Wize University Physics Textbook (Master) > Force Vectors (Advanced Info)

Scalars and Force Vectors

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Important Definitions
      Scalar: a quantity with magnitude but not direction, an amount, number, "what a scale would read"
examples:  10kg      2m      100 km/hr       3 cats         speed        time       temperature
     Vector: a quantity with both magnitude (the scalar) and direction (this will be our "unit" vector)
represented by boldface letters in textbooks, or "vector sign" above
examples:  10N @ 30°        100 km/hr North            Force         velocity     acceleration         Momentum

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Scalar: a quantity with magnitude but not direction.
Vector: a quantity with both magnitude and direction.
Properties of Vectors
Rectangular components of a 2D Vector Force, 

Fx=−Fcos⁡θ and Fy=−Fsin⁡θF_x=-F\cos\theta\quad\text{ and }\quad F_y=-F\sin\thetaFx​=−Fcosθ and Fy​=−Fsinθ

Fx=Fsin⁡θ and Fy=Fcos⁡θF_x=F\sin\theta\quad\text{ and }\quad F_y=F\cos\thetaFx​=Fsinθ and Fy​=Fcosθ

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Using a small slope triangle, 

FxF1=ac\frac{F_x}{F_1}=\frac{a}{c}F1​Fx​​=ca​
Fx=F1(ac)F_x=F_1\left(\frac{a}{c}\right)Fx​=F1​(ca​)

FyF1=bc\frac{F_y}{F_1}=\frac{b}{c}F1​Fy​​=cb​
Fy=F1(bc)F_y=F_1\left(\frac{b}{c}\right)Fy​=F1​(cb​)

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Dot product,

A⃗ ⋅B⃗ =ABcos⁡θ\vec{A}^{\,}\cdot \vec{B}^{\,}=AB\cos\thetaA⋅B=ABcosθ
A⃗ ⋅B⃗ =AxBx+AyBy+AzBz\vec{A}^{\,}\cdot \vec{B}^{\,}=A_xB_x+A_yB_y+A_zB_zA⋅B=Ax​Bx​+Ay​By​+Az​Bz​
θ=cos⁡−1(A⃗ ⋅B⃗ AB)\theta=\cos^{-1}\left(\frac{\vec{A}^{\,}\cdot \vec{B}^{\,}}{AB}\right)θ=cos−1(ABA⋅B​)

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Vector Manipulation

Things you should know:
How to break down a vector into perpendicular components
How to break down a vector into non-perpendicular component
Projecting a vector along a line
Finding the angle between two vectors
Q: How to break down a vector into perpendicular components?
A: Basic Trigonometric rules, aka. SOH CAH TOA

Q: How to break down a vector into non-perpendicular components?
A: Sine and cosine laws

Q: How to project a vector along a line
A: Using the dot product

Q: Finding the angle between two vectors
A: Modified dot product

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Scalar: a quantity with magnitude but not direction
Vector: a quantity with both magnitude and direction
Unit vector: a vector with magnitude of 1, typically used to indicate direction
Directional angles: the angles between a vector and the positive axis
Properties of Vectors

Magnitude of a vector
A= Ax2+Ay 2 +Az2A=\ \sqrt{A_x^{2_{ }}+A_{y\ }^{2\ }+}A_z^2A= Ax2​​+Ay 2 ​+​Az2​ 

Converting into unit vectors

uA⃗ =A⃗ A=AxAi +AyAj +AzAk\vec{u_A}^{\,}=\frac{\vec{A}^{\,}}{A}=\frac{A_x}{A}i\ +\frac{A_y}{A}j\ +\frac{A_z}{A}kuA​​=AA​=AAx​​i +AAy​​j +AAz​​k

Dot product

A⃗ ⋅B⃗ =ABcos⁡θ\vec{A}^{\,}\cdot \vec{B}^{\,}=AB\cos\thetaA⋅B=ABcosθ
A⃗ ⋅B⃗ =AxBx+AyBy+AzBz\vec{A}^{\,}\cdot \vec{B}^{\,}=A_xB_x+A_yB_y+A_zB_zA⋅B=Ax​Bx​+Ay​By​+Az​Bz​
θ=cos⁡−1(A⃗ ⋅B⃗ AB)\theta=\cos^{-1}\left(\frac{\vec{A}^{\,}\cdot \vec{B}^{\,}}{AB}\right)θ=cos−1(ABA⋅B​)

Can also be thought of as a measure of how parallel two vectors are

Properties of Directional Angles

Unit vector with directional angles
uA⃗ =cos⁡αi+cos⁡βj+cos⁡γk\vec{u_A}^{\,}=\cos\alpha i+\cos\beta j+\cos\gamma kuA​​=cosαi+cosβj+cosγk
Relationship between directional angles

cos⁡2α+cos⁡2β+cos⁡2γ=1\cos^2\alpha^{ }+\cos^2\beta+\cos^2\gamma=1cos2α+cos2β+cos2γ=1


Vector Manipulation

Things you should know:
How to break down a vector into perpendicular components
How to break down a vector into non-perpendicular component
Projecting a vector along a line
Finding the angle between two vectors
Q: How to break down a vector into perpendicular components?
A: Basic Trigonometric rules, aka. SOH CAH TOA

Q: How to break down a vector into non-perpendicular components?
A: Sine and cosine laws

Q: How to project a vector along a line
A: Using the dot product

Q: Finding the angle between two vectors
A: Modified dot product

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- To break down a vector into non-perpendicular components we need to the parallelogram law and trigonometry (sine and cosine laws). 
Watch Out!
DO NOT USE THE PYTHAGOREAN THEOREM. 

