WBCHSE Class 11 Physics Notes For Advantages and Disadvantages Of Friction

Disadvantages Of Friction And Their Remedies

WBBSE Class 11 Advantages of Friction Notes

Disadvantages Of Friction: Friction is disadvantageous in many fields. For example:

1. 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 the efficiency of the machine reduces.
2. 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:

Read and Learn More: Class 11 Physics Notes

1. 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.
2. 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 babbling.
3. Rolling friction is less than kinetic friction. Hence, ball bearings or roller bearings are used in machines, wherever possible.
4. 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 are driven through fluid with high speed example, aeroplanes, jets, and spacecrafts are given special shapes or streamlined to reduce fluid friction. Birds, fishes also have streamlined bodies to reduce frictional drag as they move through air or water.
5. 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 the body.

WBCHSE Class 11 Physics Notes For Advantages Of Friction

To run different machines efficiently, the frictional force between the movable parts should be minimized. However, 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:

1. Various parts of the machine are able to rotate, and hold together with nuts and bolts because of friction.
2. The tires of a vehicle are made rough to increase friction between the road and the tires and this prevents skidding.
3. Chains are attached to the tires of automobiles to increase friction while driving through snow or ice.

A few other examples from day-to-day life, where friction is involved, are given below:

1. 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.
2. Trees can hold the ground tightly with their roots.
3. Otherwise, plants could have been uprooted easily.
4. Striking ignites a matchstick.
5. Objects can be held with the fingers and palms of our hands.
6. Ladders can be supported on vertical walls.
7. 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.

WBCHSE Class 11 Physics – Disadvantages and Advantages Of Friction Numerical Examples

Disadvantages of Friction in Physics

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:

Given

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

Weight of the block, Mg = 5 x 9.8 N

∴ The normal force of the table on the block, R = 5×9.8 N

Kinetic friction against the motion of the block, f’ = μ’R = 0.2 x 5 x 9.8 = 9.8 N

Let the tension in the string be T and the acceleration of the block be a.

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 x 9.8- T = 2a [m = 2]….(2)

Adding (1) and (2) we get, 2 x 9.8-9.8 = 7 a

or, a = \(\frac{9.8}{7}\) = 1.4 m·s^-2

Substituting this value in (1), we get, T – 9.8 = 5 x 1.4 or, T= 16.8 N.

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Example 2. A block of mass m is sliding down a stationary inclined plane. The base of the inclined plane has a length of l, and the coefficient of kinetic friction between the block and the 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:

Given

A block of mass m is sliding down a stationary inclined plane. The base of the inclined plane has a length of l, and the coefficient of kinetic friction between the block and the inclined surface is 0.14.

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.

∴ AB = \(\frac{1}{2} a t^2=\frac{l}{\cos \theta}\)

or, \(\frac{1}{2} t^2 g\left(\sin \theta-\mu^{\prime} \cos \theta\right)=\frac{l}{\cos \theta}\)

or, \(t^2=\frac{2 l}{g} \frac{1}{\left(\sin \theta \cos \theta-\mu^{\prime} \cos ^2 \theta\right)}\)

= \(\frac{4 l}{g}\left[\frac{1}{\sin 2 \theta-\mu^{\prime}(1+\cos 2 \theta)}\right]\)

When t is minimum, \(t^2\) is also minimum.

∴ \(\sin 2 \theta-\mu^{\prime}(1+\cos 2 \theta)\)=x (say) is maximum.

∴ \(\frac{d x}{d \theta}=0 \quad or, 2 \cos 2 \theta-\mu^{\prime}(0-2 \sin 2 \theta)=0\),

or, \(\cos 2 \theta+\mu^{\prime} \sin 2 \theta=0\)

or, \(\tan 2 \theta=-\frac{1}{\mu^{\prime}}=-\frac{1}{0.14}=-\tan 82^{\circ}\)

or, \(\tan 2 \theta=\tan \left(180^{\circ}-82^{\circ}\right)=\tan 98^{\circ}\)

∴ \(2 \theta=98^{\circ} and \theta=49^{\circ}\).

Key Benefits of Friction in Everyday Life

Example 3 A coin slides down an inclined plane of inclination ø 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:

Given

A coin slides down an inclined plane of inclination ø at a constant speed.

Since, the coin slides down with a uniform speed, it has no acceleration along the plane. So, fμ’R = mg sinø

The downward force acting on the coin as it is pushed up, F = mg sinø + μ’R = mg sinø + mg sinø = 2mg sinø

Retardation, a = \(\frac{2 m g \sin \phi}{m}=2 g \sin \phi .\)

If the coin moves up to s, then, \(u^2=2 a s \text { or, } s=\frac{u^2}{2(2 g \sin \phi)}=\frac{u^2}{4 g \sin \phi} \text {. }\)

As the coin stops and attempts to come down, limiting friction acts on it. Downward force mg sinø along the plane is equal to μ’R which is always less than μR, as μ > μ’ (since pL 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 • s^-1. One second later, it has a velocity of 5 m • s^-1. What is the angle of the plane with respect to the horizontal?
Solution:

Given

The velocity of a 2.5 kg block sliding down an inclined plane (μ  = 0.2) is found to be 1.5 m • s^-1. One second later, it has a velocity of 5 m • s^-1.

