Answer:
a) F= 4[tex]\sqrt{2}[/tex] i + 4[tex]\sqrt{2}[/tex] j
b) Θ=11.3
c) The work done is 20[tex]\sqrt{2}[/tex]
Step-by-step explanation:
a) ||F||=8, α=π/4
Fx=||F||·sin(π/4)=8·[tex]\frac{\sqrt{2}}{2}[/tex]
Fy=||F||·cos(π/4)=8·[tex]\frac{\sqrt{2}}{2}[/tex]
F=Fx i + Fy j = 4[tex]\sqrt{2}[/tex] i + 4[tex]\sqrt{2}[/tex] j
b) We can find the value of Θ using the equation:
cos(Θ)=[tex]\frac{F.D}{||F||||D||}[/tex]
where:
D= 3 i + 2 j
F=4[tex]\sqrt{2}[/tex] i + 4[tex]\sqrt{2}[/tex] j
The dot product is defined as the sum of the products of the components of each vector as:
F · D= [tex](4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}[/tex]
||F||= 8
||D||= [tex]\sqrt{3^2+2^2} =\sqrt{13}[/tex]
Hence:
Θ=arccos([tex]\frac{20\sqrt{2} }{8\sqrt{13} }[/tex])
Θ=arccos(0.981)
Θ= 11.3°
c) Work is equal to:
F · D= [tex](4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}[/tex]=28.3
Other way of obtainig the work is:
||F||||D||cos(Θ)
where:
||F||= 8
||D||= [tex]\sqrt{3^2+2^2} =\sqrt{13}[/tex]
Θ=11.3°
So, ||F||||D||cos(Θ)=8×[tex]\sqrt{13}[/tex]×cos(11.3°)=28.3
