Comprehensive Study Notes on Simple Machines, Levers, and Pulley Systems

Fundamentals of Simple Machines

• Definition of a Machine:

  • A machine is a device by which a large resistive force (load) can be overcome by applying a relatively small force (effort) at a desired point and in a desired direction, or by which a gain in speed can be obtained.

• Mechanical Advantage ($MA$):

  • The mechanical advantage of a machine is defined as the ratio between the load ($L$) and the effort ($E$):         $MA = \frac{L}{E}$
  • Significance of $MA$ values:
    • When $MA > 1$: The machine acts as a force multiplier.
    • When $MA = 1$: The machine is used to change the direction of effort.
    • When $MA < 1$: The machine acts as a speed multiplier.

• Velocity Ratio ($VR$):

  • The velocity ratio is defined as the ratio of the velocity of the effort ($V_E$) to the velocity of the load ($V_L$):         $VR = \frac{V_E}{V_L}$
  • Expressing velocity in terms of displacement over time ($t$):         $VR = \frac{\frac{d_E}{t}}{\frac{d_L}{t}} = \frac{d_E}{d_L}$         where $d_E$ is the distance moved by the effort, and $d_L$ is the distance moved by the load in the same interval of time.
  • Significance of $VR$ values:
    • When $VR > 1$: The machine acts as a force multiplier (effort directed towards gravity).
    • When $VR = 1$: The machine changes the direction of effort.
    • When $VR < 1$: The machine acts as a speed multiplier.

• Work Input ($W_i$) and Work Output ($W_o$):

  • Work Input ($W_i$): The work done on the machine by the effort:         $W_i = E \times d_E$
  • Work Output ($W_o$): The useful work done by the machine on the load:         $W_o = L \times d_L$

• Efficiency ($\eta$):

  • Efficiency is defined as the ratio of the useful work done by the machine ($W_o$) to the work done on the machine by the effort ($W_i$), expressed as a percentage:         $\eta = \frac{W_o}{W_i} \times 100\%$
  • Mathematical derivation linking $\eta$, $MA$, and $VR$:         $\eta = \frac{L \times d_L}{E \times d_E} \times 100\%$$\eta = \frac{\frac{L}{E}}{\frac{d_E}{d_L}} \times 100\%$$\eta = \frac{MA}{VR} \times 100\%$

• Ideal Machine vs. Actual Machine:

  • Ideal Machine: A machine that is $100\%$ efficient ($\eta = 100\%$). For all ideal machines:         $MA = VR$
  • Actual (Practical) Machine: All real-world machines have an efficiency strictly less than $100\%$ ($\eta < 100\%$), meaning:         $MA < VR$
  • Reasons why practical machines are never $100\%$ efficient:
    • Some amount of energy is wasted in moving the internal parts of the machine which are not completely weightless.
    • Some amount of energy is wasted in overcoming friction between the moving contact surfaces of the machine.
    • Some amount of energy is wasted if the string used is not perfectly inelastic (i.e., energy lost due to string elasticity).

Classification and Principles of Levers

• Definition of a Lever:

  • A lever is a rigid straight or bent bar capable of turning freely about a fixed point or a fixed axis known as the fulcrum ($F$).

• Classification Criteria:

  • Levers are categorized into three distinct classes based on the relative positions of the fulcrum ($F$), the load ($L$), and the effort ($E$).

• Class 1 Levers:

  • The fulcrum ($F$) is situated between the load ($L$) and the effort ($E$).
  • Depending on the relative lengths of the effort arm ($E_{\text{arm}}$) and the load arm ($L_{\text{arm}}$), three operational regimes exist:
    • Condition 1 ($E_{\text{arm}} = L_{\text{arm}}$):
      • Mechanical Advantage: $MA = 1$
      • Primary Use: Used to change the direction of effort.
      • Example: See-saw.
    • Condition 2 ($E_{\text{arm}} < L_{\text{arm}}$):
      • Mechanical Advantage: $MA < 1$
      • Primary Use: Acts as a speed multiplier.
      • Example: Tailor\'s scissors.
    • Condition 3 ($E_{\text{arm}} > L_{\text{arm}}$):
      • Mechanical Advantage: $MA > 1$
      • Primary Use: Acts as a force multiplier.
      • Examples: Crowbar, claw hammer.

