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Friction - Factors Affecting Friction

Grade 8CBSE

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

🔑Concepts

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Friction is the force that opposes the relative motion between two surfaces in contact. It acts in a direction opposite to the applied force FappliedF_{applied}.

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The primary cause of friction is the interlocking of irregularities (microscopic hills and valleys) on the surfaces. Rougher surfaces have more irregularities, resulting in higher friction ff.

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Friction depends on the nature of the materials in contact. For instance, the friction between rubber and concrete is higher than between ice and metal.

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Friction is directly proportional to the force pressing the two surfaces together (the normal force NN). Mathematically, f∝Nf \propto N. If the weight of an object increases, the friction also increases.

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Friction is independent of the apparent area of contact. As long as the normal force NN remains constant, changing the surface area in contact does not change the frictional force.

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Static Friction (fsf_s) is the force required to overcome friction at the instant an object starts moving from rest. This is the maximum frictional force.

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Sliding Friction (fkf_k) is the force required to keep an object moving with a constant speed. Sliding friction is slightly smaller than static friction (fk<fsf_k < f_s) because the irregularities do not get enough time to interlock properly when the object is in motion.

📐Formulae

f=μNf = \mu N

N=m×gN = m \times g

fs>fkf_s > f_k

Fnet=Fapplied−fF_{net} = F_{applied} - f

💡Examples

Problem 1:

A brick of mass m=2 kgm = 2\text{ kg} is lying on the floor. If the coefficient of static friction μs\mu_s is 0.50.5 and g=10 m/s2g = 10\text{ m/s}^2, calculate the minimum force FF required to just move the brick.

Solution:

First, calculate the normal force: N=m×g=2×10=20 NN = m \times g = 2 \times 10 = 20\text{ N}. Now, use the friction formula: fs=μs×N=0.5×20=10 Nf_s = \mu_s \times N = 0.5 \times 20 = 10\text{ N}.

Explanation:

To move the brick, the applied force must be at least equal to the maximum static friction, which is 10 N10\text{ N}.

Problem 2:

Why is it easier to pull a trolley with wheels (rolling) than to drag the same trolley without wheels (sliding)?

Solution:

frolling<fslidingf_{rolling} < f_{sliding}

Explanation:

Rolling friction is much smaller than sliding friction because when an object rolls, the area of contact is minimal and the surfaces do not interlock as deeply as they do during sliding.

Problem 3:

A box is being pushed on a floor with a constant force of 25 N25\text{ N} and it moves with a constant velocity. What is the value of the sliding friction fkf_k?

Solution:

Fnet=0  ⟹  Fapplied=fk=25 NF_{net} = 0 \implies F_{applied} = f_k = 25\text{ N}

Explanation:

Since the box moves with a constant velocity, the acceleration is zero (a=0a = 0). According to Newton's law, the net force is zero, meaning the applied force is exactly balanced by the sliding friction.