FRICTION  
Friction is a force between two surfaces that are sliding, or trying to slide, across each other.  
OR; Friction – Is the force that opposes motion between any surfaces that are in contact  
For example, when you try to push a book along the floor, friction makes this difficult.  
Friction always works in the direction opposite to the direction in which the object is moving,  
or trying to move.  
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Friction always slows a moving object down.  
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The amount of friction depends on the materials from which the two surfaces are made. The  
rougher the surface, the more friction is produced.  
Friction also produces heat. If you rub your hands together quickly, you will feel them get  
warmer.  
Friction force which occurs in fluids (liquid or gas) is known as viscosity  
Causes of Friction  
Friction is caused by molecular adhesion, surface roughness and deformations  
Adhesion between body surfaces  
:
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Adhesion is the molecular force resulting when two materials are brought into close contact  
with each other. Trying to slide objects against each other requires breaking these adhesive  
bonds.  
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For years, scientists thought that friction was caused by surface roughness, but recent  
studies have shown that it is actually a result of adhesive forces between the materials  
Surface roughness:  
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When two rough surfaces come into contact with each other, they generate frictional force or  
an opposing force, which can sometimes be converted into heat.  
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All solid materials have some degree of surface roughness. If you looked at what seems to  
be a smooth surface under a high-powered microscope, you would see bumps, hills and  
valleys that could interfere with sliding motion.  
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At one time it was thought that the surface roughness of materials was the cause for friction.  
In reality, it only has a small effect on friction for most materials.  
Body deformation:  
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During motion, deformations in the body or the surface of the body in contact may cause  
friction Soft materials will deform when under pressure. This also increases the resistance  
.
to motion.  
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For example, when you stand on a rug, you sink in slightly, which causes resistance when  
you try to drag your feet along the rug's surface.  
Another example is how rubber tires flatten out at the area on contact with the road. When  
materials deform, you must "plow" through to move, thus creating a resistive force.  
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When the deformation becomes large, such that one object sinks into the other, streamlining  
can affect the friction, similar to what happens in fluid friction.  
Advantage of friction  
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It aids in walking and movement  
Helps moving body to stop by applying brakes  
Used to wear unneeded layers of some materials  
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Causes lighting in match stick  
Supports life on the earth by preventing burning asteroids  
Causes nail to stick on the wood  
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Enables bottle stopper to stick on the bottle neck  
Disadvantage of Friction  
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It causes wear and tear  
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Friction causes the wear and tear in mechanical parts of automobiles and vehicles. This is  
the reason why lubricants are used to smoothen the surfaces of brakes, clutches, or pads of  
vehicles  
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It produces heat in various machine parts causing efficiency to decrease  
Due to friction noise is produced in machine  
Causes loss of energy in form of heat and sound  
The force of friction acts in the opposite direction of motion, so friction slows down the  
motion of moving objects.  
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Forest fires are caused due to the friction between tree branches.  
Heat produced by friction can cause appliance to burn  
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It causes wounding, when skin wearing  
Methods of increasing Friction  
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Increasing the normal force by increasing the weight of the body  
Increasing the roughness of the surface  
Use materials of high coefficient of friction. Example, rubber band  
Scrubbing equipment is made rough to increase friction e.g. Steel wire for scrubbing β€œsurface”  
Methods of reducing Friction  
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The use of streamlined bodies. For objects that move in fluids such as boats, planes, cars,  
etc, the shape of their body is streamlined in order to reduce the friction between the bodies of  
the objects as the fluid.  
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By polishing the surface, as polishing makes the surface smooth and friction can be reduced.  
Lubrication. Using lubricants such as oil or grease can reduce the friction between the  
surfaces.  
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Use of ball bearings or roller bearings. When objects are rolled over the surface, the friction  
between the rolled object and surface can be reduced by using ball bearings.  
Avoiding moisture. When the moisture is present the friction is more, So, we must avoid  
moisture between the two surfaces,  
The use of alloys. Friction is reduced by lining the moving parts by an alloy because alloys  
have low coefficients of friction  
Reduce pressure or weight on the object  
Class Activity  
1. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance,  
the ball comes to rest. Explain  
2. Why it is convenient to pull luggage bags fitted with rollers?  
3. When we try to push a very heavy box kept on ground, it does not move at all. Which force is  
preventing this box to move forward? Where does this force act?  
