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Laws of Motion - Static and Kinetic Friction

Grade 11CBSEPhysics

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

🔑Concepts

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Friction is the opposing force that comes into play when two surfaces are in contact and try to move or actually move relative to each other. It acts tangentially to the interface.

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Static Friction (fsf_s) acts when there is no relative motion between the surfaces. It is a self-adjusting force that increases to match the applied force until it reaches its maximum value.

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Limiting Friction (fs,maxf_{s,max}) is the maximum value of static friction. It depends on the nature of the surfaces and the normal reaction (NN): fs,max=μsNf_{s,max} = \mu_s N.

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Kinetic Friction (fkf_k) acts when there is relative motion (sliding) between the surfaces. It is generally constant and slightly less than limiting friction: fk=μkNf_k = \mu_k N, where μk<μs\mu_k < \mu_s.

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The Coefficient of Friction (μ\mu) is a dimensionless constant. μs\mu_s represents the coefficient of static friction, and μk\mu_k represents the coefficient of kinetic friction.

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Angle of Friction (θ\theta) is the angle which the resultant of the limiting friction force (fs,maxf_{s,max}) and normal reaction (NN) makes with the direction of the normal reaction. It is given by tan⁡θ=μs\tan \theta = \mu_s.

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Angle of Repose (α\alpha) is the minimum angle of inclination of a plane with the horizontal such that a body placed on it just begins to slide down. Mathematically, tan⁡α=μs\tan \alpha = \mu_s.

📐Formulae

fs≤μsNf_s \le \mu_s N

fs,max=μsNf_{s,max} = \mu_s N

fk=μkNf_k = \mu_k N

N=mg (on a horizontal surface with no vertical external forces)N = mg \text{ (on a horizontal surface with no vertical external forces)}

μs=tan⁡θ\mu_s = \tan \theta

tan⁡α=μs\tan \alpha = \mu_s

a=Fext−fkm (Acceleration when Fext>fs,max)a = \frac{F_{ext} - f_k}{m} \text{ (Acceleration when } F_{ext} > f_{s,max} \text{)}

💡Examples

Problem 1:

A wooden block of mass m=2 kgm = 2 \text{ kg} rests on a horizontal floor. The coefficient of static friction is μs=0.4\mu_s = 0.4 and the coefficient of kinetic friction is μk=0.3\mu_k = 0.3. If a horizontal force F=6 NF = 6 \text{ N} is applied to the block, calculate the force of friction. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

N=mg=2×10=20 NN = mg = 2 \times 10 = 20 \text{ N}. The limiting friction is fs,max=μsN=0.4×20=8 Nf_{s,max} = \mu_s N = 0.4 \times 20 = 8 \text{ N}. Since the applied force F=6 NF = 6 \text{ N} is less than the limiting friction fs,max=8 Nf_{s,max} = 8 \text{ N}, the block does not move.

Explanation:

Because the block remains at rest, the static friction force must exactly balance the applied force. Therefore, the force of friction is 6 N6 \text{ N}, not 8 N8 \text{ N}.

Problem 2:

A box of mass 10 kg10 \text{ kg} is pushed across a floor with a horizontal force of 50 N50 \text{ N}. If μk=0.2\mu_k = 0.2, find the acceleration of the box. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

First, calculate the normal force: N=mg=10×10=100 NN = mg = 10 \times 10 = 100 \text{ N}. Next, find the kinetic friction: fk=μkN=0.2×100=20 Nf_k = \mu_k N = 0.2 \times 100 = 20 \text{ N}. The net force is Fnet=Fapplied−fk=50−20=30 NF_{net} = F_{applied} - f_k = 50 - 20 = 30 \text{ N}. Acceleration a=Fnetm=3010=3 m/s2a = \frac{F_{net}}{m} = \frac{30}{10} = 3 \text{ m/s}^2.

Explanation:

When the applied force exceeds the limiting static friction (assumed here), the body accelerates. The kinetic friction opposes the motion, reducing the effective force available for acceleration.

Static and Kinetic Friction - Revision Notes & Key Formulas | CBSE Class 11 Physics