• Class 2 Levers:

  • The load ($L$) is situated between the fulcrum ($F$) and the effort ($E$).
  • Operating condition: The effort arm is always greater than the load arm ($E_{\text{arm}} > L_{\text{arm}}$).
  • Mechanical Advantage: Always greater than one ($MA > 1$).
  • Primary Use: Functions as a force multiplier.
  • Examples: Nutcracker, lemon squeezer, bottle opener.

• Class 3 Levers:

  • The effort ($E$) is situated between the fulcrum ($F$) and the load ($L$).
  • Operating condition: The effort arm is always less than the load arm ($E_{\text{arm}} < L_{\text{arm}}$).
  • Mechanical Advantage: Always less than one ($MA < 1$).
  • Primary Use: Functions as a speed multiplier.
  • Examples: Sugar tongs, bread knife.

• Key Distinctions Between $MA$ and $VR$ Across Machines:

  • For any given machine, the mechanical advantage ($MA$) can vary depending on the load lifted or the effort applied (due to friction, weight of parts, and operational conditions).
  • The velocity ratio ($VR$) remains strictly constant because it is determined solely by the geometry, dimensions, and kinematic design of the machine.

Pulley Systems and Mechanisms

• Definition of a Pulley:

  • A pulley is a flat circular disc with a groove cut along its circumferential edge, capable of rotating smoothly about a fixed point passing through its center called the axle.

• Single Fixed Pulley:

  • Consists of a pulley mounted to a fixed support.
  • The load is balanced by the tension in the string ($L = T$), and the applied effort balances the same string tension ($E = T$).
  • Mechanical parameters:         $MA = \frac{L}{E} = \frac{T}{T} = 1$$VR = \frac{d_E}{d_L} = 1$$\eta = \frac{MA}{VR} \times 100\% = \frac{1}{1} \times 100\% = 100\%$
  • Reasons for using a single fixed pulley despite $MA = 1$:
    • To change the direction of the applied effort toward gravity (downward direction).
    • Applying effort downwards enables the operator to utilize their own body weight to assist in overcoming the load.

• Single Movable Pulley:

  • The pulley moves along with the load. The load is supported by two segments of the continuous string, each exerting tension $T$ ($L = 2T$), while the effort balances one segment ($E = T$).
  • Mechanical parameters:         $MA = \frac{L}{E} = \frac{2T}{T} = 2$

• Combination of a Fixed Pulley with a Movable Pulley:

  • System features two pulleys: Pulley A (fixed to a rigid support) and Pulley B (movable, supporting the load).
  • Mechanical parameters:         $MA = \frac{L}{E} = \frac{2T}{T} = 2$$VR = \frac{d_E}{d_L} = 2$$\eta = \frac{MA}{VR} \times 100\% = \frac{2}{2} \times 100\% = 100\%$
  • Functional roles of the individual pulleys:
    • Pulley A (Fixed): Serves solely to change the direction of effort applied towards gravity.
    • Pulley B (Movable): Serves to increase the mechanical advantage ($MA = 2$) so that the assembly acts as a force multiplier.

• General Rules for Determining Velocity Ratio in Pulley Systems:

  • When the effort is applied towards gravity (downward direction):         $VR = n$         where $n$ is the total number of pulleys in the system.
  • When the effort is applied against gravity (upward direction):         $VR = n + 1$         where $n$ is the total number of pulleys in the system.
  • Method for calculating tension ($T$): Divide the schematic diagram into two sides—effort side and load side—and sum the upward tensions supporting the load.

