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Simple Machines - Wheel and Axle

Grade 6ICSE

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

🔑Concepts

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The wheel and axle is a simple machine consisting of a large wheel and a small cylinder called the axle, both joined together such that they rotate about a common axis.

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The radius of the wheel is represented by RR and the radius of the axle is represented by rr.

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It acts as a modified lever where the center of the axle serves as the fulcrum. The radius of the wheel acts as the effort arm, and the radius of the axle acts as the load arm.

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When the effort is applied to the rim of the wheel, a larger force is produced on the axle to lift a heavy load (MA>1MA > 1).

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Common examples include a screwdriver, a steering wheel, a door knob, and a water well windlass.

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The Mechanical Advantage (MAMA) and Velocity Ratio (VRVR) are equal in an ideal machine where friction is neglected.

📐Formulae

Mechanical Advantage (MA)=Radius of the wheel (R)Radius of the axle (r)Mechanical\ Advantage\ (MA) = \frac{\text{Radius of the wheel (R)}}{\text{Radius of the axle (r)}}

Velocity Ratio (VR)=Radius of the wheel (R)Radius of the axle (r)Velocity\ Ratio\ (VR) = \frac{\text{Radius of the wheel (R)}}{\text{Radius of the axle (r)}}

Efficiency (η)=MAVR×100%Efficiency\ (\eta) = \frac{MA}{VR} \times 100\%

Load×r=Effort×RLoad \times r = Effort \times R

💡Examples

Problem 1:

A water well windlass has a wheel of radius 40 cm40\text{ cm} and an axle of radius 8 cm8\text{ cm}. Calculate the Mechanical Advantage of the system.

Solution:

MA=Rr=408=5MA = \frac{R}{r} = \frac{40}{8} = 5

Explanation:

By dividing the radius of the wheel (R=40 cmR = 40\text{ cm}) by the radius of the axle (r=8 cmr = 8\text{ cm}), we find that the machine multiplies the input force by 55 times.

Problem 2:

In a wheel and axle system, an effort of 20 N20\text{ N} is required to lift a load of 100 N100\text{ N}. If the radius of the axle is 5 cm5\text{ cm}, what is the radius of the wheel?

Solution:

Load×r=Effort×RLoad \times r = Effort \times R 100×5=20×R100 \times 5 = 20 \times R R=50020=25 cmR = \frac{500}{20} = 25\text{ cm}

Explanation:

Using the principle of moments for a simple machine, the product of the load and the axle radius must equal the product of the effort and the wheel radius to maintain equilibrium.

Wheel and Axle - Revision Notes & Key Formulas | ICSE Class 6 Science