Chapter 6, Problem 78RE

Solution

To find: The work done by a force $F$ of $36\text{ lb}$ by the vector $\left(3,5\right)$ in moving an object $10\text{ ft}$ from $\left(0,0\right)$ to $\left(10,0\right)$ , acting in the direction.

The work done by a force $F$ of $36\text{ lb}$ acting in the direction given by the vector $\left(3,5\right)$ in moving an object $10\text{ ft}$ from $\left(0,0\right)$ to $\left(10,0\right)$ is $185.2\text{ ft-lb}$ .

Given information:

Magnitude of force $=36\text{ lb}$

It is acting in the direction given by the vector $\left(3,5\right)$ in moving an object $10\text{ ft}$ from $\left(0,0\right)$ to $\left(10,0\right)$ .

Formula used:

If $F$ is a constant force in any direction then the work $W$ done by $F$ in moving an object from $A$ to $B$ is:

  $\begin{array}{c}W=F\cdot \overrightarrow{AB}\\ =\left|F\right|\left|\overrightarrow{AB}\right|\cos \theta \end{array}$

where $\theta$ is the angle between $F$ and $\overrightarrow{AB}$ .

Calculation:

Calculate vector $F$ .

Magnitude of force $=36\text{ lb}$

It acts in the direction $\left(3,5\right)$ then,

  $\begin{array}{c}F=36\frac{\left(3,5\right)}{\left|\left(3,5\right)\right|}\\ =\frac{36}{\sqrt{34}}\left(3,5\right)\end{array}$

The direction of motion is from $\left(0,0\right)$ to $\left(10,0\right)$ .

Let $A$ and $B$ be $\left(0,0\right)$ and $\left(10,0\right)$ respectively.

  $\overrightarrow{AB}=\left(10,0\right)$

Calculate the work done by the force.

  $\begin{array}{c}W=F\cdot \overrightarrow{AB}\\ =\frac{36}{\sqrt{34}}\left(3,5\right)\cdot \left(10,0\right)\\ =\frac{36}{\sqrt{34}}\times 30\\ \approx 185.2\end{array}$

Therefore, the work done by a force $F$ of $36\text{ lb}$ acting in the direction given by the vector $\left(3,5\right)$ in moving an object $10\text{ ft}$ from $\left(0,0\right)$ to $\left(10,0\right)$ is $185.2\text{ ft-lb}$ .