The resultant vector is 5.0 m at [tex]73.2^{\circ}[/tex] above the x-axis
Explanation:
To solve the problem, we have to resolve each vector along the x- and y-direction, and then find the x- and y- components of the resultant vector.
We have:
Magnitude of A: 3 m
Direction of A with the x-axis: [tex]20^{\circ}[/tex]
Magnitude of B: 4 m
Direction of B with the x-axis: [tex]110^{\circ}[/tex]
Resolving the vector A:
[tex]A_x = A cos \theta = (3)(cos 20)=2.82 m\\A_y = A sin \theta = (3)(sin 20)=1.03 m[/tex]
Resolving the vector B:
[tex]B_x = B cos \theta = (4)(cos 110)=-1.37 m\\B_y = B sin \theta = (4)(sin 110)=3.76 m[/tex]
Adding the components along the two directions:
[tex]R_x = A_x + B_x = 2.82 + (-1.37)=1.45 m\\R_y = A_y + B_y = 1.03+3.76=4.79 m[/tex]
Now we can find the magnitude of the resultant:
[tex]R=\sqrt{R_x^2+R_y^2}=\sqrt{1.45^2+4.79^2}=5.0 m[/tex]
And their direction is:
[tex]\theta=\tan^{-1} (\frac{R_y}{R_x})=tan^{-1}(\frac{4.79}{1.45})=73.2^{\circ}[/tex]
Learn more about vector addition:
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