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The Geometry of Power – Advanced Simple Machines - Wheel and Axle – The Steering Mastery-advanced

Grade 9CBSE

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

🔑Concepts

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A Wheel and Axle is an advanced simple machine consisting of two co-axial cylinders of different radii. The larger cylinder is the wheel (RR) and the smaller is the axle (rr). They are joined such that they rotate together around the same axis.

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The Geometry of Power refers to the mathematical relationship between the radii: the larger the radius of the wheel relative to the axle, the greater the torque multiplication.

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Mechanical Advantage (MA): In a force-multiplier setup (effort on wheel), MAMA is the ratio of the radius of the wheel to the radius of the axle. MA=RrMA = \frac{R}{r}.

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Velocity Ratio (VR): This is the ratio of the distance moved by the effort to the distance moved by the load. Since both complete one revolution together, VR=2πR2πr=RrVR = \frac{2\pi R}{2\pi r} = \frac{R}{r}.

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Steering Mastery: In a vehicle's steering system, a large steering wheel allows the driver to apply a small effort over a large distance to produce a large force on the steering column (axle), which turns the heavy wheels of the car.

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Efficiency (\eta): Real-world machines lose energy to friction. Efficiency is defined as η=Useful Work OutputTotal Work Input×100%\eta = \frac{\text{Useful Work Output}}{\text{Total Work Input}} \times 100\%, which simplifies to η=MAVR\eta = \frac{MA}{VR}.

📐Formulae

MA=Load (L)Effort (E)MA = \frac{\text{Load (L)}}{\text{Effort (E)}}

VR=Rr=DdVR = \frac{R}{r} = \frac{D}{d}

Torque (\tau)=F×r\text{Torque (\tau)} = F \times r

η=MAVR=L×rE×R\eta = \frac{MA}{VR} = \frac{L \times r}{E \times R}

WorkLost=WorkInput−WorkOutputWork_{Lost} = Work_{Input} - Work_{Output}

💡Examples

Problem 1:

An advanced steering wheel has a radius of 24 cm24 \text{ cm} and is attached to a steering column (axle) with a radius of 3 cm3 \text{ cm}. If a driver applies an effort of 15 N15 \text{ N} to the wheel, calculate the output force on the axle, assuming the system is 100%100\% efficient.

Solution:

Given: Radius of wheel (RR) = 24 cm24 \text{ cm} Radius of axle (rr) = 3 cm3 \text{ cm} Effort (EE) = 15 N15 \text{ N}

Step 1: Calculate the Mechanical Advantage (MAMA): MA=Rr=243=8MA = \frac{R}{r} = \frac{24}{3} = 8

Step 2: Calculate the Load/Output Force (LL): MA=LE⇒8=L15MA = \frac{L}{E} \Rightarrow 8 = \frac{L}{15} L=8×15=120 NL = 8 \times 15 = 120 \text{ N}

Explanation:

Because the wheel is 8 times larger than the axle, the effort applied is multiplied by 8. Thus, 15 N15 \text{ N} of input force generates 120 N120 \text{ N} of output force.

Problem 2:

A screwdriver (wheel and axle) has a handle diameter of 40 mm40 \text{ mm} and a blade width (axle) of 5 mm5 \text{ mm}. If it takes 200 J200 \text{ J} of work to turn a screw but only 160 J160 \text{ J} is used effectively to overcome friction in the wood, calculate the efficiency and the energy lost to heat.

Solution:

Given: WorkInput=200 JWork_{Input} = 200 \text{ J} WorkOutput=160 JWork_{Output} = 160 \text{ J}

Step 1: Calculate Efficiency (η\eta): η=WorkOutputWorkInput×100%=160200×100%=80%\eta = \frac{Work_{Output}}{Work_{Input}} \times 100\% = \frac{160}{200} \times 100\% = 80\%

Step 2: Calculate Work Lost using vertical subtraction: 200−16040\begin{array}{r} 200 \\ - 160 \\ \hline 40 \end{array} WorkLost=40 JWork_{Lost} = 40 \text{ J}

Explanation:

The efficiency of the screwdriver is 80%80\%. The remaining 20%20\% of the energy (40 J40 \text{ J}) is lost, primarily as heat due to friction between the screwdriver, the screw, and the wood.