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Kinetic Friction: Coefficient, Equation and Solved Examples
Friction, acting between two surfaces in contact, follows certain laws. The eminent fifteenth centuary artist and scientist, Leonardo Da Vinci was the first to establish these laws experimentally.
Laws of static friction
- Force of static friction between two surfaces in contact always acts against the force that attempts to cause relative motion.
- Force of limiting friction between two surfaces in contact is directly proportional to the normal reaction.
- When normal force remains constant, the force of friction between the surfaces is independent of the area of contact.
Laws of kinetic friction
- Force of kinetic friction between two surfaces in contact always acts against their relative motion.
- Force of kinetic friction between two surfaces in contact, is directly proportional to the normal force.
- When normal force remains constant, the force of friction between the surfaces is independent of the relative velocity of the surfaces.
Coefficient of Static Friction
On application of a horizontal force on a wooden block kept on a table, the following forces act on the block [Fig.]:
- Normal force (R) acting vertically upward
- Weight (W) of the block acting vertically downward Mil Static friction opposite to the direction of the applied force
Let the maximum value of static friction or limiting friction be denoted by fs. When the block is just about to move, let the force applied be \(\vec{F}\). Then, \(\vec{f}_s\) = –\(\vec{F}\).
According to the laws of static friction, if a body is in contact with a plane, limiting friction (fs) is directly proportional to the normal reaction (R). With the increase of normal reaction, the number of contact points also increases. So, this increases the adhesion within the area covered by the block. This in turn increases the force of friction.
Hence, fs ∝ R or, fs = µR
where, µ is a constant for that pair of surfaces. This constant is termed as coefficient of static friction or simply, coefficient of friction. The coefficient of friction depends on the materials of the two surfaces in contact and their smoothness.
Coefficient of friction does not depend on the area of the surfaces of contact.
∴ µ = \(\frac{f_s}{R}\) \(=\frac{\text { limiting frictional force }}{\text { normal force }}\)
Definition: Coefficient of static friction is the ratio between the limiting friction and the normal force between the two surfaces in contact.
Coefficient of friction (µ) being a ratio between two forces, is a dimensionless quantity, and has no unit. Value of µ is generally less than 1. In special cases, µ can be equal to or even greater than 1. µ may rise up to 10, for two scientifically cleaned metal surfaces, kept in vacuum. When highly polished and clear metal surfaces are brought together in vaccum, they instantly form a single piece of metal and cannot slide over each other due to sudden increase in frictional force. This happens because large number of atoms (lying on both surfaces) come in contact in this case. This results in a stronger adhesive force which increases friction. For example, µ ≈ 1.6 for two copper surfaces, but µ ≈ 1 if the surfaces are made of glass.
Coefficient of Kinetic Friction
As a wooden block slides over a table, two reaction forces act on the block
1. normal force (R) and
2. force of kinetic friction (fk) [Fig.].
According to the law of kinetic friction, the forces stated above are directly proportional to each other, that is,
fk ∝ R or, fk = µ’R and µ’ = \(\frac{f_k}{R}\)
The constant of proportionality µ’ is called the coefficient of kinetic friction.
This constant µ’ is a ratio between two forces. So, it is dimensionless, and has no unit. As kinetic friction is less than limiting friction, the coefficient of kinetic friction is less than that of static friction µ, i.e., µ’ < µ for a given pair of surfaces. It decreases further if the relative velocity between the surfaces becomes very high. However, for rough estimates, µ’ is often taken to be equal to µ.
Definition; For two surfaces in contact, the ratio of the kinetic friction to the normal force, is called the coefficient of kinetic friction.
Motion over a rough surface: To move a body over a rough surface, the applied force needs to be greater than the force of friction between the surfaces in contact.
Let the acceleration produced on application of a force F on mass m, be a. As the surface is not smooth, the motion of the body is opposed by kinetic friction. Let the normal force of the plane be R, and the coefficient of kinetic friction be µ’. Hence, the force of kinetic friction = µ’R.
∴ F – µ’R = ma or, a = \(\frac{F-\mu^{\prime} R}{m}\)
Numerical Examples
Example 1.
To set a body of mass 5 kg in motion over a horizontal surface, a minimum force of 30 N has to be applied. What is the value of the coefficient of friction?
Solution:
Force of limiting friction, f = 30 N;
weight of the body, mg =5 x 9.8 = 49 N
Normal force R by the plane is equal to the weight of the body as it is on a horizontal plane.
∴ R = mg = 49 N
Hence, µ = \(\frac{f}{R}\) = \(\frac{30}{49}\) = 0.612.
Example 2.
