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

`W = 25.617\,lbf\cdot ft`

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(-(7)/(6) \right) - \tan^(-1)\left(-(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`

Answer 2

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

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 * D`

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

=3i-j

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

`=√(9+1)`

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

F=
`3 / √(10) *(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)}`

`=√(36+49)`

`=√(85)`

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

`D=9 / √(85) *(6 i-7 j)`

Workdone
`W=F * D`

`= [3/ √(10) * (3i-j)] * [9/ √(85) * (6i-7j)]`

`= [3/ √(10) * 9/ √(85)] * [(3i-j)  *(6i-7j)]`

`= [3/3.1623 * 9/9.2195] * [(3*6) + ((-1) * (-7))]`

`= [0.94867 * 0.97619] * [18+7]`

`=23.15 \text { foot pounds }`

Result:

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