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The laws of Physics Topics are used to explain everything from the smallest subatomic particles to the largest galaxies.
What are the Advantages and Disadvantages of Friction with Examples?
So far we have discussed only the examples of motions where a body slides over a surface. Common examples are, pulling a chair on a floor, skiing on ice, pushing a book over a table, and so on. A characteristic feature of this sliding motion is that, the contact plane of the moving body remains unaltered. In other words, the force of friction acts on a particular surface of the sliding body all the time.
When a body rolls over a surface, the area of contact of the rolling body changes continuously with time. So, the frictional force does not act on any specific surface of the body.
Consider the motion of a wheel over a horizontal plane [Fig.]. In this case the area of contact of the two surfaces at any time during the motion is very small.
When a rigid (or hard) wheel is placed on a horizontal plane, the plane of contact practically becomes a straight line [line OA in Fig.]. Now, let a horizontal force P be applied at the centre C of the wheel. It will initiate a translatory motion of the wheel, along the horizontal surface. Then, a frictional force F will act in the backward direction at the line of contact OA.
As a result, the moment of F about C will generate a rotation of the wheel. Line OA will be displaced, and new lines like O’B will come in contact with the horizontal surface. The overall effect is that, two types of motion (translational and rotational) are set up in the wheel. Such composite motion of the wheel is called rolling.
Definition: When a body translates, as well as rotates over a surface without slipping, its motion is known as rolling; the frictional force developed opposite to the direction of motion is called rolling friction.
We already know that, sliding or kinetic friction is less than the limiting value of static friction (limiting friction). Rolling friction is much lower than even the sliding friction. So it is considerably easier to overcome the rolling friction, and thus to set a body in motion. This is how all wheeled vehicles move. Wheels and tracks are always designed so that the applied force is greater than the rolling friction, but less than the sliding friction.
Car and bicycle tyres have grooved surfaces that increase sliding friction. If, for some reason, the sliding friction suddenly drops below the applied force, wheel of the bicycle slides, or ‘skids’. Vehicles with wheels are not used at all where the sliding friction is very low—for example, flat bottomed carts are used on ice surfaces.
Source of rolling friction: When a wheel rolls on a surface, its motion gets obstructed due to deformation of the wheel, or the surface, or both, at the plane of contact.
In real life, usually a tyre and the ground are not perfectly rigid and both get deformed to some extent.
Such deformation is easily noticeable when a heavy wheel rolls over soft ground [Fig.(a)], At the point of contact of the wheel and the ground, an indentation is created on the surface due to the weight of the wheel. The ground just in front of the wheel swells up slightly.
At the point of contact, the wheel also gets deformed and a slight flattening is easily observed [Fig.(b)], These deformations at the point of contact are the source of rolling friction. When the air pressure is low in a cycle tube, the tyre is flattened, and frictional force increases. If both the surfaces in contact are rigid enough, then the deformation is negligible, and so the rolling friction is quite low. For this reason, the friction is considerably low in case of a train running over a railway track.
Note that, the deformations of the surfaces in contact, in case of a rolling motion, are temporary. As soon as the point of contact shifts, the wheel and the surface regain their original shapes and the deformation shifts to a new point of contact.
Difference between sliding friction and rolling friction:
- The surface of contact remains unchanged in case of sliding friction while in case of rolling friction, the surface of contact keeps changing.
- In case of sliding friction, two surfaces in contact are in motion with respect to each other. In case of rolling friction, there is no relative motion between the two surfaces in contact.
Frictional force opposing the sliding motion (sliding friction) is much more than the force opposing the rolling motion (rolling friction).
Disadvantages Of Friction And Their Remedies
Disadvantages of friction: Friction is disadvantageous in many fields. For example :
i) When a machine is in use, work has to be done against friction due to the relative motion among its different parts. Hence, a part of the applied energy that is spent to overcome friction changes into heat energy and efficiency of the machine reduces.
ii) Different parts of a running machine usually corrode due to friction. The machine may become useless due to heavy wear and tear.
