Chapter 6.6, Problem 12E

To determine

To Calculate: The work required to pull a chain lying on the ground to a certain height.

Answer to Problem 12E

The work to lift the chain to a height of $6m$ is $1411.2J$ .

Explanation of Solution

Given Information: Length of the chain is $10m$ . Mass of the chain is $80kg$ . The height to which one end of the chain is to be raised is $6m$ .

Concept Used: The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

If the force is constant, then work done can be expressed as $F\cdot d$ , where $F$ is the force and $d$ is the displacement.

The weight per unit length of the chain is $\frac{TotalWeight}{TotalLength}=\frac{80\times 9.8}{10}=78.4$ (where $9.8$ is the value of $g$ )

The weight of an infinitesimally short element of the chain with length $dx$ and at a distance at $x$ from the ground is

  $dm=weightperunitlength\times dx=78.4dx$

Work required to pull this infinitesimally small element from the ground to its current height $x$ is

  $dw=F\cdot d=78.4dx\cdot x=78.4xdx$

The work required to raise one end of the chain from ground to a height of $6m$ is the integral of $dw$ from $x=0$ for the lower end of the hanging part of the of the chain to $x=6m$ for the upper end of the hanging part of the chain. Therefore

  $W={{\displaystyle \int }}_0^{6}dw={{\displaystyle \int }}_0^{6}78.4xdx$

  $={\frac{78.4}{2}{x}^2]}_0^6$

  $=\frac{78.4}{2}\left(6^2-0^2\right)$

  $=\frac{78.4}{2}\times 36=1411.2$

Therefore, the force required to raise one end of the chain by $6m$ is $1411.2J$