FRICTION
RON NIÑO Q. ABIERA
• force that resists the sliding or
rolling of one solid object over
another.
• oppose to motion.
• is the resistance to motion of one
object moving relative to another.
• is the force exerted by a surface as
an object moves across it or makes
an effort to move across it.
TYPES OF FRICTION
Kinetic friction (sliding friction), the
frictional force resists the motion of an
object.
Static friction is friction between two
or more solid objects that are not
moving relative to each other.
Rolling friction is the frictional force that
occurs when one object rolls on another,
like a car's wheels on the ground.
Ways of Reducing Friction
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Polishing
Lubricating (oil, waxing grease)
Used ball bearings, wheel, rollers
Powder
Used air cushion
Coefficient of Friction
FRICTION
𝑓𝑓= 𝑢𝑢𝑢𝑢𝑁𝑁
Where:
or
𝑢𝑢 = 𝑓𝑓/𝐹𝐹𝑁𝑁
𝑓𝑓 = Friction force (N)
𝑢𝑢 = Coefficient of Friction
𝐹𝐹𝑁𝑁 = Normal Force (N)
1. A 5 kg box on a horizontal table is
pushed by a horizontal force
of 15 N as shown on the right.
If the coefficient of friction is 0.4, will
the box move?
Given:
m = 5kg.
F = 15 N
𝑢𝑢 = 0.4
to solve for 𝑓𝑓:
𝑓𝑓 = 𝑢𝑢𝐹𝐹𝑁𝑁
= (0.4)(5kg)(9.8m/s2)
= 19.6 N
*the box will not moved because only
15N is applied on the box.
2. A car of mass 1200 kg is travelling along a
straight horizontal road with a velocity of 20
m/s. When it brakes sharply then skids.
Friction
brings
the
car
to
rest.
If the coefficient of friction between the tires
and road is 0.8, calculate deceleration and
distance of the car.
Given:
m = 1200kg
v = 20 m/s
to solve for a, we used;
𝑢𝑢 = 0.8
a = F/m
= (0.8)(1200kg)(9.8m/s2)
* this will be the deceleration since “𝑓𝑓 ”
opposes motion.
to solve for 𝑓𝑓:
𝑓𝑓 = 𝑢𝑢𝐹𝐹𝑁𝑁
= 9408 N
a = -9408N / 1200kg
a = -7.84 m/s2
*the car travels 25.51 m before coming to rest.
3. A 248 kg object moving at 19 m/s comes to
stop over a distance of 38 m. What is the
coefficient of kinetic friction between the
surfaces?
Given:
m = 248 kg.
v = 19 m/s
d = 38m
μk = ?
to solve for a we used kinematic
equation:
vf2 = vi2 + 2ad (we derived)
a = vf2 – vi2/2d
= 0 – (19 m/s)2 / (2)(38m)
= (361 m2/s2) / (76m)
= 4.75 m/s2
to solve for 𝑢𝑢, we derived; 𝑓𝑓= 𝑢𝑢𝑢𝑢𝑁𝑁
𝑓𝑓= 𝑢𝑢𝑢𝑢𝑁𝑁
ma = 𝑢𝑢*mg
a = 𝑢𝑢*g
(w e divide both sides by “m”)
(w e divide both sides by “g”)
𝑢𝑢 = a /g
= (4.75 m/s2)/(9.8m/s2)
= 0.48 is the coefficient of kinetic friction
REFERENCES
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https://www.britannica.com/science/friction
https://www.livescience.com/37161-what-is-friction.html
https://www.youtube.com/watch?v=HRe90ySP38U
https://www.nuffieldfoundation.org/sites/default/files/files/FSMA%20Solve
%20friction%20problems%20student.pdf
Santos, Gil Nonato et.al. I – Physics (Investigatory Physics) 2006.
Cruz, Carmelita. Contemporary Physics. 2000
Cordero – Navaza, Delia et.al. Physics (You and the Natural World). 2011
Cordero – Navaza, Delia et.al. Physics (You and the Natural World). 1996
Van Heuvelen, Alan. Physics, A General Introduction 2nd Ed. 1986
Haliday, David et.al. Fundamental of Physics. 6th Ed. 2001
Microsoft Encarta 2009.
Encyclopedia Britannica 2012.