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Simple Machines - Pulley

Grade 6ICSE

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

🔑Concepts

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A pulley is a simple machine consisting of a circular disc (wheel) with a grooved rim to hold a rope or a chain.

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A Single Fixed Pulley has its axis of rotation fixed in space. Its primary purpose is to change the direction of the applied force (Effort) from upwards to downwards, which is more convenient.

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In an ideal Single Fixed Pulley, the Mechanical Advantage (MAMA) and Velocity Ratio (VRVR) are both equal to 11.

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A Single Movable Pulley has one end of the rope fixed to a support while the pulley itself moves with the load. It acts as a force multiplier.

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For an ideal Single Movable Pulley, the Mechanical Advantage (MAMA) is 22 and the Velocity Ratio (VRVR) is 22.

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Mechanical Advantage (MAMA): It is the ratio of the Load (LL) to the Effort (EE).

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Velocity Ratio (VRVR): It is the ratio of the distance moved by the effort (dEd_E) to the distance moved by the load (dLd_L).

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Efficiency (η\eta): It is the ratio of the work done on the load to the work done by the effort. In terms of MAMA and VRVR, η=MAVR\eta = \frac{MA}{VR}.

📐Formulae

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

VR=Distance moved by effort (dE)Distance moved by load (dL)VR = \frac{\text{Distance moved by effort } (d_E)}{\text{Distance moved by load } (d_L)}

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

For Single Fixed Pulley: MA=1,VR=1\text{For Single Fixed Pulley: } MA = 1, VR = 1

For Single Movable Pulley: MA=2,VR=2 (Ideal case)\text{For Single Movable Pulley: } MA = 2, VR = 2 \text{ (Ideal case)}

💡Examples

Problem 1:

A worker uses a single fixed pulley to lift a bucket of cement weighing 150 N150\text{ N}. If the efficiency of the pulley is 100%100\%, calculate the effort required.

Solution:

Given: Load L=150 NL = 150\text{ N}, Efficiency η=100%\eta = 100\%. For a single fixed pulley, MA=1MA = 1. Since MA=LEMA = \frac{L}{E}, we have 1=150 NE1 = \frac{150\text{ N}}{E}. Therefore, E=150 NE = 150\text{ N}.

Explanation:

In an ideal single fixed pulley, the effort applied is exactly equal to the load because the mechanical advantage is 11. The pulley only helps by changing the direction of force.

Problem 2:

Using a single movable pulley, an effort of 60 N60\text{ N} is used to lift a load. Calculate the maximum load that can be lifted if the pulley is ideal.

Solution:

Given: Effort E=60 NE = 60\text{ N}. For an ideal single movable pulley, MA=2MA = 2. Using the formula MA=LEMA = \frac{L}{E}, we get 2=L60 N2 = \frac{L}{60\text{ N}}. Therefore, L=2×60 N=120 NL = 2 \times 60\text{ N} = 120\text{ N}.

Explanation:

A single movable pulley acts as a force multiplier with a MAMA of 22, meaning it can lift a load twice the value of the effort applied.

Problem 3:

If the effort in a pulley system moves 4 m4\text{ m} to lift a load by 2 m2\text{ m}, find the Velocity Ratio (VRVR).

Solution:

Given: dE=4 md_E = 4\text{ m} and dL=2 md_L = 2\text{ m}. Using the formula VR=dEdLVR = \frac{d_E}{d_L}, we get VR=4 m2 m=2VR = \frac{4\text{ m}}{2\text{ m}} = 2.

Explanation:

The Velocity Ratio is the ratio of the displacement of the effort to the displacement of the load. A VRVR of 22 indicates this is likely a single movable pulley system.

Pulley - Revision Notes & Key Formulas | ICSE Class 6 Science