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Law of Interaction

Newton’s Laws of Motion 

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Reaction Forces

• Normal Force

 – is a reaction force, perpendicular to the surface(surface being a plane) of contact, of the contact

force exerted on an object by, – for example, the surface of a floor or wall, preventing

the object from penetrating the surface.

• Frictional Force, friction

 – is a reaction force that is parallel to the surface contactand is always opposite of motion

https://en.wikipedia.org/wiki/Contact_forcehttps://en.wikipedia.org/wiki/Contact_forcehttps://en.wikipedia.org/wiki/Contact_forcehttps://en.wikipedia.org/wiki/Contact_force

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Friction Force vs. Weight

W

Ff

Cause of Friction: the microscopic roughness between

surfaces…like two gears lockingtogether.

Cause of Friction

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Friction due to the Surface

Corrugations in the surfaces grind when things slide.

Lubricants fill in the gaps and let things slide more easily.

How does the corrugated surface change Friction?

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  Factors that affect Sliding Friction:

a. The normal Force acting on an object. F N (Fg only equals F N in aspecial circumstance, ie. lab) 

 b.  The coefficient of Friction ( µk  < µs, unitless) The texture of the surfaces. The

stickiness between surfaces.

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Relation of Normal Force and Friction

•    =  •    =  

•    ≤  

•   =  

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Coefficient of Friction

Material on Material s = static friction k  = kinetic friction

steel / steel 0.6 0.4

add grease to steel 0.1 0.05

metal / ice 0.022 0.02

 brake lining / iron 0.4 0.3

tire / dry pavement 0.9 0.8

tire / wet pavement 0.8 0.7

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W

FN

Ff

FA

A father was playing with his daughter inthe snow. The father was exerting a 100 Nforce at a 30 degree angle with respect tothe horizontal to drag the 60 kg mass of

both his daughter and the sled. Find theacceleration of the sled if the coefficient offriction between the snow and sled is 0.15for static and 0.10 for kinetic

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W

FN

Ff

FA

FA(x)

FA(y)

 = ()+=  

 = () +  + =  

() = cos 30 = 100 30  

= 86.6  

() = sin(30) = 100 30  

= 50  

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 = ()+=  

 = () +  + =  

= 50 +  + 60(−9.8) =  

50 +  − 588 = 60(0) 

 − 538 = 0 → = 538 

 =  = (538) = (0.15)(538)   = 80.7 () >   

=  = 0.10 538 = 53.8   = ()+=  86.6 + −53.8 =  

32.8 = 60() 

 = 0.547/2 

From Previous page

() = cos 30 = 

100 30 

= 86.6  () = sin(30) = 

100 30  = 50  

W

FNFA

FA(x)

FA(y)Ff

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Calvin and Hobbes were moving at 10 m/swhen they encountered a patch of snowsuch that they slowed down to stop 20 m

away. If the mass of Calvin, Hobbes andtheir sled is 50 kg, find:

Normal Force presentFrictional Force presentCoefficient of friction (  )

:

∆S =  −

2   → 

a =  −

2∆S

  = 0 − (10/)

2(20)= −2.5/ 

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W

FNFf

 =  =  = 50(−2.5 ) 

 = −125 

 =  + =  

=  + 50 −9.8   = 0   − 490 = 0  = 490 

 =  = 490 → 

=  = 125490 = 0.255 

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A hockey puck was sliding in alayer of air in an air hockey table.

It has a mass of 0.125 kg and

moving at 5 m/s when themachine was turned off causingthe puck to slide in the table.If the coefficient of friction is0.65,

How much time will elapsebefore the puck stops?

 =  =  

 =  + =  

=  + 0.125 −9.8   = 0   − 1.225 = 0 

 = 1.225 

 =  = 0.65 1.225 = 0.79625  

 =  =  −0.79625 = 0.125  

 = −0.79625

0.125

  = −6.37/ W

FNFf

V = 5m/s

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A hockey puck was sliding in alayer of air in an air hockey table. Ithas a mass of 0.125 kg and moving

at 5 m/s when the machine wasturned off causing the puck to slidein the table.If the coefficient of friction is 0.65,How much time will elapse beforethe puck stops?

From Kinematics:

=  −   → 

=  −

 

= 0/ − 5/

−6.37/2  

= 0.785  W

FNFf

V = 5m/s