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Work, Energy and Simple Machines - Explain mechanical advantage of levers, pulleys, and inclined planes

Grade 9CBSE

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

🔑Concepts

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Mechanical Advantage (MAMA): It is the ratio of the load (output force) to the effort (input force). It indicates how much a machine multiplies the input force. MA=Load (L)Effort (E)MA = \frac{\text{Load (L)}}{\text{Effort (E)}}.

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Velocity Ratio (VRVR): The ratio of the distance moved by the effort to the distance moved by the load. VR=dEdLVR = \frac{d_E}{d_L}.

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Efficiency (η\eta): The ratio of work output to work input, often expressed as η=MAVR×100%\eta = \frac{MA}{VR} \times 100\%. For an ideal machine, efficiency is 100%100\%.

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Levers: A rigid bar capable of rotating about a fixed point called the fulcrum. According to the principle of moments, for a lever in equilibrium, Effort×Effort Arm=Load×Load Arm\text{Effort} \times \text{Effort Arm} = \text{Load} \times \text{Load Arm}.

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Inclined Plane: A sloping surface used to lift heavy loads by applying less force over a longer distance. The MAMA increases as the slope becomes gentler.

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Pulleys: A simple machine consisting of a wheel on an axle or shaft that may have a groove between two flanges around its circumference. In a single fixed pulley, MA=1MA = 1. In a single movable pulley, MA=2MA = 2 (assuming no friction).

📐Formulae

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

MAlever=Effort ArmLoad ArmMA_{\text{lever}} = \frac{\text{Effort Arm}}{\text{Load Arm}}

MAinclined plane=Length of Plane (l)Height of Plane (h)MA_{\text{inclined plane}} = \frac{\text{Length of Plane (l)}}{\text{Height of Plane (h)}}

Work Done (W)=F×s\text{Work Done (W)} = F \times s

Efficiency (\eta)=Work OutputWork Input=MAVR\text{Efficiency (\eta)} = \frac{\text{Work Output}}{\text{Work Input}} = \frac{MA}{VR}

💡Examples

Problem 1:

A crowbar of length 120 cm120\text{ cm} has its fulcrum situated at a distance of 20 cm20\text{ cm} from the load. Calculate the Mechanical Advantage (MAMA) of the crowbar.

Solution:

Given: Total length = 120 cm120\text{ cm}, Load arm (dLd_L) = 20 cm20\text{ cm}. Therefore, Effort arm (dEd_E) = 120 cm−20 cm=100 cm120\text{ cm} - 20\text{ cm} = 100\text{ cm}. Using the formula MA=dEdLMA = \frac{d_E}{d_L}, we get MA=10020=5MA = \frac{100}{20} = 5.

Explanation:

Since the effort arm is five times longer than the load arm, the lever multiplies the input force by a factor of 55.

Problem 2:

A heavy box weighing 500 N500\text{ N} is pushed up an inclined plane of length 10 m10\text{ m} to a height of 2 m2\text{ m}. Find the Mechanical Advantage and the effort required (assuming no friction).

Solution:

Given: L=500 NL = 500\text{ N}, l=10 ml = 10\text{ m}, h=2 mh = 2\text{ m}. MA=lh=102=5MA = \frac{l}{h} = \frac{10}{2} = 5. To find effort: MA=LE  ⟹  5=500E  ⟹  E=5005=100 NMA = \frac{L}{E} \implies 5 = \frac{500}{E} \implies E = \frac{500}{5} = 100\text{ N}.

Explanation:

The inclined plane provides a mechanical advantage of 55, meaning only 1/51/5th of the load's weight is required as effort to move it up the slope.

Problem 3:

Calculate the work done in lifting a load of 200 N200\text{ N} through a height of 5 m5\text{ m} using a single fixed pulley with an efficiency of 80%80\%.

Solution:

Work Output = Load×height=200×5=1000 J\text{Load} \times \text{height} = 200 \times 5 = 1000\text{ J}. Given efficiency η=80%=0.8\eta = 80\% = 0.8. Work Input=Work Outputη=10000.8=1250 J\text{Work Input} = \frac{\text{Work Output}}{\eta} = \frac{1000}{0.8} = 1250\text{ J}.

Explanation:

Due to friction and the weight of the pulley, more work (1250 J1250\text{ J}) must be put into the system than the actual useful work performed (1000 J1000\text{ J}).