Question

Starting at the origin of coordinates, the following displacements are made in the xy-plane. 60mm in the +y
direction, 30mm in the -x direction, 40mm at 150 degrees, and 50mm at 240 degrees. Find the resultant vector
both graphically and algebraically.

Answer 1

The vectors addition  allows finding that the answer for the addition of a series of vectors by vector and analytical methods is

• R = 96.86 mm

• θ = 157.26º

Vectors are magnitudes that have modulus and direction, so the addition must be done with vector algebra.

There are graphical and analytical methods to perform the vectors addition

• A graphic method of adding vectors is to start a vector and place each of the other vectors at the tip of the previous one and the resulting vector is drawn by drawing a result vector from the origin of the first vector to the tip of the last, in the attachment we can see a diagram of this method.

• The analytical method consists of decomposing each vector into a coordinate system, summing each component and then constructing the resulting vector.

We decompose the vectors in the coordinate system that we see as the adjoint.

Vector A

             $A_y$ = 60 j ^ mm

            Aₓ = 0  

Vector B

             Bₓ = -30 i ^ mm

            $B_y$ = 0

Vector C

Module   C = 40 mm

Angle      θ = 150º

let's use trigonometry

            cos 150 = $\frac{C_x}{C}$

            sin 150 = $\frac{C_y}{C}$

            Cₓ = C cos 150

            $C_y$ = C sin 150

            Cₓ = 40 cos 150 = -34.64 mm

            $C_y$ = 40 without 150 = 20 mm

Vector D

 Module D = 50 mm

 Angle   θ = 240º

           cos 240 = $\frac{D_x}{D}$

           sin 240 = $\frac{D_y}{D}$

           Dₓ = D cos 240

           $D_y$ = D sin 240

           Dₓ = 50 cos 240 = -25 mm

           $D_y$ = 50 without 240 = -43.30 mm

We perform the sum of each component

x-axis

            Rₓ = Aₓ + Bₓ + Cₓ + Dₓ

            Rₓ = 0 -30 -34.64 -25

            Rₓ = -89.64 mm

y-axis

           $R_y = A_y + B_y + C_y + D_y$

          $R_y$ = 60 + 0 + 20 -43.30

          $R_y$ = 36.7 mm

We construct the resulting vector

For the module we use the Pythagoras' theorem

           R = $\sqrt{R_x^2 +R_y^2}$

           R = $\sqrt{89.64^2 + 36.7^2}$

           R = 96.86 mm

For the angle we use trigonometry

           tab θ’= $\frac{R_y}{R_x}$

           θ'= tan⁻¹ $\frac{R_y}{R_x}$

           θ’= tan⁻¹ $\frac{36.7}{89.64}$

           θ ’= 22.26º

Since the x component is negative, the angle is in the second quadrant

To measure from the positive side of the x axis

            θ = 180 - θ '

            θ = 180 -22.26

            θ = 157.26º

In conclusion using the vectors addition we can find that the answer for the sum of a series of vectors by vector and analytical methods is

      R = 96.86 mm          

       θ = 157.26º

Learn more about vector addition here:

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