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Laws of Motion - Friction

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Friction is an opposing force that comes into play when two surfaces in contact move or tend to move relative to each other. It acts tangentially to the surfaces in contact.

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Static Friction (fsf_s) is a self-adjusting force that acts when there is no relative motion. Its maximum value is called Limiting Friction (flimf_{lim}).

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Kinetic Friction (fkf_k) acts when there is relative motion between surfaces. It is generally slightly less than limiting friction (fk<flimf_k < f_{lim}).

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Laws of Limiting Friction: (i) It depends on the nature of surfaces in contact. (ii) It acts tangentially and opposite to the direction of motion. (iii) It is independent of the area of contact. (iv) It is directly proportional to the normal reaction (NN).

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Angle of Friction (θ\theta): The angle which the resultant of limiting friction (flimf_{lim}) and normal reaction (NN) makes with the normal reaction. It is given by tan⁡θ=μs\tan \theta = \mu_s.

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Angle of Repose (α\alpha): The minimum angle of inclination of a plane with the horizontal such that a body placed on it just begins to slide down. Numerically, the angle of repose is equal to the angle of friction (α=θ\alpha = \theta).

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Friction as a centripetal force: For a vehicle taking a turn on a level road, the static friction provides the necessary centripetal force mv2r\frac{mv^2}{r} to prevent skidding.

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Banking of Roads: To reduce reliance on friction and prevent wear and tear, roads are tilted at an angle θ\theta. The horizontal component of the normal reaction helps provide centripetal force.

📐Formulae

fs≤μsNf_s \leq \mu_s N

fk=μkNf_k = \mu_k N

μ=tan⁡θ\mu = \tan \theta

N=mgcos⁡ϕ (on an inclined plane at angle ϕ)N = mg \cos \phi \text{ (on an inclined plane at angle } \phi)

a=g(sin⁡ϕ−μkcos⁡ϕ) (acceleration down an incline)a = g(\sin \phi - \mu_k \cos \phi) \text{ (acceleration down an incline)}

vmax=μsrg (maximum speed on a level circular road)v_{max} = \sqrt{\mu_s rg} \text{ (maximum speed on a level circular road)}

voptimum=rgtan⁡θ (on a banked road without friction)v_{optimum} = \sqrt{rg \tan \theta} \text{ (on a banked road without friction)}

vmax=rg(μs+tan⁡θ1−μstan⁡θ) (on a banked road with friction)v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)} \text{ (on a banked road with friction)}

💡Examples

Problem 1:

A block of mass 10 kg10 \text{ kg} is placed on a horizontal surface. If the coefficient of static friction μs=0.4\mu_s = 0.4 and g=10 m/s2g = 10 \text{ m/s}^2, find the minimum horizontal force required to just move the block.

Solution:

The normal reaction N=mg=10×10=100 NN = mg = 10 \times 10 = 100 \text{ N}. The minimum force required to move the block is equal to the limiting friction flimf_{lim}. Using flim=μsNf_{lim} = \mu_s N, we get F=0.4×100=40 NF = 0.4 \times 100 = 40 \text{ N}.

Explanation:

To initiate motion, the applied force must overcome the maximum value of static friction, which is proportional to the normal reaction.

Problem 2:

Calculate the angle of repose for a block on a plane if the coefficient of friction is 0.5770.577.

Solution:

The angle of repose α\alpha is given by tan⁡α=μ\tan \alpha = \mu. Given μ=0.577\mu = 0.577, we have tan⁡α=0.577≈13\tan \alpha = 0.577 \approx \frac{1}{\sqrt{3}}. Therefore, α=tan⁡−1(13)=30∘\alpha = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30^\circ.

Explanation:

The angle of repose is the specific angle of inclination where the component of weight down the plane exactly equals the limiting frictional force.

Problem 3:

A car of mass 1000 kg1000 \text{ kg} negotiates a curve of radius 50 m50 \text{ m} on a level road. If the coefficient of friction between the tyres and road is 0.50.5, what is the maximum speed the car can achieve without skidding? (g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

The maximum speed is given by vmax=μrgv_{max} = \sqrt{\mu rg}. Substituting the values: vmax=0.5×50×9.8=245≈15.65 m/sv_{max} = \sqrt{0.5 \times 50 \times 9.8} = \sqrt{245} \approx 15.65 \text{ m/s}.

Explanation:

On a level road, the friction force provides the centripetal force. If the required centripetal force exceeds the maximum available friction, the car skids.

Friction - Revision Notes & Key Formulas | ICSE Class 11 Physics