4. Explain why most of vehicles have their engines directly over the drive wheel?  
5. In which case will there be more friction between the truck and the road: when the truck is empty  
or when it is loaded?  
6. Jack has to push a lighter box and Joseph has to push a similar heavier box on the same floor.  
Who will have to apply larger force and why?  
7. You might have noticed that when used for a long time, slippers with rubber soles become  
slippery. Explain the reason.  
8. We use ball bearings between the hub and axle of ceiling fan and bicycles. Why?  
9. Imagine that friction suddenly vanishes. How would life be affected? List ten such situations.  
Normal Force and Friction  
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Let us consider a body of weight mg lying on a horizontal surface as shown in the figure. When  
a body presses against a surface, the surface deforms even if it appears to be rigid. The  
deformed surface pushes the body with a normal force R that is perpendicular to the surface.  
This is called normal reaction force which is equal and opposite to the weight of the body.  
ν‘–ν‘’, . ν‘Ή = ν’Žν’ˆ  
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If you push on an object with a force that is smaller than friction force, then the static friction force  
is exactly equal and opposite to your force ν’Šν’†, . 푭푨 = ν‘­ν‘­), and the object stays at rest.  
But if you increase your force so that it exceeds ν‘­ν‘­, then the maximum friction force is not  
(
enough to keep the object at rest. So it will move, and the friction force will abruptly drop to the  
kinetic value of Β΅kN  
.
(It turns out that Β΅k is always less than or equal to Β΅s). At this situation  
FF β‰  FA so we have to find the net force, ie,. FNET = FA – FF  
NB:  
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When the body starts to move static friction force is equal to limiting friction then the minimum  
force applied tends to start the motion  
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When the body starts to move kinetic friction force is not equal to the minimum force applied,  
then the body tends to start the motion  
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If limiting friction is less than the force applied, the body will move  
If limiting friction is greater than the force applied, then the body cannot move  
Limiting friction: Is the maximum possible value of static friction  
Laws of friction forces  
1. Frictional force is directly proportional to the normal force between the two surfaces in contact.  
(
Fr Ξ± R)  
2. Friction depends on the nature (roughness) of surfaces in contacts.  
3. Friction does not depend on the surface areas in contact. For example If you slide a book lying  
flat or turn it on edge, the force of friction is the same.  
4. Frictional force is independent of the speed once an object has been set in motion  
Types of Friction  
o
o
Solid surfaces are subjected to two types of friction: Static Friction and Kinetic (Dynamic)  
Friction.  
Imagine, for example, trying to slide a heavy box across a concrete floorβ€”you might push very  
hard on the box and not move it at all. This means that the static friction responds to what  
you do β€” it increases to be equal to and in the opposite direction of your push.  
o
If you finally push hard enough, the box seems to slip suddenly and starts to move. Now static  
friction gives way to kinetic friction. Once in motion, it is easier to keep it in motion than it  
was to get it started, indicating that the kinetic frictional force is less than the static  
frictional force.  
Static Friction  
β€’
Static friction – Is the friction which occurs when the two objects are not moving relative to each  
other  
ν‘­
From: ν‘­ ∝ ν‘Ή β†’ ν‘­ = 흁 ν‘Ή  
β†’ 흁 =  
ν’”
ν’”
ν‘Ή
Whereby; 흁풔 – is the coefficient of static friction  
∴
Coefficient of static friction is the ratio of limiting friction force to the normal reaction  
=
(ν‘΅ν’ν’“ν’Žν’‚ν’ ν’“ν’†ν’‚ν’„ν’•ν’Šν’ν’)  
(ν’”ν’•ν’‚ν’•ν’Šν’„ ν’‡ν’“ν’Šν’„ν’•ν’Šν’ν’ν’‚ν’ 풇풐풓풄풆) = (ν‘­)  
ν’Šν’†, ,. νν’”  
ν‘Ή
Dynamic (kinetic) Friction  
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Kinetic friction – Is the friction that occurs when objects are moving relative to each other  
and rub  
against each other  
ν‘­
=
From: ν‘­ ∝ ν‘Ή β†’ ν‘­ = νν’Œ ν‘Ή  
β†’ νν’Œ  
ν‘Ή
Whereby; νν’Œ – is the coefficient of kinetic friction  
∴
Coefficient of kinetic friction is the ratio of kinetic friction force to the  
ν‘²ν’Šν’ν’†ν’•ν’Šν’„ ν’‡ν’“ν’Šν’„ν’•ν’Šν’ν’ 푭풐풓풄풆  
= ( ν‘­  
)
Normal reaction. ν’Šν’†, . , νν’Œ  
=
ν‘΅ν’ν’“ν’Žν’‚ν’ ν‘Ήν’†ν’‚ν’„ν’•ν’Šν’ν’  
ν‘Ή
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Kinetic friction is always slightly less than the limiting friction. This is because once, the motion  
starts actually; inertia of rest has been overcome. Also, when motion has actually started,  
irregularities of one surface have little time to get locked again into the irregularities of other  
surface.  