Remedy to minimise the disadvantages caused by friction:
i) If the contact surfaces are very smooth, then friction is less. The surfaces can be made slippery by lubrication. This deposits a thin layer of oil between the two surfaces. Different mineral oils, vaseline, graphite, wax and fat are a few commonly used lubricants.
ii) The coefficient of friction between two steel surfaces is experimentally found to be greater than that between steel and, for example, an alloy of lead and antimony. So, these types of alloys are used in machines made of steel. These are called antifriction alloys. The process of decreasing friction by using antifriction alloys was invented by the scientist Babbit, and so the process is known as babbiting.
iii) Rolling friction is less than kinetic friction. Hence, ball bearings or roller bearings are used in machines, wherever possible.
iv) Streamlining is another method of reducing frictional drag when an object moves through a fluid and the fluid closest to the object opposes that motion. Vehicles driven through fluid with high speed e.g. aeroplanes, jets, spacecrafts are given special shape or streamlined to reduce fluid friction. Birds, fishes also have stream lined bodies to reduce frictional drag as they move through air or water.
v) A large frictional force is encountered by a spacecraft during its high speed journey through the atmosphere, which produces excessive heat on its outer surface. To protect its body against this enormous heat, a special type of thermal barrier is used to cover its body.
Advantages of Friction
To run different machines efficiently, frictional force between the movable parts should he minimised. But it is wrong to think that machines would have been more efficient if friction was totally absent.
As an illustration, we may cite the following examples:
- Various parts of the machine are able to rotate, held together with nuts and bolts because of friction.
- The tyres of a vehicle are made rough to increase friction between the road and the lyres and this prevents skidding.
- Chains are attached to the res of automobiles to increase frictioñ while driving through snow or ice.
A few other examples from day to day life, where friction is involved, are given below:
- Human beings and animals can walk on the ground. A force is exerted obliquely. While walking. the reaction force generated due to friction makes it possible for a man to move forward without slipping.
- Trees can hold the ground tightly with their roots. Otherwise, plants could have been uprooted easily.
- Striking ignites a matchstick.
- Objects can be held with the fingers and palms of our hands.
- Ladders can be supported on vertical walls.
- Sand is thrown on tracks covered with snow to increase friction so that driving or walking on snow becomes safer. Similarly, on a rainy day sand is thrown on the slippery ground to increase the friction between our feet and the ground and thus the chances of slipping is reduced.
Hence, in spite of the inconveniences caused by friction, it plays an important role in nature and in our day-to-day lives.
Numerical Examples
Example 1.
A block of mass 5 kg is kept on a horizontal table. It is connected to a weight of mass 2 kg by a weightless string passing over a smooth pulley. The part of the string on the table Is horizontal, and the weight is hanging freely from the pulley. If the coefficient of kinetic friction between the table and the block is 0.2, find the acceleration of the block. What will be the tension in the string?
Solution:
Weight of the block,
Mg = 5 × 9.8 N
∴ Normal force of the table on the block,
R = 5 × 9.8N
Kinetic friction against the motion of the block,
f’ = µ’R = 0.2 × 5 × 9.8 = 9.8 N
Let the tension in the string be T and acceleration of the block be a [Fig.].
Hence, for the motion of the block,
T – f’ = Ma
or, T – 9.8 = 5a ……. (1)
The weight of mass m (say) moves downwards with the same acceleration.
∴ mg – T = ma
or, 2 × 9.8 – T = 2a [∵ m = 2]
Adding (1) and (2) we get,
2 × 9.8 – 9.8 = 7a
or, a = \(\frac{9.8}{7}\) = 1.4 m ᐧ s-2
Substituting this value in (1), we get,
T – 9.8 = 5 × 1.4 or, T = 16.8 N.
Example 2.
