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Wize University Statics Textbook (Master) > Force Vectors

Forces Cartesian Vector Form (2D)

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We can represent the x and y components of a force in terms of Cartesian unit vectors i and j. These are called unit vectors because they have a dimensionless magnitude of 1. They are used to assigned the directions along the x and y axes.

For a Force (F) below, the Cartesian vector for F is,


F=Fxi+Fyj\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}


The resultant for co-planar forces (forces in the same plane) can be determine by using the Cartesian vector format. For the figure below, each for can be represented in its Cartesian vector,



F1=F1xi+F1yjF2=−F2xi+F2yjF3=−F3xi−F3yj\begin{array}{c}{\mathbf{F}_{1}=F_{1 x} \mathbf{i}+F_{1 y} \mathbf{j}} \\ {\mathbf{F}_{2}=-F_{2 x} \mathbf{i}+F_{2 y} \mathbf{j}} \\ {\mathbf{F}_{3}=-F_{3 x} \mathbf{i}-F_{3 y} \mathbf{j}}\end{array}








We combine the i and j components for each force to determine the vector resultant,


FR=F1+F2+F3=F1xi+F1yj−F2xi+F2yj−F3xi−F3yj=(F1x−F2x−F3x)i+(F1y+F2y−F3y)j=(FRx)i+(FRy)j\begin{array}{c}{\mathbf{F}_{\mathrm{R}}=\mathbf{F}_{1}+\mathbf{F}_{2}+\mathbf{F}_{3}} \\ {=F_{1 x} \mathbf{i}+F_{1 y} \mathbf{j}-F_{2 x} \mathbf{i}+F_{2 y} \mathbf{j}-F_{3 x} \mathbf{i}-F_{3 y} \mathbf{j}} \\ {=\left(F_{1 x}-F_{2 x}-F_{3 x}\right) \mathbf{i}+\left(F_{1 y}+F_{2 y}-F_{3 y}\right) \mathbf{j}} \\ {=\left(F_{R x}\right) \mathbf{i}+\left(F_{R y}\right) \mathbf{j}}\end{array}

Therefore we can represent the resultant force for several co-planar forces by taking the sum of the x and y components for all the forces,


(FR)x=ΣFx(FR)y=ΣFy\begin{array}{l}{\left(F_{R}\right)_{x}=\Sigma F_{x}} \\ {\left(F_{R}\right)_{y}=\Sigma F_{y}}\end{array}

The magnitude of the resultant force is arrived at by using the following expression,

FR=(FR)x2+(FR)y2F_{R}=\sqrt{\left(F_{R}\right)_{x}^{2}+\left(F_{R}\right)_{y}^{2}}


The direction of the resultant force is determined from trigonometry,


θ=tan⁡−1∣(FR)y(FR)x∣\theta=\tan ^{-1}\left|\frac{\left(F_{R}\right)_{y}}{\left(F_{R}\right)_{x}}\right|



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Example

The eye-bolt illustrated is subjected to three cable forces, as indicated. Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

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We will apply the following principles,
(FR)x=ΣFx(FR)y=ΣFy\begin{array}{l}{\left(F_{R}\right)_{x}=\Sigma F_{x}} \\ {\left(F_{R}\right)_{y}=\Sigma F_{y}}\end{array}

FR=(FR)x2+(FR)y2F_{R}=\sqrt{\left(F_{R}\right)_{x}^{2}+\left(F_{R}\right)_{y}^{2}}

θ=tan⁡−1∣(FR)y(FR)x∣\theta=\tan ^{-1}\left|\frac{\left(F_{R}\right)_{y}}{\left(F_{R}\right)_{x}}\right|
1) Sum the Cartesian force components along x and y axes,



→+(FR)x=ΣFx:F1x+F2x−F3x=5kN+6kNcos⁡(45)−10kNsin⁡(15)=6.6544kN→↑+(FR)y=ΣFy:F2y+F3y=6kNsin⁡(45)+10kNcos⁡(15)=13.902kN↑\begin{array}{l}{\rightarrow+\left(F_{R}\right)_{x}=\Sigma F_{x} : F_{1 x}+F_{2 x}-F_{3 x}=5 k N+6 k N \cos (45)-10 k N \sin (15)=6.6544 \mathrm{kN} \rightarrow} \\ {\uparrow+\left(F_{R}\right)_{y}=\Sigma F_{y} : F_{2 y}+F_{3 y}=6 k N \sin (45)+10 k N \cos (15)=13.902 \mathrm{kN} \uparrow}\end{array}

2) Magnitude FRF_R ,

FR=(FR)x2+(FR)y2=6.65442+13.9022=15.41kNF_{R}=\sqrt{\left(F_{R}\right)_{x}^{2}+\left(F_{R}\right)_{y}^{2}}=\sqrt{6.6544^{2}+13.902^{2}}=15.41 \mathrm{kN}

3) Directional angle θ\theta for FRF_R measured counterclockwise from the positive x axis,

θ=tan⁡−1[(FR)y(FR)x]=tan⁡−1(13.9026.6544)=64.4∘\theta=\tan^{-1}\left[\frac{\left(F_R\right)_y}{\left(F_R\right)_x}\right]=\tan^{-1}\left(\frac{13.902}{6.6544}\right)=64.4^{\circ}

Practice

Determine the single resultant of the four forces applied to the bracket is FRF_Rand it's direction counterclockwise from the positive x-axis.




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Represent the following forces in Cartesian vector form.



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