NB:  
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When a body rolls on the surface of another, the form of kinetic friction that exists between the  
surfaces is called β€˜ROLLING FRICTION’’ For example, when a wheel, a circular disc or a ring or  
a sphere or a cylinder rolls over a surface, the force opposing it is the rolling friction  
.
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Sliding Friction is the kind of kinetic friction that is caused by two bodies rubbing or sliding  
against each other. For example, when a flat block is moved over that flat surface of a table, the  
opposing force is called sliding friction  
.
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It is easy to roll a body than to slide it on the ground ,This is because Rolling friction is always  
less than Sliding friction  
The coefficient of kinetic friction is always less than the coefficient of static friction  
Worked Examples:  
1. A block of mass 500g is pulled along a horizontal surface. If the coefficient of kinetic friction  
between the block and the surface is 0.8. What is the friction force acting on the block as it  
slides?  
Soln:  
Given: m = 500 g = 0.5 kg, νν’Œ = ퟎ. νŸ–, ν‘­ =?  
ν‘­
=
F
rom: νν’Œ  
β†’ ν‘­ = νν’Œν‘Ή = νν’Œν’Žν’ˆ = ퟎ. νŸ“ Γ— ퟎ. νŸ– Γ— ퟏퟎ = νŸ’ν‘΅  
ν‘Ή
2. A block of mass 1000 kg lying steady on the horizontal surface of a table needs 200 N horizontal  
force to come into motion. What is the coefficient of static friction between the block and the surface  
of table?  
ν‘­
ퟐퟎퟎ  
[ANSW: 흁 =  
=
= 0.02]  
νŸνŸŽνŸŽνŸŽΓ—νŸνŸŽ  
ν’Žν’ˆ  
3. A box of 12 kg is being pulled across a horizontal floor by a force of 60 N. If the acceleration of  
the box is 2m/s2, what is the force of friction acting between the box and the floor?  
ANS: m = 12g, FA = 60 N, a = 2m/s2, ma = 12 x 2 = 24 N  
From; FNET = FAPP – FF β†’ ν’Žν’‚ = 60 – FF  
β†’
FF = FA – ma = 60 – 24 = 36N  
4. The brakes of a car moving at 20m/s along a horizontal road are suddenly applied and it  
comes to rest after travelling some distance. If the coefficient of friction between the tyres  
and the road is 0.90 and it is assumed that all four tyres behave identically, find the  
shortest distance the car would travel before coming to a stop.  
ANS; u = 20m/s, v = 0 m/s, νœ‡ = 0.9, s =?  
From; retardation, ν‘Ž = νœ‡ν‘” = 0.9 Γ— 10 = βˆ’9m/s2  
ퟐ
ퟐ
ퟐ
ퟐ
ν’— βˆ’ν’–  
ퟎ βˆ’νŸνŸŽ  
Also; v2 = u2 + 2as β†’ ν’” =  
=
=
βˆ’νŸ’νŸŽνŸŽ = ퟐퟐ.2 m  
νŸν’‚  
νŸΓ—βˆ’νŸ—  
βˆ’νŸνŸ–  
Alternatively;  
k.e = work done against friction  
β†’
ퟏ ν’Žν’—2 = ν‘­ν’”, whereby F = νν’Žν’ˆ  
ퟐ
ퟐ
ퟐ
ν’—
ퟐퟎ  
ퟏ ν’Žν’—2 = νν’Žν’ˆν’” β†’ ν’” =  
=
=
νŸ’νŸŽνŸŽ =22.2 m  
ퟐ
νŸν’ˆν  
νŸΓ—νŸνŸŽΓ—νŸŽ.νŸ—  
νŸνŸ–  
5. Consider an object of mass 50 kg at rest on the floor. A Force of 5 N is applied on the object but  
it does not move. What is the frictional force that acts on the object?  