• Block and Tackle (Sheaf) System:

  • Consists of two blocks: a fixed upper block and a movable lower tackle block carrying the load.
  • 6-Pulley System (3 pulleys in fixed block, 3 in movable block; effort directed downwards):$L = 6T$$E = T$$MA = \frac{L}{E} = \frac{6T}{T} = 6$$VR = \frac{d_E}{d_L} = 6$$\eta = \frac{6}{6} \times 100\% = 100\%$
  • 5-Pulley System (3 pulleys in fixed block, 2 in movable block; effort directed downwards):$L = 5T$$E = T$$MA = \frac{L}{E} = \frac{5T}{T} = 5$$VR = \frac{d_E}{d_L} = 5$$\eta = \frac{5}{5} \times 100\% = 100\%$

Solved Numerical Problems

• Problem 1: 5-Pulley System with $90\%$ Efficiency

  • Problem Statement: A pulley system has $5$ pulleys in all and is $90\%$ efficient. Calculate:
    • (a) The mechanical advantage ($MA$)
    • (b) The effort required to lift a load of $1000\,\text{N}$
    • (c) The resistance due to the movable parts of the machine and friction
  • Solution:
    • Assumption: The effort is applied towards gravity.
    • Total number of pulleys: $n = 5$
    • Velocity ratio: $VR = 5$
    • Efficiency: $\eta = 90\%$
    • (a) Finding Mechanical Advantage:$\eta = \frac{MA}{VR} \times 100\%$$90\% = \frac{MA}{5} \times 100\%$$MA = \frac{90 \times 5}{100} = 4.5$
    • (b) Finding Required Effort:$MA = \frac{L}{E}$$4.5 = \frac{1000\,\text{N}}{E}$$E = \frac{1000}{4.5} = 222.22\,\text{N}$
    • (c) Finding Resistive Force ($x$):
      • Let $x$ represent the resistive force due to friction and the weight of movable parts:                 $x = E \times (VR - MA)$$x = 222.22\,\text{N} \times (5 - 4.5)$$x = 222.22\,\text{N} \times 0.5 = 111.11\,\text{N}$

• Problem 2: 5-Pulley System with Effort Directed Upward

  • Problem Statement: A pulley system has $5$ pulleys in all: $2$ in the movable block (tackle) and $3$ in the fixed block, configured such that effort is applied in the upward direction. A load of $60\,\text{kgf}$ is attached to the movable block. Calculate:
    • (i) The velocity ratio ($VR$)
    • (ii) The mechanical advantage ($MA$), assuming an ideal machine
    • (iii) The magnitude of the effort for the ideal case
    • (iv) The effort required if the efficiency drops to $75\%$
  • Solution:
    • (i) Finding Velocity Ratio:
      • Because effort is applied in the upward direction (against gravity):                 $VR = n + 1 = 5 + 1 = 6$
    • (ii) Finding Ideal Mechanical Advantage:
      • For an ideal machine, $\eta = 100\%$:                 $MA = VR = 6$
    • (iii) Finding Ideal Effort:$MA = \frac{L}{E}$$6 = \frac{60\,\text{kgf}}{E}$$E = \frac{60}{6} = 10\,\text{kgf}$
    • (iv) Finding Effort at \eta = 75\%$:* \n * The velocity ratio remains unchanged: VR = 6.\n            \eta = \frac{MA}{VR} \times 100\%\n            75\% = \frac{MA}{6} \times 100\%\n            MA = \frac{75 \times 6}{100} = 4.5\n * Calculating the new required effort:\n                MA = \frac{L}{E}\n                4.5 = \frac{60\,\text{kgf}}{E}\n                E = \frac{60}{4.5} = 13.33\,\text{kgf}\n\n* **Problem 3: Two-Pulley System with Upward Effort**\n * **Problem Statement:** A system comprising 2$pulleys has effort applied upwards lifting a load of$30\,\text{kgf}. Calculate:\n * (i) Velocity ratio (VR)\n * (ii) Ideal mechanical advantage (MA)\n * (iii) Effort for ideal conditions\n * (iv) Effort if the efficiency is 60\%\n * **Solution:**\n * *(i) Finding Velocity Ratio:* \n * With effort acting against gravity:\n                VR = 2 + 1 = 3\n * *(ii) Finding Ideal Mechanical Advantage:* \n * At \eta = 100\%:\n                MA = VR = 3\n * *(iii) Finding Ideal Effort:* \n            MA = \frac{L}{E}\n            3 = \frac{30\,\text{kgf}}{E}\n            E = \frac{30}{3} = 10\,\text{kgf}\n * *(iv) Finding Effort at \eta = 60\%$:$\eta = \frac{MA}{VR} \times 100\%$$60\% = \frac{MA}{3} \times 100\%$$MA = \frac{60 \times 3}{100} = 1.8$
      • Calculating effort:                 $MA = \frac{L}{E}$$1.8 = \frac{30\,\text{kgf}}{E}$$E = \frac{30}{1.8} = 16.67\,\text{kgf}$

