Question

A 3-lb force acting in the direction of the vector (3,-1) moves an object just over 9 ft from point (0,5) to (6,-2). Find the work done to move the object to the nearest foot-pound

Answer 1

Answer: The Work done to move the object is  $=23.15 \text { foot pounds }$

Step-by-step explanation:

Given;

Force F=3 $l b s$

Vectors of F= (3,-1)

Moves object to a distance=9 $f t s$

Let A and B be the displacements Vectors

A= (0,-5)

B= (6,-2)

To Find:

Work done in foot-pounds

Solution:

Work done $W=F \times D$

Direction of force vector= (3,-1)  (i, j)  

                                        =3i-j

Unit of force vector $=\sqrt{3^{2}+\left(-1^{2}\right)}$

                                 $=\sqrt{9+1}$

Force vector=(Force/Unit Vector)Direction of  force vectors

                   F=$3 / \sqrt{10} \times(3 i-j)$

Direction of motion vector= (B-A)  

                                          = (6,-2)-(0,-5) x i,j)

                                          =(6-0),(-2-5) x (i, j)

                                          =(6,-7) x (i, j)

                                          =6i-7j

Unit of motion vector $=\sqrt{6^{2}+\left(-7^{2}\right)}$

                                   $=\sqrt{36+49}$

                                   $=\sqrt{85}$

Motion Vector= (Distance moved by the object/Unit motion vector) × (Direction of motion vectors)

                        $D=9 / \sqrt{85} \times(6 i-7 j)$

Workdone $W=F \times D$

                  $= [3/ \sqrt{10} \times (3i-j)] \times [9/ \sqrt{85} \times (6i-7j)]$

                  $= [3/ \sqrt{10} \times 9/ \sqrt{85}] \times [(3i-j) \times(6i-7j)]$

                  $= [3/3.1623 \times 9/9.2195] \times [(3\times6) + ((-1) \times (-7))]$

                  $= [0.94867 \times 0.97619] \times [18+7]$

                  $=23.15 \text { foot pounds }$

Result:

   Work done to move an object $=23.15 \text { foot pounds }$

Answer 2

Answer:

$W = 25.617\,lbf\cdot ft$

Step-by-step explanation:

The work done to move the object is:

$W = F \cdot s \cdot \cos \alpha$

The angle of the force with respect to change in the position vector is:

$\alpha = \tan^{-1}\left(-\frac{7}{6} \right) - \tan^{-1}\left(-\frac{1}{3} \right)$

$\alpha \approx 18.415^{\textdegree}$

The magnitude of the work done is:

$W = (3\,lbf)\cdot (9\,ft)\cdot \cos 18.415^{\textdegree}$

$W = 25.617\,lbf\cdot ft$