Answer:
a) [tex]R_{A+B+C} =\sqrt{(30.3603)^{2}+(-23.2963)^{2} } =38.2683 m[/tex]
b) Θ=[tex]tan^{-1} (\frac{R_{y} }{R_{x} } )=tan^{-1} (\frac{-23.2963}{30.3603} )=142.5[/tex]º
c) [tex]R_{A-B+C} =\sqrt{(126.953)^{2}+(2.5856)^{2} } =126.97 m[/tex]
d) β=[tex]tan^{-1} (\frac{R_{y} }{R_{x} } )=tan^{-1} (\frac{2.5856}{126.953} )=1.16[/tex]º
e) [tex]D=\sqrt{(-5)^{2} +(12.059)^{2} } =13.0545m[/tex]
f) ∅=[tex]tan^{-1} (\frac{R_{y} }{R_{x} } )=tan^{-1} (\frac{12.059}{-5} )=111.66[/tex]º
Explanation:
First of all, we need to establish our vectors and it's directions:
A=50 m, 30º
B=50 m, 195º
C=50 m, 315º
Now that we have the three vectors, we need to calculate the x and y components:
[tex]A_{x} =50cos(30)=25\sqrt{3}m[/tex]
[tex]A_{y} =50sin(30)=25m[/tex]
[tex]B_{x} =50cos(195)=-48.2963 m[/tex]
[tex]B_{y} =50sin(195)=-12.941 m[/tex]
[tex]C_{x}=50cos(315)=25\sqrt{2}m[/tex]
[tex]C_{y}=50sin(315)=-25\sqrt{2}m[/tex]
Now, that we have the components, we can calculate the resultant's components:
[tex]R_{x} =A_{x} +B_{x} +C_{x}=25\sqrt{3} +(-48.2963)+25\sqrt{2} =30.3603m[/tex]
[tex]R_{y} =A_{y} +B_{y} +C_{y}=25 +(-12.941)+(-25\sqrt{2}) =-23.2963m[/tex]
To find the resultant of the vector A+B+C we need to do the following steps:
a) [tex]R_{A+B+C} =\sqrt{(30.3603)^{2}+(-23.2963)^{2} } =38.2683 m[/tex]
To find the angle it's necessary to use [tex]tan^{-1}[/tex]:
b) Θ=[tex]tan^{-1} (\frac{R_{y} }{R_{x} } )=tan^{-1} (\frac{-23.2963}{30.3603} )=142.5[/tex]º
To find the resultant of the vector A-B+C we need to do the following steps:
[tex]R_{x} =A_{x} -B_{x} +C_{x}=25\sqrt{3} -(-48.2963)+25\sqrt{2} =126.953m[/tex]
[tex]R_{y} =A_{y} -B_{y} +C_{y}=25 -(-12.941)+(-25\sqrt{2}) =2.5856m[/tex]
c) [tex]R_{A-B+C} =\sqrt{(126.953)^{2}+(2.5856)^{2} } =126.97 m[/tex]
To find the angle it's necessary to use [tex]tan^{-1}[/tex]:
d) β=[tex]tan^{-1} (\frac{R_{y} }{R_{x} } )=tan^{-1} (\frac{2.5856}{126.953} )=1.16[/tex]º
To find D=A+B it's important to follow the following steps:
[tex]R_{x} =A_{x} +B_{x} =25\sqrt{3} +(-48.2963)=-5m[/tex]
[tex]R_{y} =A_{y} +B_{y} =25 +(-12.941)=12.059m[/tex]
e) [tex]D=\sqrt{(-5)^{2} +(12.059)^{2} } =13.0545m[/tex]
f) ∅=[tex]tan^{-1} (\frac{R_{y} }{R_{x} } )=tan^{-1} (\frac{12.059}{-5} )=111.66[/tex]º
