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Force and Laws of Motion - The Force of Friction: Often

Grade 9CBSE

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

🔑Concepts

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Force of Friction: It is a contact force that opposes the relative motion between two surfaces in contact. It acts along the surfaces of contact and in the direction opposite to the motion or the intended motion.

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Static Friction (fsf_s): The frictional force that acts between two surfaces when there is no relative motion between them, despite an applied force. The maximum value of static friction is called limiting friction.

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Sliding (Kinetic) Friction (fkf_k): The frictional force that acts when one surface slides over another. It is generally slightly less than the limiting static friction (fk<fsf_k < f_s).

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Rolling Friction: The resistance offered when a body rolls over a surface. Rolling friction is much smaller than sliding friction, which is why wheels and ball bearings are used in machinery.

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Factors Affecting Friction: Friction depends on the nature of the surfaces (smoothness or roughness) and the normal force (NN) pressing the two surfaces together. It is independent of the area of contact (for a given normal force).

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Friction as a Necessary Evil: Friction is essential for walking, writing, and braking vehicles, but it causes wear and tear and produces heat which leads to energy loss.

📐Formulae

fs≤μsNf_s \leq \mu_s N

fk=μkNf_k = \mu_k N

N=mg(on a horizontal surface)N = m g \quad \text{(on a horizontal surface)}

Fnet=Fapplied−fkF_{net} = F_{applied} - f_k

a=Fapplied−fkma = \frac{F_{applied} - f_k}{m}

💡Examples

Problem 1:

A wooden block of mass m=5 kgm = 5\text{ kg} is kept on a horizontal floor. The coefficient of static friction μs\mu_s between the block and the floor is 0.40.4. Calculate the minimum force FF required to start moving the block. (Assume g=10 m/s2g = 10\text{ m/s}^2)

Solution:

First, calculate the normal force (NN): N=m×g=5×10=50 NN = m \times g = 5 \times 10 = 50\text{ N} Next, calculate the limiting static friction (fsf_s): fs=μs×N=0.4×50=20 Nf_s = \mu_s \times N = 0.4 \times 50 = 20\text{ N} To move the block, the applied force must overcome the limiting friction. Therefore, Fmin=20 NF_{min} = 20\text{ N}.

Explanation:

The block remains at rest as long as the applied force is less than or equal to the limiting friction. Once the force exceeds 20 N20\text{ N}, the block begins to move.

Problem 2:

A box of mass 10 kg10\text{ kg} is being pushed with a force of 50 N50\text{ N} across a floor. If the force of kinetic friction acting on the box is 20 N20\text{ N}, what is the acceleration of the box?

Solution:

The net force (FnetF_{net}) acting on the box is: Fnet=Fapplied−fkF_{net} = F_{applied} - f_k Fnet=50−20=30 NF_{net} = 50 - 20 = 30\text{ N} Using Newton's second law (F=maF = m a): a=Fnetm=3010=3 m/s2a = \frac{F_{net}}{m} = \frac{30}{10} = 3\text{ m/s}^2

Explanation:

The net force is the difference between the pushing force and the opposing frictional force. This net force is responsible for the acceleration of the object according to the second law of motion.