A block of mass m is sliding down a stationary inclined plane. The base of the inclined plane has a length l, and the coefficient of kinetic friction between the block and inclined surface is 0.14. What should be the inclination of the plane so that the block can slide down to the ground in minimum time?
Solution:
Acceleration of the block along the plane,
a = g(sinθ – µ’cosθ)
Let the time required to slide down from A to B be t [Fig.].
when t is minimum, t2 is also minimum.
∴ sin2θ – µ’(1 + cos2θ) = x (say) is maximum.
∴ \(\frac{d x}{d \theta}\) = 0 or, 2cos2θ – µ'(0 – 2sin2θ) = 0,
or, cos 2θ + µ’sin2θ = 0
or, tan2θ = –\(\frac{1}{\mu^{\prime}}\) = –\(\frac{1}{0.14}\) = -tan 82°
or tan2θ = tan(180° – 82°) = tan98°
∴ 2θ = 98° and θ = 49°
Example 3.
A coin slides down an inclined plane of inclination at a constant speed. Prove that if the coin is pushed up with a velocity u on that plane, it can rise up to \(\frac{u^2}{4 g \sin \phi}\), and from there it will not slide down again.
Solution:
Since, the coin slides down [Fig.] with a uniform speed, it has no acceleration along the plane. So,
µ’R = mg sinϕ
Downward force acting on the coin as it is pushed up,
F = mg sinϕ + µ’R = mg sinϕ + mg sinϕ = 2mg sinϕ
Retardation, a = \(\frac{2 m g \sin \phi}{m}\) = 2gsinϕ.
If the coin moves up to s, then,
u2 = 2as or, s = \(\frac{u^2}{2(2 g \sin \phi)}\) = \(\frac{u^2}{4 g \sin \phi}\)
As the corn stops and attempts to come down, limiting friction acts on it. Downward force mg sin along the plane is equal to µ‘R which is always less than µR, as µ > µ’ (since µ is the coefficient of static friction).
Hence, the coin cannot slide down again.
Example 4.
The velocity of a 2.5 kg block sliding down an inclined plane (µ = 0.2) is found to be 1.5 m ᐧ s-1. One second later, it has a velocity of 5 m ᐧ s-1. What is the angle of the plane with respect to the horizontal? [WBCHSE Sample Question ‘13]
Solution:
Downward resultant force on the block [Fig.] along the inclined plane
= mgsinθ – µN = mgsinθ – µmgcosθ
= mg(sinθ – µcosθ)
Downward acceleration,
a = g(sinθ – µmgcosθ) = 9.8(sinθ – 0.2cos)m ᐧ s-1
From the relation, v = u + at,
a = \(\frac{v-u}{t}\) = \(\frac{5-1.5}{1}\) = 3.5 m ᐧ s-1
∴ 9.8(sinθ – 0.2cosθ) = 3.5
or, sinθ – 0.2 cosθ = \(\frac{3.5}{9.8}\) = \(\frac{5}{14}\)
Solving this equation we get, θ = 32°
It is the angle of the plane with respect to the horizontal.
Example 5.
A, B and C are the three blocks of masses 3kg, 4kg and 8kg respectively.
The blocks are placed over one another [Fig.]. Coefficient of friction between each pair of surfaces in contact is 0.25. A is connected to the wall by a massless rigid rod, B and C are connected by an inextensible thread passing over a rigidly fixed smooth pulley. Find the force F required to pull C at a constant speed.
Solution:
Let the weights of the blocks A, B and C be denoted by WA, WB and WC respectively. When the block C moves to the left with uniform velocity, B moves to the right and A remains stationary. Hence, the tension on the thread connected to B is equal to the sum of the forces of friction acting on the upper and lower surfaces of B.
∴ From the free body diagram[Fig.] of B, tension, T = fs + \(f_{k_1}\) = µ × WA + µ(WA + WB)
= 0.25(3 + 3 + 5) × g = 2.5 × g
For C to move at a constant speed, T and frictions on the upper and lower surfaces of C, together, should be equal to F.
