Chapter 6.6, Problem 11E

(a).

To determine

To Calculate: The work done in pulling a rope hanging over the edge of a building to the top the building

Answer to Problem 11E

The work done is $1500lb\cdot ft$

Explanation of Solution

Given Information: Length of the rope is $50ft$ . Weight of the rope is $0.5lbperft$ . The height of the building is $120ft$ .

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 of an infinitesimally short elementof the rope with length $dx$ and at a distance of $x$ from the top of the building is

  $dm=weightperunitlength\times dx=0.5dx$

Work required to pull this infinitesimally small element to the top of the building is

  $dw=F\cdot d=0.5dx\cdot x=0.5xdx$

The work required to pull the entire rope to the top of the building is the integral of $dw$ from $x=0$ for the top of the rope to $x=50ft$ for the bottom of the rope. Therefore

  $W={{\displaystyle \int }}_0^{50}dw={{\displaystyle \int }}_0^{50}0.5xdx$

  $={\frac{0.5}{2}{x}^2]}_0^{50}$

  $=\frac{0.5}{2}\left({50}^2-0^2\right)$

  $=\frac{0.5}{2}\times 2500=625$

Therefore, the force required to pull the entire rope to the top of the building is $625 lb\cdot ft$

(b).

To determine

To Calculate: The work done in pulling up half of a rope hanging over the edge of a building.

Answer to Problem 11E

Explanation of Solution

Given Information: Length of the rope is $50ft$ . Weight of the rope is $0.5lbperft$ . The height of the building is $120ft$ .

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.

UPPER HALF:

Each part of the upper half of the rope travels a different distance. So, use integration to calculate the work done on the upper half as follows:

The weight of an infinitesimally short element of the rope with length $dx$ and at a distance at $x$ from the top of the building is

  $dm=weightperunitlength\times dx=0.5dx$

Work required to pull this infinitesimally small element to the top of the building is

  $dw=F\cdot d=0.5dx\cdot x=0.5xdx$

The work required to pull the upper half of the rope to the top of the building is the integral of $dw$ from $x=0$ for the top of the rope to $x=25ft$ for the mid-point of the rope. Therefore

  $W_1={{\displaystyle \int }}_{x=0}^{x=25}dw={{\displaystyle \int }}_{x=0}^{x=25}0.5xdx$

  $={\frac{0.5}{2}{x}^2]}_0^{25}$

  $=\frac{0.5}{2}\left({25}^2-0^2\right)$

  $=\frac{0.5}{2}\times 625=156.25$

Therefore, the force required to pull the upper half of the rope to the top of the building is $625 lb\cdot ft$ .

LOWER HALF:

The weight of an infinitesimally short element of the rope with length $dx$ and at a distance at $x$ from the top of the building is again given by

  $dm=weightperunitlength\times dx=0.5dx$

Now, each part of the lower half of the rope moves up by exactly a distance of half the length of the rope.

So, the work done to pull this infinitesimally small length by a distance of half the length of the rope is given by

  $dw=F\cdot d=dm\cdot \frac{l}{2}$

  $\Rightarrow dw=0.5dx\times \frac{l}{2}=0.5dx\times \frac{50}{2}=12.5dx$

The total work done one the lower half of the rope is the integral of $dw$ from $x=25$ to $x=50$ . So

  $W_2={{\displaystyle \int }}_{x=25}^{x=50}dw={{\displaystyle \int }}_{x=25}^{x=50}12.5dx$

  $\Rightarrow W_2=12.5\cdot \left(50-25\right)=12.5\times 25=$ 312.5

Therefore, the total work $W$ is given by $W_1+W_2=156.25+312.5=468.75lb\cdot ft$