An iron block of mass 10 kg is kept on a horizontal floor. The block is pulled by a rope at an angle 30° with the floor. What should be the minimum force necessary to set the block in motion. Given µ = 0.5
Solution:
Let the minimum force applied by the rope be T which makes an angle θ with the horizontal. Horizontal and vertical components of T are T cos θ and T sin θ respectively. T sinθ acts in the direction of the normal force (R) as shown in the diagram [Fig.], In this problem, θ = 30°.
As the block is about to move, the frictional force becomes equal to the force of limiting friction. The resultant of forces acting on the body is zero, as the block is still stationary.
According to the figure,
net horizontal force = Tcosθ – µR
net vertical force = R + Tsinθ – W
As the block is stationary,
∴ Tcosθ – µR = 0 or µR = Tcosθ …….. (1)
and R + Tsinθ – W = 0 or, R = W – Tsinθ ….. (2)
Dividing (1) by (2), we get µ = \(\frac{T \cos \theta}{W-T \sin \theta}\)
or, T = \(\frac{\mu W}{\mu \sin \theta+\cos \theta}\) = \(\frac{0.5 \times 10 \times 9.8}{0.5 \times \frac{1}{2}+\frac{\sqrt{3}}{2}}\)
= \(\frac{49}{0.25+0.866}\) = 43.9 N
Example 3.
A body moving on the surface of the earth at 14 m ᐧ s-1 comes to rest due to friction after covering 50 m. Find the coefficient of friction between the body and the earth’s surface. Given, acceleration due to gravity = 9.8 m ᐧ s-2
Solution:
Initial velocity, u = 14m ᐧ s-1; final velocity, v = 0; displacement, s = 50 m and mass of the body = m. If retardation is a, then using the formula v2 = u2 – 2as,
∴ Kinetic friction, fk = ma = 1.96m N and normal force, R = mg = 9.8m N
∴ Coefficient of friction,
µ’ = \(\frac{f_k}{R}\) = \(\frac{1.96 m}{9.8 m}\) = 0.2
Example 4.
A man holds a book vertically between his two palms so that It does not fall. The mass of book Is 1 kg and the force exerted by each palm is 2.5 kg × g. Find the coefficient of friction between the book and the palm.
Solution:
Weight of the book is 1 kg × g, which acts vertically downwards. As two surfaces of the book are pressed by two hands, downward force on each surface of the book
= \(\frac{1 \mathrm{~kg} \times g}{2}\) = 0.5 kg × g.
As the book is at rest, the frictional force acts upwards to balance this downward force of 0.5 kg × g on each surface. So the limiting force of friction, f = 0.5 kg × g.
Force exerted by each hand = normal force (R) = 2.5 kg × g
∴ Coefficient of friction.
µ = \(\frac{f}{R}\) = \(\frac{0.5 \mathrm{~kg} \times g}{2.5 \mathrm{~kg} \times g}\) = 0.2
Example 5.
A block of mass 0.1 kg is kept pressed onto a wall by applying a horizontal force of 5 N. If the coefficient of friction between the block and the wall is 0.5 ,find the force of friction on the block.
Solution:
Downward force on the block = weight of the block = mg = 0.1 × 9.8 = 0.98 N
Since the block is at rest the down-ward force must be balanced by the upward frictional force.
∴ Frictional force on the block = 0.98N.
Example 6.
Part of a uniform chain of length L is hanging out of a table. If the coefficient of friction between the chain and the table is µ, estimate the maximum length of the chain that can hang without slipping.
Solution:
Let the maximum length of the chain that can hang out without slipping be l.
Length of the chain on the table = L – l.
If m is mass per unit length of the chain, weight of the chain resting on the table = (L – l)mg
∴ Normal force of the table, R = (L – l) mg
Downward force on the hanging part = weight of the hanging part = l mg
To avoid slipping this downward force has to be balanced by the limiting friction f, i.e., f = lmg Hence, coefficient of friction,
µ = \(\frac{f}{R}\) = \(\frac{l m g}{(L-l) m g}\) = \(\frac{l}{L-l}\)
or, l = µL – µl or, l(1 + µ) = µL
∴ l = \(\frac{\mu L}{1+\mu}\)
Example 7.
A tram is moving with an acceleration of 49 cm ᐧ s-2 using 50 % of its motor power; the remaining 50 % is used up to overcome friction. Find the coefficient of friction between the wheel and the tram line.
Solution:
Let the power of the motor be P, and the mass of the tram be m. Let the distance covered by the tram in time t be s.
As per given conditions, work against friction per second,
\(\frac{\mu m g \times s}{t}\) = P × 50% = \(\frac{P}{2}\)
and rate of work done for the accelerated motion,
\(\frac{m a \times s}{t}\) = P × 50% = \(\frac{P}{2}\) [where a = acceleration] ……… (2)
Now dividing (1) by (2) we get,
∴ \(\frac{\mu g}{a}\) = 1 or, \(\frac{a}{g}\) = \(\frac{49}{980}\) = 0.05