Parallelogram Law and Resultants.  
Two or “component” forces add according to the parallelogram law, yielding a resultant force which forms the diagonal of the parallelogram.


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Example
Determine the components of the force along lines AC and BC:


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Begin by reconstructing into a triangle:

By Sine law:

500sin⁡80=ACsin⁡60=BCsin⁡40\frac{500}{\sin80}=\frac{AC}{\sin60}=\frac{BC}{\sin40}sin80500​=sin60AC​=sin40BC​

500sin⁡80=BCsin⁡40\frac{500}{\sin80}=\frac{BC}{\sin40}sin80500​=sin40BC​

BC = 326 NBC\ =\ 326\ NBC = 326 N

500sin⁡80=ACsin⁡60\frac{500}{\sin80}=\frac{AC}{\sin60}sin80500​=sin60AC​

AC = 440 NAC\ =\ 440\ NAC = 440 N

The screw eye illustrated is subjected to two cable forces, as indicated. Determine the equivalent force consisting of a single resultant force.

The magnitude of the resultant vector in N = (answer in N, don't include units)

I don't know

Practice Question
Determine the resultant force along lines AC and BC:

I don't know

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Rectangular Components of a Force. 
- Can be represented in two ways: Scalar or Cartesian Vector notation. 
- To find rectangular components we use the parallelogram law where we resolve the magnitude force into its respective x and y components. 
Example (we will come back to this later!) 
Resolve F2 and F3 into their respective x and y components. 

For force vector F2 = 175N, 

Figure (a), 

F2x = 175N sin⁡(35) = 100.38 NF_{2x}\ =\ 175N\ \sin\left(35\right)\ =\ 100.38\ NF2x​ = 175N sin(35) = 100.38 N
F2y = 175N cos⁡(35) = 143.35 NF_{2y}\ =\ 175N\ \cos\left(35\right)\ =\ 143.35\ NF2y​ = 175N cos(35) = 143.35 N

Figure (b), 

F2x = 175N cos⁡(55) = 100.38 NF_{2x}\ =\ 175N\ \cos\left(55\right)\ =\ 100.38\ NF2x​ = 175N cos(55) = 100.38 N
F2y = 175N sin⁡(55) = 143.35 NF_{2y}\ =\ 175N\ \sin\left(55\right)\ =\ 143.35\ NF2y​ = 175N sin(55) = 143.35 N

Figure (c), 

F2x = 175N sin⁡(35) = 100.38 NF_{2x}\ =\ 175N\ \sin\left(35\right)\ =\ 100.38\ NF2x​ = 175N sin(35) = 100.38 N
F2y = 175N cos⁡(35) = 143.35 NF_{2y}\ =\ 175N\ \cos\left(35\right)\ =\ 143.35\ NF2y​ = 175N cos(35) = 143.35 N

For force vector F3 = 135N, 
We use similar triangles of the given slopes, 

Given Similar Triangles (based on the figure above.)

F310=F3y6→135N10=F3y6\frac{F_3}{10}=\frac{F_{3y}}{6}\rightarrow\frac{135N}{10}=\frac{F_{3y}}{6}10F3​​=6F3y​​→10135N​=6F3y​​

Cross multiply and solve for F3y,

F3y=81.0 NF_{3y}=81.0\mathrm{\ N}F3y​=81.0 N

F310=F3x8→135N10=F3x8\frac{F_3}{10}=\frac{F_{3x}}{8}\rightarrow\frac{135N}{10}=\frac{F_{3x}}{8}10F3​​=8F3x​​→10135N​=8F3x​​

Cross multiply and solve for F3x,

F3x=108.0 NF_{3x}=108.0\ \mathrm{N}F3x​=108.0 N

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Example
Represent the 325N force into its x and y components. 

        Fxy=32513 →  Fx=125N\frac{Fx}{y}=\frac{325}{13}\ \rightarrow\ \ Fx=125NyFx​=13325​ →  Fx=125N

         Fy12=32513  →   Fy=300N\frac{Fy}{12}=\frac{325}{13}\ \ \rightarrow\ \ \ Fy=300N12Fy​=13325​  →   Fy=300N

Practice Question
The following bracket illustrated is subjected to three cable forces. Resolve each force into its corresponding x and y components.

F1x

F1y

F2x

F2y

F3x

F3y

I don't know