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The Geometry of Power – Advanced Simple Machines - Introduction-advanced

Grade 9CBSE

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

🔑Concepts

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A simple machine is a mechanical device that changes the direction or magnitude of a force. The 'Geometry of Power' relates to how the spatial configuration of the machine determines its efficiency and mechanical benefit.

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Mechanical Advantage (MAMA) is defined as the ratio of the output force (Load) to the input force (Effort). It is expressed as MA=FloadFeffortMA = \frac{F_{load}}{F_{effort}}.

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Velocity Ratio (VRVR) is the ratio of the distance moved by the effort (dEd_E) to the distance moved by the load (dLd_L) in the same time interval. It is determined solely by the geometry of the machine: VR=dEdLVR = \frac{d_E}{d_L}.

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The Principle of Work states that for an ideal machine (where friction is zero), the work input is equal to the work output: Win=WoutW_{in} = W_{out}, which implies FE×dE=FL×dLF_E \times d_E = F_L \times d_L.

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Efficiency (η\eta) is the measure of how much input work is converted into useful output work. In real machines, efficiency is always less than 100%100\% due to energy losses like friction. It is calculated as η=MAVR\eta = \frac{MA}{VR}.

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Power (PP) in the context of machines is the rate at which work is done. It can be expressed as P=F×vP = F \times v, where FF is the force and vv is the velocity.

📐Formulae

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

VR=dEdLVR = \frac{d_E}{d_L}

η=Work OutputWork Input=MAVR\eta = \frac{\text{Work Output}}{\text{Work Input}} = \frac{MA}{VR}

Work=F⋅s⋅cos⁡(θ)\text{Work} = F \cdot s \cdot \cos(\theta)

For an Inclined Plane: VR=1sin⁡(θ)=lh\text{For an Inclined Plane: } VR = \frac{1}{\sin(\theta)} = \frac{l}{h}

Power (P)=Wt=F×v\text{Power } (P) = \frac{W}{t} = F \times v

💡Examples

Problem 1:

A crowbar of length 120 cm120\text{ cm} is used as a lever of the first order to lift a heavy rock. The fulcrum is placed at a distance of 20 cm20\text{ cm} from the rock. If the rock weighs 600 N600\text{ N}, calculate the effort required to lift it, assuming 100%100\% efficiency.

Solution:

Given: Total length of lever =120 cm= 120\text{ cm} Load arm (dLd_L) =20 cm= 20\text{ cm} Effort arm (dEd_E) =120 cm−20 cm=100 cm= 120\text{ cm} - 20\text{ cm} = 100\text{ cm} Load (LL) =600 N= 600\text{ N}

For an ideal lever: E×dE=L×dLE \times d_E = L \times d_L E×100=600×20E \times 100 = 600 \times 20 E=12000100E = \frac{12000}{100} E=120 NE = 120\text{ N}

Explanation:

According to the principle of moments for a lever, the product of effort and effort arm equals the product of load and load arm. By placing the fulcrum closer to the load, we increase the mechanical advantage.

Problem 2:

An inclined plane has a height of 3 m3\text{ m} and a length of 5 m5\text{ m}. A worker uses it to push a crate of 1000 N1000\text{ N} with an effort of 750 N750\text{ N}. Calculate the Velocity Ratio (VRVR) and the Efficiency (η\eta) of the machine.

Solution:

Given: Height (hh) =3 m= 3\text{ m} Length (ll) =5 m= 5\text{ m} Load (LL) =1000 N= 1000\text{ N} Effort (EE) =750 N= 750\text{ N}

  1. Velocity Ratio (VRVR): VR=lh=53≈1.67VR = \frac{l}{h} = \frac{5}{3} \approx 1.67

  2. Mechanical Advantage (MAMA): MA=LE=1000750=43≈1.33MA = \frac{L}{E} = \frac{1000}{750} = \frac{4}{3} \approx 1.33

  3. Efficiency (η\eta): η=MAVR×100%\eta = \frac{MA}{VR} \times 100\% η=4/35/3×100%=45×100%=80%\eta = \frac{4/3}{5/3} \times 100\% = \frac{4}{5} \times 100\% = 80\%

Explanation:

The Velocity Ratio is determined by the geometry (slope length divided by height). The Efficiency is lower than 100%100\% because the actual Mechanical Advantage is reduced by friction along the plane.

Introduction-advanced Class 9 Notes & Examples