drawing force diagrams

drawing force diagrams

  • frictional force is from the centre of the object, along the bottom of the object
  • weight is from center of object downward

Newtons laws

  1. An object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity, unless acted upon by an unbalanced external force.
  2. The acceleration of an object depends directly on the net force acting upon it and inversely on its mass.
  1. For every action, there is an equal and opposite reaction.
    • F⃗A→B=−F⃗B→A\vec{F}_{A \to B} = -\vec{F}_{B \to A}
    • if Object A exerts a force on Object B, Object B simultaneously exerts a force of equal magnitude and in opposite direction on Object A.

Types of forces

Contact

  • Normal
  • Frictional / fluid resistance
  • Tensional
  • Elastic spring force

Non-contact

  • Gravitational
  • Electrostatic

Equations

Newton's second law

\begin{align*} F &= ma \\ & \hspace{-5em} \rlap{\text{\textbf{where}}}\phantom{F} & \\ & \hspace{-5em} \rlap{\quad \begin{array}{rl} F & \text{resultant force in newtons ($N$)} \\ m & \text{mass in kilograms ($kg$)} \\ a & \text{acceleration in metres per second squared ($m/s^2$)} \end{array}}\phantom{F} & \end{align*}

weight

\begin{align*} W &= mg \\ & \hspace{-5em} \rlap{\text{\textbf{where}}}\phantom{W} & \\ & \hspace{-5em} \rlap{\quad \begin{array}{rl} W & \text{weight in newtons ($N$)} \\ m & \text{mass in kilograms ($K$)} \\ g & \text{acceleration due to gravity in $m/s^2$} \end{array}}\phantom{W} & \end{align*}

Fluid resistance

\begin{align*} F &= kv^2 \\ & \hspace{-5em} \rlap{\text{\textbf{where}}}\phantom{F} & \\ & \hspace{-5em} \rlap{\quad \begin{array}{rl} F & \text{force of fluid resistence in newtons ($N$)} \\ k & \text{drag coefficient} \\ v & \text{velocity of object in $ms^{-1}$} \end{array}}\phantom{F} & \end{align*}

Hooke's law

F=−kxwhereFElastic force in newtons (N)kspring constant in (N/M)xdeformation from the equilibrium position in (m)\begin{align*} F &= -kx \\ &\hspace{-5em} \begin{array}{l} \text{\textbf{where}} \\ \quad \begin{array}{rl} F & \text{Elastic force in newtons ($N$)} \\ k & \text{spring constant in ($N/M$)} \\ x & \text{deformation from the equilibrium position in ($m$)} \end{array} \end{array} \end{align*}

Momentum

p=mvwherepmomentum in Newton seconds (N⋅s)mmass in kgvvelocity in ms−1\begin{align*} p &= mv \\ &\hspace{-5em} \begin{array}{l} \text{\textbf{where}} \\ \quad \begin{array}{rl} p & \text{momentum in Newton seconds ($N \cdot s$)} \\ m & \text{mass in $kg$} \\ v & \text{velocity in $ms^{-1}$} \end{array} \end{array} \end{align*}

idk problems

Phases of object dropped in air

object experiences two forces

  • Weight (WW): Pulling it down due to gravity (W=mgW = mg).
  • Drag / Air Resistance (FF): Pushing it up because the object is pushing through air particles F=kv2F = kv^2
  • Fres=W−FF_{\text{res}} = W - F
  • a=Fresma = \frac{F_{\text{res}}}{m}

Just released

  • vv is very small, thus FF is very small, thus FresF_{\text{res}} is very large, thus aa is at its highest

As speed increases

  • vv increases, thus FF increases, thus FresF_{\text{res}} decreases, thus aa decreases

Terminal velocity

  • F=WF = W
  • FF has increased such that it is now equal to WW, thus FresF_{\text{res}} is 0, thus aa is 0 and the object's speed stops changing

objects touching each other

Three blocks of masses m1m_1, m2m_2, and m3m_3 are placed in a straight line on a smooth, frictionless horizontal surface, touching each other in order: m1=2.0 kgm_1 = 2.0\text{ kg}, m2=3.0 kgm_2 = 3.0\text{ kg}, m3=5.0 kgm_3 = 5.0\text{ kg}

A constant horizontal push force of F=20 NF = 20\text{ N} is applied to the left side of block 1 (m1m_1), pushing the entire system to the right.

find

  1. The acceleration (aa) of the whole system.
    • Treat the whole system as one object
F=ma20N=(2.0kg+3.0kg+5.0kg)aa=2.0m/s2\begin{align*} F &= ma \\ 20N &= (2.0kg + 3.0kg + 5.0kg)a \\ a &= 2.0m/s^2 \end{align*}
  1. The contact force (normal force) between block 1 and block 2 (F12F_{12}).
    • block 1 experiences the pushing force from the right and the contact force of block 2 pushing back on block 1
\begin{align*} \text{net force} &= 20N - F_{12} \\ m_{\text{block1}} \cdot a &= 20N - F_{12} \\ (2.0kg)(2.0m/s^2) &= 20N - F_{12} \\ f_{12} &= 16N \end{align*}$$ 3. The contact force between block 2 and block 3 ($F_{23}$). - block 2 experiences the pushing force of block 1, $F_{12}$, and the contact force of block 3 pushing back on block 2

\begin{align*} \text{net force} &= F_{12} + F_{23} \ m_{\text{block2}} \cdot a &= 16N + F_{23} \ (3.0kg)(2.0m/s^2) &= 16N + F_{23} \ F_{23} &= 10N \end{align*}$$

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