Solution  
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When the object is at rest, force applied and the static frictional force are equal and  
opposite. The magnitudes of these two forces are equal, ν’Šν’†, . 푭푨 = ν‘­ν‘­ = νŸ“ν‘΅  
Therefore, the static frictional force acting on the object is, ν‘­ν‘Ί = νŸ“ ν‘΅  
β€’
Class Activity  
1. A block of mass 270kg is pulled along a horizontal surface. If the coefficient of kinetic friction  
between the block and the surface is 0.4. What is the friction force acting on the block as it  
slides?  
(ANS: Fr = 1, 080N)  
2. A box of mass 2kg rest on a horizontal surface, a force of 4.4 N is required to just start  
the box moving. What is the coefficient of static friction between the block and the  
surface?  
(ANS: ΞΌ = 0.22)  
3. A box weighing 2000 N is sliding across a cement floor. The force pushing the box is 500N, and  
the coefficient of sliding friction between the box and the floor is 0.20. What is the acceleration  
of the box? [ANS; a = 0.49m/s2]  
4. An alluminium block of mass 2.1kg rests on a steel platform. A horizontal force of 15N is applied  
to the block  
(a) Given that coefficient of limiting friction 0.6, will the block move?  
(b)  
If will move, what will be its acceleration. Given that coefficient of kinetic friction is 0.47  
(
ANS: (a) Since: FA > FF, hence the car will move  
(b) a=2.44 m/s2)  
5. A brick starts sliding with 6m/s across a concrete horizontal surface floor and the coefficient of  
friction between the two surfaces is 0.4. How far will it travel before coming to rest?  
(ANS:  
S = 4.5 m)  
6. Find the static friction between a block of wood of mass 10kg placed on a table. A minimum  
force of 50N is required to make the block just move on the top. (ANS: ΞΌ = 0.5)  
Friction force at Inclined Plane  
β€’
Consider the diagram below, a body of mss,  
m
sliding down the inclined plane at  
휽
β€’
When the object begins to slide (from rest) there will be kinetic friction between the object and  
the incline .Thus the net force will be:  
푭풏풆풕 = 푭푨 βˆ’ ν‘­ν‘Ή  
Whereby: 푭푨 = ν’Žν’ˆν’”ν’Šν’νœ½ and; ν‘­ν‘Ή = ν‘­ν’Œ = νν’Œν‘Ή = νν’Œν’Žν’ˆν’„ν’ν’”νœ½  
푭풏풆풕 = ν’Žν’ˆν’”ν’Šν’νœ½ βˆ’ νν’Œν’Žν’ˆν’„ν’ν’”νœ½ = ν’Žν’ˆ(ν’”ν’Šν’νœ½ βˆ’ νν’Œν’„ν’ν’”νœ½)  
Thus acceleration will be given as: From  
푭풏풆풕 = ν’Žν’‚ = ν’Žν’ˆ(ν’”ν’Šν’νœ½ βˆ’ νν’Œν’„ν’ν’”νœ½)  
β€’
Therefore, the acceleration is given by ,  
(
)
ν’‚ = ν’ˆ ν’”ν’Šν’νœ½ + νν’Œν’„ν’ν’”νœ½ , For upward motion  
β€’
When Fr = 0 (when the incline is frictionless) , 풕풉풆풏, ν’‚ = ν’ˆν’”ν’Šν’νœ½  
At constant speed (at rest), a = 0 m/s2,  
β€’
ν‘­ = ν’Žν’ˆν’”ν’Šν’νœ½ ν‘Žν‘›ν‘‘ ν‘Ή = ν’Žν’ˆν’„ν’ν’”νœ½  
ν‘­
ν‘Ή
ν’Žν’ˆν’”ν’Šν’νœ½  
ν’Žν’ˆν’„ν’ν’”νœ½  
ν‘­ = 흁푹  
β†’ 흁 =  
=
= ν’•ν’‚ν’νœ½  
∴ 푨풕 풄풐풏풔풕풂풏풕 ν’”ν’‘ν’†ν’†ν’… (ν’‚ν’• ν’“ν’†ν’”ν’•), 흁 = ν’•ν’‚ν’νœ½  
Worked Examples:  
1. A block of wood of 4kg just slides without acceleration down an inclined plane of 400 to the  
horizontal. What is the coefficient of dynamic friction?  
