The addition of vectors allows to find the vector that the equilibrium is
F = (40.02 i ^ - 49.95 j ^) N
Parameters given
- Vector value A = 64.0 N and tea = 128.7º
To find
- The vector that allows equilibrium
The force is a vector magnitude so the sum of the force must be done using the methods to add vectors.
One of the easiest methods to perform the addition of vectors is the analytical method where each vector is decomposed in a Cartesian system and the components added using algebraic summation and then the resulting vector is constructed.
We decompose the vector
cos θ = [tex]\frac{A_x}{A}[/tex]Ax / A
sin θ = [tex]\frac{A_y}{A}[/tex]
Aₓ = A cos θ
[tex]A_y[/tex]= A sin θ
Aₓ = 64 cos 128.7
[tex]A_y[/tex] = 64 sin 128.7
Aₓ = -40.02 N
[tex]A_y[/tex] = 49.95 N
To find the vector that allows equilibrium, we work each axis independently
X axis
Aₓ + Fₓ = 0
Fₓ = - Aₓ
Fₓ = 40.02 N
Y axis
[tex]A_y + F_y =0 \\F_y = - A_y\\F_y = - 49.95 N[/tex]
We can write the resulting vector in two ways
1) F = (40.02 i ^ - 49.95 j ^) N
2) in the form of module and angle
Let's find the module with the Pythagoras' Theorem
F =[tex]\sqrt{F_x} ^2 + F_y^2)\\F = \sqrt{40.02^2 + 49.95^2 }[/tex]
F = 64 N
Angles
tan θ = [tex]\frac{F_y}{F_x}[/tex]
θ = tan⁻¹ [tex]\frac{F_y}{F_x}[/tex]
θ = tan⁻¹ [tex]\frac{-49.95}{40.02}[/tex]
θ = -51.3º
This angle is measured clockwise from the positive side of the x-axis
In conclusion using the sum of vectors we can find the vector that allows the equilibrium is
F = (40.02 i^ - 49.95 j^ ) N
Learn more about adding vectors here:
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