Downward resultant force on the block along the inclined plane

= mgsinθ – μN = mgsinθ – μmgcosθ = mg(sinθ – μcosθ)

Downward acceleration, a = g(sinθ- μcosθ) =9.8(sinθ-0.2cosθ)m · s^-1

From the relation, v = u+at, a = \(\frac{v-u}{t}=\frac{5-1.5}{1}=3.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ \(9.8(\sin \theta-0.2 \cos \theta)=3.5\)

or, \(\sin \theta-0.2 \cos \theta=\frac{3.5}{9.8}=\frac{5}{14}\)

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 3 kg, 4 kg, and 8kg respectively. The blocks are placed over one another. The coefficient of friction between each pair of surfaces in contact is 0.25. A is connected to the wall by a massless rigid rod, and 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:

Given

A, B, and C are the three blocks of masses 3 kg, 4 kg, and 8kg respectively. The blocks are placed over one another. The coefficient of friction between each pair of surfaces in contact is 0.25. A is connected to the wall by a massless rigid rod, and B and C are connected by an inextensible thread passing over a rigidly fixed smooth pulley.

Let the weights of the blocks A, B, and C be denoted by W_A, W_B, and W_C respectively. When 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.

Negative Impacts of Friction: Class 11 Notes

∴ From the free body diagram of B, tension, T =f_s + f_k1 =μ X W_A + μ(W_A+ W_B)

= 0.25(3 + 3 + 4) x g = 2.5 xg

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 of C,

F = T+f_k1+f_k2

= T+μ(W_A+W_B) + μ(W_A+W_B+W_C)

= 2.5 x g+ 0.25(3 + 4) x g+ 0.25(3 + 4 + 8) x g = 8×9.8 = 78.4N.

Example 6. A block of mass M is at rest on a table. The 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:

Given

A block of mass M is at rest on a table. The coefficient of friction between the block and the table is μ.

Let the tension in the thread in part OC be T. Resolving the tension T into components, we get,

1. Tcosθ along AO and
2. Fsinθ at right angles to AO along BO.

Let the maximum permitted weight of B = W

In equilibrium, T sinθ = W, T cosθ = μMg

⇔ \(\frac{T \sin \theta}{T \cos \theta}=\frac{W}{\mu M g} \text { or, } W=\mu M g \tan \theta \text {. }\)

Applications of Friction: Pros and Cons

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 μ.
Solution:

Let the applied force be F, acting at an angle θ with the horizontal.

f_s = force of limiting friction;

mg = the weight of the body and R = normal force.

For equilibrium, R+ F sinθ = mg or, R = mg-F sinθ

and F cosθ = f_s = μR = μ(mg- F sinθ)

or, F cos# + ftF sin# = μmg

∴ F = \(\frac{\mu m g}{\cos \theta+\mu \sin \theta}\)

The value of F will be minimum when (cosθ + μsinθ) is maximum.

∴ \(\frac{d}{d \theta}(\cos \theta+\mu \sin \theta)=0\)

or, \(-\sin \theta+\mu \cos \theta=0 \text { or, } \tan \theta=\mu\)

∴ \(\sin \theta=\frac{\mu}{\sqrt{1+\mu^2}}, \text { and } \cos \theta=\frac{1}{\sqrt{1+\mu^2}}\)

∴ \(F_{\min }=\frac{\mu m g}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu m g}{\sqrt{1+\mu^2}} \text { and } \theta=\tan ^{-1} \mu\)

Short Answer Questions on Friction’s Effects

Example 8. A block of mass 4 kg is kept on a smooth horizontal table surface. Another body of mass 1 kg is placed over one end of the block. The length of the block is 150 cm. The 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:

Given

A block of mass 4 kg is kept on a smooth horizontal table surface. Another body of mass 1 kg is placed over one end of the block. The length of the block is 150 cm. The coefficient of friction between the block and the body is 0.1. If a force of 106 dyn is applied on the block

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 μmg.

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,

∴ \(a_2-a_1 =\frac{f}{m}-\frac{F-f}{M}\)

= \(\frac{\mu m g}{m}-\frac{F-\mu m g}{M}=\frac{\mu g(m+M)-F}{M}\)

Substituting the values, \(a_2-a_1=\frac{0.1 \times 9.8(1+4)-10}{4}\)

= \(-\frac{5.1}{4}=-1.275 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

As, \(s=\frac{1}{2} a t^2\),

t = \(\sqrt{\frac{2 s}{a}}=\sqrt{\frac{2 \times 1.5}{1.275}}\)

[s=150 cm =1.5 m and a = 1.275 m · s^-2]

=1.53 s

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

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 = mg sinθ

Again F = μR = μmgcosθ

∴ mg sinθ = μmg cosθ

or, tanθ = μ

Now, from the triangle OBC, we have OC = OB cosθ = r cosθ

∴ h = AC = OA- OC = r- r cosθ

= r(1 – cosθ)

As \(\tan \theta=\mu\), we have \(\cos \theta=\frac{1}{\sqrt{1+\mu^2}}\)

∴ h = \(r\left(1-\frac{1}{\sqrt{1+\mu^2}}\right)\)