• Problem 4: Pulley System Kinematics and Energy Calculations

  • Problem Statement: Given a pulley system with $VR = 3$, lifting a load of $60\,\text{kgf}$ through a distance $d_L = 20\,\text{m}$, driven by an effort of $25\,\text{kgf}$ applied in the upward direction ($g = 10\,\text{N/kg}$). Calculate:
    • (i) The distance through which effort is applied ($d_E$)
    • (ii) The work done by the effort ($W_E$)
    • (iii) Mechanical advantage ($MA$)
    • (iv) Efficiency ($\eta$)
    • (v) Total number of pulleys in the fixed and movable blocks
  • Solution:
    • (i) Finding Effort Displacement ($d_E$):$VR = \frac{d_E}{d_L}$$3 = \frac{d_E}{20\,\text{m}}$$d_E = 3 \times 20\,\text{m} = 60\,\text{m}$
    • (ii) Finding Work Done by Effort ($W_E$):$E = 25\,\text{kgf} = 25 \times 10\,\text{N} = 250\,\text{N}$$W_E = E \times d_E = 250\,\text{N} \times 60\,\text{m} = 15000\,\text{N\,m} = 15000\,\text{J}$
    • (iii) Finding Mechanical Advantage ($MA$):$MA = \frac{L}{E} = \frac{60\,\text{kgf}}{25\,\text{kgf}} = 2.4$
    • (iv) Finding Efficiency ($\eta$):$\eta = \frac{MA}{VR} \times 100\% = \frac{2.4}{3} \times 100\% = 80\%$
    • (v) Finding Number of Pulleys:
      • Since effort is in the upward direction:                 $\text{Total pulleys} = VR - 1 = 3 - 1 = 2$
      • Distribution: $1$ pulley in the fixed block and $1$ pulley in the movable block.

• Problem 5: Single Movable Pulley System with Load of $20\,\text{kgf}$

  • Problem Statement: A single movable pulley system has a load of $20\,\text{kgf}$ attached to it with upward-directed effort. Determine:
    • (a) The velocity ratio ($VR$)
    • (b) The ideal mechanical advantage ($MA$) and required effort at $\eta = 100\%$
    • (c) The mechanical advantage and required effort if efficiency is $\eta = 80\%$
  • Solution:
    • (a) Finding Velocity Ratio:$VR = 1 + 1 = 2$
    • (b) Ideal Operating Parameters ($\eta = 100\%$):$MA = VR = 2$$MA = \frac{L}{E} \implies 2 = \frac{20\,\text{kgf}}{E} \implies E = 10\,\text{kgf}$
    • (c) Operating Parameters at $\eta = 80\%$:$\eta = \frac{MA}{VR} \times 100\%$$80\% = \frac{MA}{2} \times 100\%$$MA = \frac{80 \times 2}{100} = 1.6$
      • Calculating actual effort:                 $MA = \frac{L}{E}$$1.6 = \frac{20\,\text{kgf}}{E}$$E = \frac{20}{1.6} = 12.5\,\text{kgf}$

