Given
A = 5 m/s
B = 7 m/s at 25° NE
C = 3 m/s at 45° NW
To find the resultant vector for these given vectors,
first we need to resolve the vector B and C into X and Y direction respectively.
On resolution, vector gets distributed into its X-axis and Y-axis components.
Vector B has BCos25° as its X component and BSin25° as its Y component
Similarly, Vector C has CCos45° ac its X-axis component and CSin45° as its Y-axis component.
Now, the algebraic sum of the resolved parts of a series of forces in a particular direction is equal to the resolved part of their resultant in the same direction.
Σ along X-axis
= 5 + BCos25° - CCos45°
= 5 + 7(0.92) - 3(0.76)
= 5 + 6.44 - 2.28
Fx = 9.16 m/s
Σ along Y-axis
= BSin25° + CSin45°
= 7(0.38) + 3(0.64)
= 2.66 + 1.92
Fy = 4.58 m/s
The direction of the resultant vector is given by calculating Tan[tex]\alpha[/tex]
Here Tan[tex]\alpha[/tex] = [tex]\frac{P}{B\\}[/tex]
Tan[tex]\alpha[/tex] = [tex]\frac{4.58}{9.16}[/tex]
Tan[tex]\alpha[/tex] = 0.5
[tex]\alpha[/tex] = Tan⁻¹(0.5)
[tex]\alpha \\[/tex] = 29.5° with the X-axis
Magnitude of the resultant vector can be calculated by
= [tex]\sqrt{Fx^{2} +Fy^{2} +FxFyCos\alpha }[/tex]
= [tex]\sqrt{9.16^{2} +4.58^{2}+ 9.16*4.58*Cos29.5 }[/tex]
= 11.93 m/s
Therefore the resultant vector will be having 11.93 m/s as magnitude directing 29.5° from East to North.
To know more about resultant vector,
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