# Stress, strain and modulus - log hung from 2 wires

• PirateFan308
In summary, Homework Equations:-The force of wire B is 58134.579x greater than the force of wire A.-The distance from wire A to the center of gravity (com) is 1.26 greater than the distance from wire B to the center of gravity.
PirateFan308

## Homework Statement

There is a 104kg uniform log hung on two steel wires, A and B, both of radius 1.00mm. Initially, wire A was 2.70m long and 2.00mm shorter than wire B. The log is now horizontal. Note that the centre of gravity (com) is not necessarily centered between the wires and the modulus of steel is 20x1010Pa.
a) What is the magnitude of the force on the log from wire B?
b) What is the ratio of distance from wire A to com/distance from wire B to com?

## Homework Equations

$$modulus = \frac{stress}{strain}$$
$$modulus_B = \frac{F_B~/~m^{2}}{Δl~/~l_{o}}$$
$$F_B = \frac{(Δl)(area)(modulus)}{l_{o}}$$

## The Attempt at a Solution

I'm assuming the left wire is wire A
length of A = 2.70 m
length of B = 2.702 m
area of wire = $\pi d^2 ÷ 4$
let x = change in length of B
so that x+0.002 = change in length of A
$F_A$ = force of wire A
$F_B$ = force of wire B
total weight of block = (104kg)(9.81) = 1020.24N
$F_A + F_B$ = 1020.24 N
distance from wire A = d_A
distance from wire B = d_B

a)
$$F_B = \frac{(Δl)(area)(modulus)}{l_{o}}$$
$$F_B = \frac{(x)(7.85398*10^{-7}m^{2})(20*10^{10}Pa}{2.702m}$$
$$F_B = 58134.579x$$

$$F_A = \frac{(Δl)(area)(modulus)}{l_{o}}$$
$$F_A = \frac{(x+0.002)(7.85398*10^{-7}m^{2})(20*10^{10}Pa}{2.70m}$$
$$F_A = 58177.64x + 116.355$$

$$F_A + F_B = 1020.24 N$$
$$(58177.64x)+(58130x)+(116.355)=1020.24 N$$
$$x(58177.64+58130)=1020.24-116.355$$
$$x=7.77*10^{-3} m$$

$$F_{B}=(area)(modulus)(strain)$$
$$F_{B}=(7.85398*10^{-7})(20*10^{10})(X ÷ 2.702) = 4.51775*10^{2} N$$
$$F_{A}=5.6846*10^{2} N$$b)
$$\sum\tau=0$$
$$(d_A)(F_A)-(d_B)(F_B)=0$$
$$\frac{d_A}{d_B} = \frac{F_B}{F_A}$$
$$\frac{d_A}{d_B} = \frac{4.51775*10^2}{5.6846*10^2}$$
$$\frac{d_A}{d_B} = 1.25829 = 1.26$$

Unfortunately, the website said that my answers for a) and b) are wrong. I know that since a) is wrong, that automatically makes b) wrong. Any help would be appreciated!

In part a, your approach is excellent. I didn't check all your math, but I notice you have calculated the area incorrectly. If the wire has a 1mm radius, it's area in x-section is ∏r^2 = 3.14E-06 m^2. Looks like you used d =1 mm instead of r = 1mm. Redo the math and you should get the correct answer.

In part b, aside from making the necessary correction in numbers, it's da/db, but your result calculates db/da.

Otherwise, you've done yourself proud

Wow, what a simple mistake! Thanks for catching that - I would have been trying to figure it out for hours!

## What is stress?

Stress is a measure of the external force applied to an object per unit area. It is typically represented by the symbol σ and is calculated by dividing the applied force by the cross-sectional area of the object.

## What is strain?

Strain is a measure of the deformation or change in shape of an object due to stress. It is typically represented by the symbol ε and is calculated by dividing the change in length of the object by its original length.

## What is modulus?

Modulus, also known as the Young's modulus or elastic modulus, is a measure of a material's stiffness or resistance to deformation. It is typically represented by the symbol E and is calculated by dividing stress by strain.

## How is a log hung from 2 wires used to study stress and strain?

A log hung from 2 wires is a common experiment used to study stress, strain, and modulus. By measuring the amount of stretch in the wires (strain) and the weight of the log (stress), the modulus of the material can be determined using the formula E = σ/ε.

## What are the limitations of using a log hung from 2 wires to study stress, strain, and modulus?

While the log hung from 2 wires experiment can provide a general understanding of stress, strain, and modulus, it does have limitations. These include assumptions made about the material's uniformity and the wires' ability to perfectly support the weight of the log without any deformation. Additionally, the experiment does not account for other factors that may affect stress and strain, such as temperature or time.

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