∴ From the free body diagram [Fig.] of C,
F = T + \(f_{k_1}\) + \(f_{k_2}\)
= T + µ(WA + WB) + µ(WA + WB + WC)
= 2.5 × g + 0.25(3 + 4) × g + 0.25(3 + 4 + 8) × g
= 8 × 9.8 = 78.4 N.
Example 6.
A block of mass M is at rest on a table [Fig.]. Coefficient of friction between the block and the table is µ.
What can be the maximum weight of B so that the system remains in equilibrium?
Solution:
Let the tension in the thread in part OC be T.
Resolving the tension T into components, we get,
- Tcosθ along AO and
- Tsinθ at right angles to AO along BO.
Let the maximum permitted weight of B = W
In equilibrium, T sinθ = W, T cosθ = µMg
∴ \(\frac{T \sin \theta}{T \cos \theta}\) = \(\frac{W}{\mu M g}\) or, W = µMg tanθ.
Example 7.
Calculate the minimum force required to drag a body of mass m, resting on a horizontal surface. The coefficient of friction between the body and the surface is p. [HS ‘05]
Solution:
Let the applied force be F, acting at an angle θ with the horizontal [Fig.].
fs = force of limiting friction;
mg = weight of the body and R = normal force.
For equilibrium,
R + Fsinθ = mg or, R = mg – F sinθ
and F cosθ = fs = µR = µ(mg – F sinθ)
or, Fcosθ + µF sinθ = µmg
∴ F = \(\frac{\mu m g}{\cos \theta+\mu \sin \theta}\)
Value of F will be minimum when (cosθ + µ sinθ) is maximum.
Example 8.
A block of mass 4 kg is kept on a smooth horizontal table surface [Fig.].
Another body of mass 1 kg is placed over one end of the block. Length of the block is 150 cm. Coefficient of friction between the block and the body is 0.1. If a force of 106 dyn is applied on the block, when does the body fall off the block?
Solution:
Let the force applied on the block, so that the smaller body begins to slide be F (= 106 dyn = 10 N). Let the mass of the block be M and that of the smaller body be m. Hence, the friction f acting in the direction of F and opposing the slipping of the smaller body, is equal to µ mg.
Let the acceleration of masses M and m be a1 and a2 (both taken in the direction of F) respectively, with respect to the table.
Hence, for the 4 kg block,
F – f = Ma1 …….. (1)
and for the 1 kg body,
f = ma2 …….(2)
From equations (1) and (2), we get the acceleration of the body with respect to the block,
If calculations show that a2 – a1 is positive, i.e., a2 > a1, it means that F is less than the limiting friction. In that case, the force of friction actually adjusts itself, and remains less than the limiting value, such that a2 = a1. Then, both the blocks move together without any sliding between them.
Example 9.
At what maximum height, with respect to the lowest point of a hollow sphere of radius r, can a particle stay at rest inside it? Given the coefficient of friction between the sphere and the particle is μ.
Solution:
Let m = mass of the particle, h = AC = maximum height of the point B where the particle can stay at rest [Fig.]
Normal force of the sphere on the particle, R = mg cosθ = component of the particle’s weight in the radial direction.
mgsinθ = tangential component of this particle’s weight = force responsible for this particle’s motion.
For the highest equilibrium point B, this force is equal and opposite to the force of limiting friction, F.
∴ F = mgsinθ
Again F = µR = μmgcosθ
∴ mgsinθ = μmgcosθ
or, tanθ = μ
Now, from the triangle OBC, we have
OC = OB cosθ = r cosθ
∴ h = AC = OA – OC = r – rcosθ
= r(1 – cosθ)
As tanθ = µ, we have cosθ = \(\frac{1}{\sqrt{1+\mu^2}}\)
∴ h = r(1 – \(\frac{1}{\sqrt{1+\mu^2}}\))