Action and Reaction

Storyboard

Newton's third principle defines that forces have to be generated in pairs so that their sum is zero. This implies that before an action there is always a reaction of equal magnitude but in the opposite direction.

>Model

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Mechanisms

Iframe

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Code
Concept
Example: Boosting
Force on a wall
Walking

Mechanisms

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Boosting

Concept

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When a swimmer pushes off, she exerts a force of ($$) on the pool wall, which in turn generates a force of ($$) on her body, propelling her movement:

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Force on a wall

Description

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If we attempt to exert force against a wall, we will notice that the main limitation is determined by the adherence of our shoes to the floor. If the floor is smooth, we will typically begin to slip, thereby limiting the force we are capable of exerting.

It is interesting to note that if we push in a non-horizontal manner, the vertical component will affect our vertical force against the floor. In other words, the vertical reaction to our action against the wall will result in an increase (if we are pushing more upward) or a decrease (if we are pushing more downward) in our weight.

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Walking

Image

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Every time we walk, we need to propel our body with each step. To do this, we place our foot on the ground, and assuming it doesn\'t slide due to friction, our muscles exert a force on our body that propels it forward and transfers the reaction to the foot, which in turn transmits it to the ground (the planet):

Since the planet is enormous, we don\'t directly observe the effect of this reaction. However, if we are standing on a smaller object like a cylinder, we can induce its rolling motion by moving relative to our position on the cylinder while it rolls in the opposite direction.

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Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$m_i$
m_i
Inertial Mass
kg

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$F_A$
F_A
Action force
N
$\Delta p_A$
Dp_A
Momentum variation in action
N/m^2
$\Delta p_R$
Dp_R
Momentum variation in the reaction
N/m^2
$F_R$
F_R
Reaction force
N
$\Delta v_A$
Dv_A
Speed difference after action
m/s
$\Delta t$
Dt
Time elapsed
s
$\Delta v_R$
Dv_R
Variation of speed in reaction to action
m/s

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ \Delta p_A = m_i \Delta v_A $

Dp = m_i * Dv


$ \Delta p_R = m_i \Delta v_R $

Dp = m_i * Dv


$ F_A \equiv\displaystyle\frac{ \Delta p_A }{ \Delta t }$

F = Dp / Dt


$ F_R \equiv\displaystyle\frac{ \Delta p_R }{ \Delta t }$

F = Dp / Dt


$ F_R =- F_A $

F_R =- F_A

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Action and reaction

Equation

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An important aspect of force is that it cannot be created out of nothing. Every time we attempt to generate a action force ($F_A$) (an action), a reaction force ($F_R$) will inevitably be generated with the same magnitude but opposite direction:

$ F_R =- F_A $

$F_A$
Action force
$N$
9790
$F_R$
Reaction force
$N$
9789

In other words, forces always occur in pairs, and the sum of these pairs always equals zero.

ID:(10984, 0)



Medium force (1)

Equation

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The force ($F$) is defined as the momentum variation ($\Delta p$) by the time elapsed ($\Delta t$), which is defined by the relationship:

$ F_A \equiv\displaystyle\frac{ \Delta p_A }{ \Delta t }$

$ F \equiv\displaystyle\frac{ \Delta p }{ \Delta t }$

$F$
$F_A$
Action force
$N$
9790
$\Delta p$
$\Delta p_A$
Momentum variation in action
$kg m/s$
10278
$\Delta t$
Time elapsed
$s$
5103

ID:(3684, 1)



Medium force (2)

Equation

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The force ($F$) is defined as the momentum variation ($\Delta p$) by the time elapsed ($\Delta t$), which is defined by the relationship:

$ F_R \equiv\displaystyle\frac{ \Delta p_R }{ \Delta t }$

$ F \equiv\displaystyle\frac{ \Delta p }{ \Delta t }$

$F$
$F_R$
Reaction force
$N$
9789
$\Delta p$
$\Delta p_R$
Momentum variation in the reaction
$kg m/s$
10279
$\Delta t$
Time elapsed
$s$
5103

ID:(3684, 2)



Variation of moment with constant mass (1)

Equation

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In the case where the inertial Mass ($m_i$) is constant, the momentum variation ($\Delta p$) is proportional to the speed Diference ($\Delta v$):

$ \Delta p_A = m_i \Delta v_A $

$ \Delta p = m_i \Delta v $

$m_i$
Inertial Mass
$kg$
6290
$\Delta p$
$\Delta p_A$
Momentum variation in action
$kg m/s$
10278
$\Delta v$
$\Delta v_A$
Speed difference after action
$m/s$
10280

As the momentum variation ($\Delta p$) is with the inertial Mass ($m_i$) and the speed Diference ($\Delta v$) equal to

$ p = m_i v $



for the case where mass is constant, the change in momentum can be written with the moment ($p$) and the initial moment ($p_0$), which, combined with the speed ($v$) and the initial Speed ($v_0$), yields

$\Delta p = p - p_0 = m_i v - m_i v_0 = m_i ( v - v_0 ) = m_i \Delta v$



where the speed Diference ($\Delta v$) is computed with:

$ \Delta v \equiv v - v_0 $



thus resulting in

$ \Delta p = m_i \Delta v $

ID:(13998, 1)



Variation of moment with constant mass (2)

Equation

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In the case where the inertial Mass ($m_i$) is constant, the momentum variation ($\Delta p$) is proportional to the speed Diference ($\Delta v$):

$ \Delta p_R = m_i \Delta v_R $

$ \Delta p = m_i \Delta v $

$m_i$
Inertial Mass
$kg$
6290
$\Delta p$
$\Delta p_R$
Momentum variation in the reaction
$kg m/s$
10279
$\Delta v$
$\Delta v_R$
Variation of speed in reaction to action
$m/s$
10281

As the momentum variation ($\Delta p$) is with the inertial Mass ($m_i$) and the speed Diference ($\Delta v$) equal to

$ p = m_i v $



for the case where mass is constant, the change in momentum can be written with the moment ($p$) and the initial moment ($p_0$), which, combined with the speed ($v$) and the initial Speed ($v_0$), yields

$\Delta p = p - p_0 = m_i v - m_i v_0 = m_i ( v - v_0 ) = m_i \Delta v$



where the speed Diference ($\Delta v$) is computed with:

$ \Delta v \equiv v - v_0 $



thus resulting in

$ \Delta p = m_i \Delta v $

ID:(13998, 2)