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Forces and Motion - Friction and Air Resistance

Grade 7IB

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

🔑Concepts

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Friction is a contact force that acts between two surfaces sliding, or trying to slide, across each other. It always acts in the direction opposite to the direction of motion.

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The magnitude of friction depends on two main factors: the nature of the surfaces in contact (roughness) and the normal force (FNF_N) pressing the surfaces together.

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Air resistance, also known as drag, is a type of friction that occurs when an object moves through the air. It is caused by the collision of the object's surface with air molecules.

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The size of air resistance depends on the object's speed, its cross-sectional area, and its shape. Streamlining is the process of shaping an object to reduce drag.

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Terminal velocity is reached when the downward force of gravity (WW) is exactly balanced by the upward force of air resistance (FdragF_{drag}), resulting in a net force of 0 N0\text{ N} and zero acceleration (a=0 m/s2a = 0\text{ m/s}^2).

📐Formulae

Fnet=Fapplied−FfrictionF_{net} = F_{applied} - F_{friction}

W=m⋅gW = m \cdot g

a=Fnetma = \frac{F_{net}}{m}

At terminal velocity: Fdrag=W\text{At terminal velocity: } F_{drag} = W

💡Examples

Problem 1:

A wooden crate with a mass of 20 kg20\text{ kg} is pushed across a floor with an applied force of 100 N100\text{ N}. If the force of friction between the crate and the floor is 30 N30\text{ N}, calculate the net force (FnetF_{net}) and the resulting acceleration (aa). Assume g=9.8 m/s2g = 9.8\text{ m/s}^2.

Solution:

Fnet=100 N−30 N=70 NF_{net} = 100\text{ N} - 30\text{ N} = 70\text{ N}

a=70 N20 kg=3.5 m/s2a = \frac{70\text{ N}}{20\text{ kg}} = 3.5\text{ m/s}^2

Explanation:

The net force is the difference between the applied force and the friction opposing it. Using Newton's Second Law (F=maF = ma), we find the acceleration by dividing the net force by the mass.

Problem 2:

A skydiver of mass 70 kg70\text{ kg} is falling through the air. At a certain point, the air resistance acting on the skydiver is 700 N700\text{ N}. If g=10 m/s2g = 10\text{ m/s}^2, what is the skydiver's acceleration?

Solution:

First, calculate the weight (WW): W=70 kg×10 m/s2=700 NW = 70\text{ kg} \times 10\text{ m/s}^2 = 700\text{ N}

Calculate Net Force: Fnet=W−Fdrag=700 N−700 N=0 NF_{net} = W - F_{drag} = 700\text{ N} - 700\text{ N} = 0\text{ N}

Therefore: a=0 m/s2a = 0\text{ m/s}^2

Explanation:

Since the weight (downward force) and air resistance (upward force) are equal and opposite, the net force is zero. The skydiver has reached terminal velocity and is no longer accelerating.

Friction and Air Resistance - Revision Notes & Key Formulas | IB Grade 7 Science