Soln: Given: m = 4 kg,  
Consider the fig below  

= 400  
(
)
From: 푨풕 풄풐풏풔풕풂풏풕 ν’”ν’‘ν’†ν’†ν’… ν’‚ = ퟎ , 흁 = ν’•ν’‚ν’νœ½ ∴ ν = ν’•ν’‚ν’νœ½ = ν’•ν’‚ν’νŸ’νŸŽ = ퟎ. νŸ–νŸ’  
2.  
A
rectangular box of mass 10 kg rests on an incline with coefficients of static and kinetic friction  
of 0.55 and 0.25 respectively.  
(a) At what angle will the box begin to slide?  
(b) If the incline is kept at that angle after the box begins to slide, what will be the box’s  
acceleration?  
ANS;  
(a)  
(b)  
0
From: ν‘‘ν‘Žν‘›νœƒ = νœ‡ β†’ νœƒ = ν‘‘ν‘Žν‘›βˆ’1 0.55 = 28.8 β‰ˆ 29  
(
)
2
(
)
ν’‚ = ν’ˆ ν‘Ίν’Šν’νœ½ βˆ’ 흁ν‘ͺν’ν’”νœ½ = ퟐ. νŸ”νŸ”m/s  
3. A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination  
with the horizontal reaches 300 the box starts to slip and slides 4.0 m down the plank in 4.0s. What  
will be the coefficients of static and kinetic friction between the box and the plank, respectively?  
Solution:  
흁S tan = 0.5774,  
=

from; s=ut +1at2, β†’ 4 = 0 Γ— 4 + 1 Γ— ν‘ŽΓ— 42 β†’ ν‘Ž = 0.5ν‘š/s2,also;a=g(Sinνœƒ - νœ‡kCosνœƒ)  
2
2
ퟎ.νŸŽνŸ“βˆ’ν‘Ίν’Šν’νŸ‘νŸŽ  
=
=
(
)
0.5 = 10 ν‘Ίν’Šν’νŸ‘νŸŽ βˆ’ νν’Œν‘ͺν’ν’”νŸ‘νŸŽ β†’ νν’Œ  
βˆ’ν‘ͺν’ν’”νŸ‘νŸŽ  
4. A block of mass 1 kg slides down on a rough inclined plane of inclination 600starting from its top.  
If the coefficient of kinetic friction is 0.5 and length of the plane is 1m, what is the value of work  
done against friction?  
Soln;  
Wd =Force x distance = friction force x length of the plane  
But; F=rictional force, F = 흁푹 = νν’Žν’ˆν‘ͺ풐풔  
,
lengtof plane, s = 1m  
Therefore; Work done = F  
Γ—
S =  
=
Wd = νν’Žν’ˆν‘ͺ풐풔 = ퟎ. νŸ“ Γ— ퟏ Γ— ퟏퟎ Γ— ν‘ͺν’ν’”νŸ”νŸŽ Γ— ퟏ = ퟐ. νŸ“ν‘±  
5.  
Starting from rest, the time taken by a body to slide down a 45Β° inclined plane with  
friction, is twice the time it takes to slide down the same distance in the absence of  
friction. Determine the coefficient of friction between the body and the inclined Plane  
Soln: u = 0m/s,  

=450, t1 = 2t2 = 2t, t2 = t  
Consider the various forces acting on the body have been shown in the figure below.  
