Tug of war

Description

Let’s say we have a tug of war between six people, where the rope is loaded as in the figure. The rope has a cross-sectional area $A=100 \ \rm{mm}^2$.

Determine the normal stress $\sigma$ in the three sections I-III.

 

 

Answer

  • $\sigma_\rm{I} = 3\MPa$
  • $\sigma_\rm{II} = 9\MPa$
  • $\sigma_\rm{III} = 7\MPa$

 

Solution

First, start by checking that we have equilibrium (meaning no one is actually winning):

$\eqright -300\Newton -250\Newton-350\Newton + 200\Newton + 350\Newton +350\Newton=0$ Equilibrium fulfilled!

The stress in each section is determined as $\sigma = \frac{N}{A}$, where the normal force is determined using the method of sections.

Equilibrium

Make a free-body diagram through each section and consider the part of the rope to the left.

Section I

$$
\eqright N_\rm I – 300\Newton =0 \gives N_\rm I = 300\Newton
$$
$$
\gives \sigma_\rm I = \frac{300}{100\cdot 10^6} = 3\cdot 10^6 \Pa = 3\MPa
$$
Section II

$$
\eqright N_\rm{II} – 300\Newton – 250\Newton – 350\Newton =0 \gives N_\rm{II} = 900\Newton
$$
$$
\gives \sigma_\rm{II} = \frac{900}{100\cdot 10^6} = 9\MPa
$$

Section III

For variation, we consider the part to the right of the section this time (one is always free to choose).

$$
\eqright -N_\rm{II} + 350\Newton + 250\Newton =0 \gives N_\rm{II} = 700\Newton
$$
$$
\gives \sigma_\rm{III} = \frac{700}{100\cdot 10^6} = 7\MPa
$$

 

 

Leave a Reply