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F = ma - Vector Notation Simplified Revision Notes

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3.2.5 F = ma - Vector Notation

infoNote

Newton's second law of motion is fundamental in mechanics, expressing the relationship between the force applied to an object, its mass, and the resulting acceleration. In vector notation, this law is expressed as:

F=ma\mathbf{F} = m\mathbf{a}

Where:

  • F\mathbf{F} is the force vector (in Newtons, N\text{N} ).
  • mm is the mass of the object (in kilograms, kg\text{kg} ).
  • a\mathbf{a} is the acceleration vector (in metres per second squared, m/s2\text{m/s}^2 ).

1. Understanding the Vector Notation

  • Vector Quantities: Both force and acceleration are vector quantities, meaning they have both magnitude and direction.
  • Direction: The direction of the force vector F\mathbf{F} is the same as the direction of the acceleration vector a\mathbf{a} .
  • Magnitude: The magnitude of the force is proportional to the mass of the object and the magnitude of the acceleration.

2. Breaking Down the Equation

In vector notation, F=ma\mathbf{F} = m\mathbf{a} applies in multiple dimensions (typically 2D or 3D). The vectors can be broken down into their components along the Cartesian axes:

infoNote

In 2D (Two Dimensions)

Let:

  • F=Fxi^+Fyj^\mathbf{F} = F_x \hat{i} + F_y \hat{j}
  • a=axi^+ayj^\mathbf{a} = a_x \hat{i} + a_y \hat{j} Then:

Fxi^+Fyj^=m(axi^+ayj^)F_x \hat{i} + F_y \hat{j} = m(a_x \hat{i} + a_y \hat{j})

This equation can be split into two separate equations for each component:

  • Along the x-axis: Fx=maxF_x = m a_x
  • Along the y-axis: Fy=may F_y = m a_y
infoNote

In 3D (Three Dimensions)

Let:

  • F=Fxi^+Fyj^+Fzk^\mathbf{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}
  • a=axi^+ayj^+azk^\mathbf{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} Then:

Fxi^+Fyj^+Fzk^=m(axi^+ayj^+azk^)F_x \hat{i} + F_y \hat{j} + F_z \hat{k} = m(a_x \hat{i} + a_y \hat{j} + a_z \hat{k})

This results in three separate equations:

  • Along the xx-axis: Fx=maxF_x = m a_x
  • Along the yy-axis: Fy=mayF_y = m a_y
  • Along the z z-axis: Fz=mazF_z = m a_z

3. Example Problem Using Vector Notation

infoNote

📝Problem: A 5 kg object is subjected to two forces: F1=10 ,N ,i^\mathbf{F_1} = 10 \ , \text{N} \ , \hat{i} and F2=20 ,N ,j^\mathbf{F_2} = 20 \ , \text{N} \ , \hat{j} . Find the acceleration vector a\mathbf{a} of the object.

Solution:

  1. Find the Net Force:

Fnet=F1+F2=10 N i^+20 N j^\mathbf{F_{\text{net}}} = \mathbf{F_1} + \mathbf{F_2} = 10 \, \text{N} \, \hat{i} + 20 \, \text{N} \, \hat{j}

So, Fnet=10 ,i^+20 ,j^ ,N\mathbf{F_{\text{net}}} = 10 \ , \hat{i} + 20 \ , \hat{j} \ , \text{N} . 2. Apply Newton's Second Law:

Fnet=ma\mathbf{F_{\text{net}}} = m\mathbf{a}

10 i^+20 j^=5 a10 \, \hat{i} + 20 \, \hat{j} = 5 \, \mathbf{a}

  1. Solve for the Acceleration Vector a\mathbf{a} :

a=15(10 i^+20 j^)=2 i^+4 j^ m/s2\mathbf{a} = \frac{1}{5}(10 \, \hat{i} + 20 \, \hat{j}) = 2 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}^2

The acceleration vector is a=2 i^+4 j^ m/s2\mathbf{a} = 2 \, \hat{i} + 4 \, \hat{j} \, \text{m/s}^2 .

4. Summary

infoNote

In vector notation, Newton's second law F=ma\mathbf{F} = m\mathbf{a} highlights the directional nature of forces and accelerations. By breaking down vectors into their components along the axes, you can solve for unknown quantities in both 2D and 3D problems. This approach is essential for accurately analysing the motion of objects under the influence of multiple forces.

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