NB:  
β€’
The force on the body down the inclined plane in presence of friction, is given by; ν‘šν‘Ž =  
ν‘šν‘”ν‘Ίν’Šν’νœ½ βˆ’ ν’‡ = ν‘šν‘”ν‘Ίν’Šν’νœ½ βˆ’ νν’Žν’ˆν‘ͺν’ν’”νœ½  
ν‘‘β„Žν‘–ν‘  ν‘”ν‘–ν‘£ν‘’ν‘ , ν‘Ž = ν‘”(ν‘Ίν’Šν’νœ½ βˆ’ 흁ν‘ͺν’ν’”νœ½)  
β€’
The force on the body down the inclined plane in absence of friction, is given by; ν‘šν‘Ž =  
ν‘šν‘”ν‘Ίν’Šν’νœ½ β†’ ν’‚ = ν’ˆν‘Ίν’Šν’νœ½  
Since the block is at rest thus initial velocity u = 0, and S1 = S2 =S  
1 ν‘‘2  
1 ν‘Žν‘‘2  
Then, ν‘  = ν‘’ν‘‘ + ν‘Ž  
β†’ ν‘  =  
2
Thus; time taken to slide down the plane for both cases is given by  
νŸν’”  
√
ν’ˆ(ν‘Ίν’Šν’νœ½βˆ’νν‘ͺν’ν’”νœ½)  
= βˆšνŸν’”  
=
S = ퟏ ν’‚ν’•  
ퟐ
----(i),  
β†’ ν’•νŸ  
ퟏ
ν’‚
ퟐ
ퟐ
S = ퟏ ν’‚ν’•  
β†’ ν’•  
----(ii) (when no friction)  
2ν‘   
= βˆšνŸν’” =  
√
ퟐ
ퟐ
ν‘”ν‘†ν‘–ν‘›νœ½  
ퟐ
ν’‚
Now; compare equation (i) and (ii) ie,.. ν‘‘1 = 2ν‘‘2  
2ν‘   
2ν‘   
4
1
= 2 Γ—  
√
√
β†’
=
ν’ˆν‘Ίν’Šν’νœ½  
ν‘”(ν‘Ίν’Šν’νœ½βˆ’νν‘ͺν’ν’”νœ½)  
ν‘”(ν‘Ίν’Šν’νœ½βˆ’νν‘ͺν’ν’”νœ½)  
ν‘”ν‘†ν‘–ν‘›νœ½  
4ν‘”ν‘Ίν’Šν’νœ½ βˆ’ νŸ’νν‘ͺν’ν’”νœ½ = ν‘”ν‘†ν‘–ν‘›νœ½  
β†’ νŸ’νν‘ͺν’ν’”νœ½ = νŸ‘ν‘”ν‘Ίν’Šν’νœ½  
3 ν‘†ν‘–ν‘›νœƒ  
3
3
3
∴ 흁 =  
= ν‘‘ν‘Žν‘›νœƒ = ν‘‘ν‘Žν‘›45 = = ퟎ.75  
4 νΆν‘œν‘ νœƒ  
4
4
4
Class Activity  
1. A mass is placed on an inclined plane such that it can move at constant speed, when slightly  
tapped. If the angle of the plane makes with the horizontal plane is 300. Find the coefficient of  
kinetic friction. (ANS: ΞΌ = 0.56)  
A mass of 5 kg is placed on a plane inclined at an angle of 300 to the horizontal. What is the  
accelerating force required to pull the mass up the plane if the coefficient of friction is 0.5?  
(ANS: FA = 46.65N)  
2.  
3. Block of mass 10kg is moving on inclined plane with constant velocity 10 m/s. What is the  
coefficient of kinetic friction between incline plane and block [ANS; 흁 = ν’•ν’‚ν’νœ½ =  
]
4. A block of wood of mass 5kg is placed on a rough plane inclined at 600. Calculate its  
acceleration down the plane if coefficient of friction between the block and the plane is  
0.32 (ANS: a = 7.1 ms-2)  
5. Why mountain roads are winded up rather than keeping them straight?  
6. You are playing with your younger sister in the snow. She is sitting on a sled and asking you to  
slide her across a flat, horizontal field. You have a choice of pushing her from behind, by  
applying a force at 300 below the horizontal or attaching a rope to the front of the sled and  
pulling with a force at 300 above the horizontal. Which would be easier for you and why?  
