
Further explanation
Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.
[tex]\boxed {F = k \times \Delta x}[/tex]
F = Force ( N )
k = Spring Constant ( N/m )
Δx = Extension ( m )
The formula for finding Young's Modulus is as follows:
[tex]\boxed {E = \frac{F / A}{\Delta x / x_o}}[/tex]
E = Young's Modulus ( N/m² )
F = Force ( N )
A = Cross-Sectional Area ( m² )
Δx = Extension ( m )
x = Initial Length ( m )
Let us now tackle the problem !
Given:
x₀ = 3 m
d = 40 cm = 0,4 m
m = 235000 kg
E = 3 × 10¹⁰ N/m²
Unknown:
Δx = ?
Solution:
[tex]F = w[/tex]
[tex]F = m \times g[/tex]
[tex]F = 235000 \times 9,8[/tex]
[tex]\boxed {F = 2303000 ~ Newton}[/tex]
[tex]A = \frac{1}{4} \pi d^2[/tex]
[tex]A = \frac{1}{4} \pi 0.4^2[/tex]
[tex]\boxed {A = 0.04 \pi ~ m^2}[/tex]
[tex]E = \frac{F / A}{\Delta x / x_o}[/tex]
[tex]3 \times 10^{10} = \frac{2303000 / ( 0.04 \pi )}{\Delta x / 3}[/tex]
[tex]\Delta x / 3 = \frac{2303000 / ( 0.04 \pi )}{3 \times 10^{10}}[/tex]
[tex]\Delta x = 3 \times \frac{2303000 / ( 0.04 \pi )}{3 \times 10^{10}}[/tex]
[tex]\large {\boxed {\Delta x \approx 1.8 \times 10^{-3} ~ m} }[/tex]
Learn more
- Young's modulus : https://brainly.com/question/6864866
- Young's modulus for aluminum : https://brainly.com/question/7282579
- Young's modulus of wire : https://brainly.com/question/9755626
Answer details
Grade: College
Subject: Physics
Chapter: Elasticity
Keywords: Elasticity , Diameter , Concrete , Column , Load , Compressed , Stretched , Modulus , Young
The Young modulus E is given by:
[tex]E= \frac{F L_0}{A \Delta L} [/tex]
where
F is the force applied
A is the cross-sectional area perpendicular to the force applied
[tex]L_0[/tex] is the initial length of the object
[tex]\Delta L [/tex] is the increase (or decrease) in length of the object.
In our problem, [tex]L_0 = 3.0 m[/tex] is the initial length of the column, [tex]E=3.0 \cdot 10^{10}N/m^2[/tex] is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
[tex]r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m [/tex]
and the cross-sectional area is
[tex]A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2[/tex]
The force applied to the column is the weight of the load:
[tex]W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N [/tex]
Now we have everything to calculate the compression of the column:
[tex]\Delta L = \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m[/tex]
So, the column compresses by 1.83 millimeters.
[tex]E= \frac{F L_0}{A \Delta L} [/tex]
where
F is the force applied
A is the cross-sectional area perpendicular to the force applied
[tex]L_0[/tex] is the initial length of the object
[tex]\Delta L [/tex] is the increase (or decrease) in length of the object.
In our problem, [tex]L_0 = 3.0 m[/tex] is the initial length of the column, [tex]E=3.0 \cdot 10^{10}N/m^2[/tex] is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
[tex]r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m [/tex]
and the cross-sectional area is
[tex]A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2[/tex]
The force applied to the column is the weight of the load:
[tex]W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N [/tex]
Now we have everything to calculate the compression of the column:
[tex]\Delta L = \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m[/tex]
So, the column compresses by 1.83 millimeters.