• Problem 6: System with Velocity Ratio $6$ and Downward Effort

  • Problem Statement: A pulley system with $VR = 6$ lifts a load of $180\,\text{kgf}$ through a vertical height of $30\,\text{m}$. The effort applied in the downward direction is $50\,\text{kgf}$ ($g = 10\,\text{N/kg}$). Calculate:
    • (a) Distance through which effort is applied ($d_E$)
    • (b) Work done by effort ($W_E$)
    • (c) Mechanical advantage ($MA$)
    • (d) Efficiency ($\eta$)
    • (e) Total number of pulleys in the movable and fixed blocks
  • Solution:
    • Given values: $VR = 6$, $L = 180\,\text{kgf}$, $d_L = 30\,\text{m}$, $E = 50\,\text{kgf}$, $g = 10\,\text{N/kg}$.
    • (a) Finding Distance Moved by Effort:$VR = \frac{d_E}{d_L}$$6 = \frac{d_E}{30\,\text{m}}$$d_E = 6 \times 30\,\text{m} = 180\,\text{m}$
    • (b) Finding Work Done by Effort:$E = 50\,\text{kgf} \times 10\,\text{N/kg} = 500\,\text{N}$$W_E = E \times d_E = 500\,\text{N} \times 180\,\text{m} = 90000\,\text{J}$
    • (c) Finding Mechanical Advantage:$MA = \frac{L}{E} = \frac{180\,\text{kgf}}{50\,\text{kgf}} = 3.6$
    • (d) Finding Efficiency:$\eta = \frac{MA}{VR} \times 100\% = \frac{3.6}{6} \times 100\% = 60\%$
    • (e) Finding Number and Distribution of Pulleys:
      • Because effort is downward, total pulleys $n = VR = 6$.
      • Distribution: $3$ pulleys in the fixed block and $3$ pulleys in the movable block.

• Problem 7: Upward-Driven Pulley System with Velocity Ratio $3$

  • Problem Statement: A pulley system with $VR = 3$ lifts a load of $60\,\text{kgf}$ through a vertical displacement of $20\,\text{m}$. The effort force is applied in the upward direction with a magnitude of $25\,\text{kgf}$ ($g = 10\,\text{N/kg}$). Calculate:
    • (a) Distance through which effort moves ($d_E$)
    • (b) Work done by effort ($W_E$)
    • (c) Mechanical advantage ($MA$)
    • (d) Efficiency ($\eta$)
    • (e) Total number of pulleys in the movable and fixed blocks
  • Solution:
    • Given values: $VR = 3$, $L = 60\,\text{kgf}$, $d_L = 20\,\text{m}$, $E = 25\,\text{kgf}$, $g = 10\,\text{N/kg}$.
    • (a) Finding Effort Displacement:$VR = \frac{d_E}{d_L}$$3 = \frac{d_E}{20\,\text{m}}$$d_E = 60\,\text{m}$
    • (b) Finding Work Done by Effort:$W_E = E \times d_E = (25\,\text{kgf} \times 10\,\text{N/kg}) \times 60\,\text{m} = 250\,\text{N} \times 60\,\text{m} = 15000\,\text{J}$
    • (c) Finding Mechanical Advantage:$MA = \frac{L}{E} = \frac{60\,\text{kgf}}{25\,\text{kgf}} = 2.4$
    • (d) Finding Efficiency:$\eta = \frac{MA}{VR} \times 100\% = \frac{2.4}{3} \times 100\% = 80\%$
    • (e) Finding Pulley Count in Blocks:
      • Since effort is applied in the upward direction, total pulleys $= VR - 1 = 3 - 1 = 2$.
      • Distribution: $1$ pulley in the fixed block and $1$ pulley in the movable block.