Overall summary;  
1. Acceleration of a block (body) against Friction  
(a) Acceleration of a block on horizontal surface  
When a body is moving under application of force P, then kinetic friction opposes its motion, if  
a
is the net acceleration of the body, then  
ν‘·βˆ’ν‘­  
ν’Œ
ν’Žν’‚ = ν‘· βˆ’ ν‘­ν’Œ β†’ ν’‚ =  
ν’Ž
ν‘­
When, P = 0 N, ν’Žν’‚ = ν‘­ β†’ ν’‚ =  
=
νν’Žν’ˆ = νν’ˆ  
ν’Ž
ν’Ž
(b) Acceleration of a block down a rough inclined plane  
From; ma = mgsin - F, but F = νœ‡ν‘šν‘”νΆν‘œν‘ νœƒ  

From; ν‘šν‘Ž = ν‘šν‘”ν‘†ν‘–ν‘›νœƒ βˆ’ νœ‡ν‘šν‘”νΆν‘œν‘ νœƒ = ν‘šν‘” (ν‘†ν‘–ν‘›νœƒ βˆ’ νœ‡νΆν‘œν‘ νœƒ)  
∴ ν’‚ = ν’ˆ (ν‘Ίν’Šν’νœ½ βˆ’ 흁ν‘ͺν’ν’”νœ½)  
(c) Retardation of a block up a rough inclined plane  
ma = mgsin  

+ F, but F = νœ‡ν‘šν‘”νΆν‘œν‘ νœƒ  
ν‘šν‘Ž = ν‘šν‘”ν‘†ν‘–ν‘›νœƒ + νœ‡ν‘šν‘”νΆν‘œν‘ νœƒ = ν‘šν‘” (ν‘†ν‘–ν‘›νœƒ + νœ‡νΆν‘œν‘ νœƒ)  
∴ ν’‚ = ν’ˆ (ν‘Ίν’Šν’νœ½ + 흁ν‘ͺν’ν’”νœ½)  
See the figure below  
(d) For smooth inclined plane, (ie,..when  
흁
= 0)  
∴
a = ν’ˆν‘Ίν’Šν’νœ½  
(e) When the top object keeps sliding with constant velocity, the tangent of that angle is equal to  
흁 (ie,.. ν’•ν’‚ν’νœ½ = νν’Œ)  
2. The work done against friction  
(a) Work done over a rough inclined surface  
β€’
If a body of mass  
m
is moved up on a rough inclined plane through distance s, then;  
Work done (Wd) =Fs = ma x s= mg (Sin 훉 + νν’Œν‘ͺν’ν’”νœ½) S  
β€’
For body of mass  
Work done (Wd) =Fs = ma x s = mg (Sin 훉 βˆ’ νν’Œν‘ͺν’ν’”νœ½) x S  
(b) Work done in sliding a body over a horizontal surface is given by;  
Work done = FF x S = νν’Œν’Žν’ˆ  
m
moves down rough inclined plane through distance s, then;  
S
Self Assessment  
1. Calculate the coefficient of kinetic friction between the surface of a table and a block of wood  
when 5 kg block of wood is moving on the table and experiencing a frictional of 5 N.  
(
ANS: 흁 = ퟎ. ퟏ)  
2. A box weighing 2 kg is at rest on a wooden floor. The coefficient of static friction is 0.6 and  
the coefficient of kinetic friction is 0.35.  
a) What minimum force is required to start the box sliding?  
b) What minimum force is required to keep it sliding at a constant velocity?  
3. A 12 kg box is being pulled across a level floor by a force of 60 N. If the acceleration of the  
box is 2 ms-2 ,What is the force of friction between the box and the floor  
4. A 0.5 kg object is given an initial velocity of 3 m/s after which it slides a distance of 8 m  
across a level floor. What is the coefficient of kinetic friction between the object and the floor?  
5. The coefficient of kinetic friction between a block of wood and a wooden inclined plane at an  
angle of 400 is 0.126. If the friction acting on the sliding prism is 42 N, calculate the mass of  
the prism.(ANS: mass = 43.4 kg)  
6. Calculate the friction force acting on a carton box of mass 9 kg which is moving over a  
surface .The coefficient of kinetic friction between the two surfaces is given as 0.45.  
(ANS: FR = 40.5 N)  
7.  
The coefficient of friction between a particle of mass 8 kg, and a rough horizontal plane is 0.4.  
Given that a horizontal force of 29 N acts on the particle as shown in the figure below. Would  
it start to move (ANS: Since FA(29 N) < FR(32N), No motion)