Respuesta :
Answer: The Work done to move the object is [tex]=23.15 \text { foot pounds }[/tex]
Step-by-step explanation:
Given;
Force F=3 [tex]l b s[/tex]
Vectors of F= (3,-1)
Moves object to a distance=9 [tex]f t s[/tex]
Let A and B be the displacements Vectors
A= (0,-5)
B= (6,-2)
To Find:
Work done in foot-pounds
Solution:
Work done [tex]W=F \times D[/tex]
Direction of force vector= (3,-1) (i, j)
=3i-j
Unit of force vector [tex]=\sqrt{3^{2}+\left(-1^{2}\right)}[/tex]
[tex]=\sqrt{9+1}[/tex]
Force vector=(Force/Unit Vector)Direction of force vectors
F=[tex]3 / \sqrt{10} \times(3 i-j)[/tex]
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 [tex]=\sqrt{6^{2}+\left(-7^{2}\right)}[/tex]
[tex]=\sqrt{36+49}[/tex]
[tex]=\sqrt{85}[/tex]
Motion Vector= (Distance moved by the object/Unit motion vector) × (Direction of motion vectors)
[tex]D=9 / \sqrt{85} \times(6 i-7 j)[/tex]
Workdone [tex]W=F \times D[/tex]
[tex]= [3/ \sqrt{10} \times (3i-j)] \times [9/ \sqrt{85} \times (6i-7j)][/tex]
[tex]= [3/ \sqrt{10} \times 9/ \sqrt{85}] \times [(3i-j) \times(6i-7j)][/tex]
[tex]= [3/3.1623 \times 9/9.2195] \times [(3\times6) + ((-1) \times (-7))][/tex]
[tex]= [0.94867 \times 0.97619] \times [18+7][/tex]
[tex]=23.15 \text { foot pounds }[/tex]
Result:
Work done to move an object [tex]=23.15 \text { foot pounds }[/tex]
Answer:
[tex]W = 25.617\,lbf\cdot ft[/tex]
Step-by-step explanation:
The work done to move the object is:
[tex]W = F \cdot s \cdot \cos \alpha[/tex]
The angle of the force with respect to change in the position vector is:
[tex]\alpha = \tan^{-1}\left(-\frac{7}{6} \right) - \tan^{-1}\left(-\frac{1}{3} \right)[/tex]
[tex]\alpha \approx 18.415^{\textdegree}[/tex]
The magnitude of the work done is:
[tex]W = (3\,lbf)\cdot (9\,ft)\cdot \cos 18.415^{\textdegree}[/tex]
[tex]W = 25.617\,lbf\cdot ft[/tex]
