Question

When a 75 kg man sits on the stool, by what percent does the length of the legs decrease? assume, for simplicity, that the stool's legs are vertical and that each bears the same load?

Answer 1

First, let's find the total force exerted by the man on the stool.

F = mg
F = (75 kg)(9.81 m/s²)
F = 735.75 N

Next, we must know the maximum load the stool can take. Suppose, the it can take up to 700 N. Since each leg of the stool takes an equal amount of force, this is uniformly distributed. So, the solution is as follows:

Percentage = [(735.75- 700)/700]*100 = 5.1%

Answer 2

The percentage decrease in the length of the stool’s legs is $\boxed{3.84 \times {{10}^{-3}\%}}$.

 

Further explanation:

Here, we have to calculate the percentage of change in the length of the stool leg.

Consider the stool’s legs are made up of wood.

Given:

The mass the man is $75{\text{ kg}}$.

Diameter of the leg of the stool is $2.5{\text{ cm}} = 0.025{\text{ m}}$

Young’s modulus $(E)$ of the wood is $1.3 \times10^{10}\text{ N/m}^2}$.

Formula and concept used:

Number of the legs $\left( n \right)$ in the stool is $3{\text{ legs}}$.

Weight of the man can be calculated as,

$\boxed{W = mg}$

Here, $m$ is the mass of the man, $g$ is the acceleration due to gravity.

So, weight of the man will be,

$\begin{aligned}W&=75 \times 9.8\\W&=735{\text{ N}}\\\end{aligned}$

Since, each leg is vertical and bears the same load.

So, the load on each leg will be,

$\begin{aligned}F&=\frac{W}{3}\\F&=\frac{{735}}{3}\\F&=245{\text{ N}}\\\end{aligned}$

 

Now the longitudinal stress in the stool’s leg can be calculated as,

$\boxed{\sigma= \frac{F}{A}}$                                                     …… (1)

Here, $A$ is the cross-sectional area of the leg.

Cross-sectional area of the leg cab be calculated as,

$A=\dfrac{\pi }{4}{d^2}$

Substitute this value of $A$ in equation (1).

$\sigma= \dfrac{{4F}}{{\pi {d^2}}}$                                         …… (2)

Now, linear strain in the leg will be,

$\varepsilon= \dfrac{\sigma }{E}$

Substitute the value of \sigma  from equation (2) in the above equation.

$\varepsilon=\dfrac{{4F}}{{\pi {d^2}E}}$                                     …… (3)

Calculation:

Substitute the value of $F$ as $245{\text{ N}}$, value of $d$ as $0.025{\text{ m}}$ and the value of $E$ as $1.3 \times {10^{10}}\text{ N/m}^2$ in equation (3).

$\begin{aligned}\varepsilon&=\frac{{4\left( {245} \right)}}{{\pi {{\left( {0.025} \right)}^2}\left( {1.3\times{{10}^{10}}}\right)}}\\&=\frac{{980}}{{25.525\times {{10}^6}}}\\&=3.84\times{10^{-5}}\\\end{aligned}$

 

Strain can be expressed as,

$\varepsilon=\dfrac{{{l_2} - {l_1}}}{{{l_1}}}$

Here, ${l_2}$ is less than ${l_1}$ due to compression of the legs.

So, above equation can be written as,

$\begin{aligned}\frac{{{l_2} - {l_1}}}{{{l_1}}}&=3.84 \times {10^{ - 5}}\\\left( {\frac{{{l_2} - {l_1}}}{{{l_1}}}} \right)\%&=3.84 \times {10^{ - 5}} \times 100\\&=\boxed{3.84 \times {{10}^{ - 3}}\% }\\\end{aligned}$

Thus, the percentage decrease in the length of the stool’s legs is $\boxed{3.84 \times {{10}^{-3}\%}}$.

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3. Two children fight over a 200g stuffed bear https://brainly.com/question/6268248

Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Stress and strain

Keywords:

Stool, 3 legged, man, length of the leg, decreases, percentage decrease in length, force, stress, strain, young modulus, legs are vertical, same load.