Machines

Class 10 Physics • Chapter 03 (Deep Detail)

1. Basic Concepts

Machine: A device which enables us to obtain a gain in force or gain in speed or change the direction of force.

Functions:

Force Multiplier: Lifts heavy load with small effort (e.g., Jack, Lever).
Speed Multiplier: Gains speed (e.g., Scissors with long blades).
Changing Direction: Makes work convenient (e.g., Single fixed pulley).

IMPORTANT A machine CANNOT allow us to gain both force and speed simultaneously. $P_{out}$ is always $\le P_{in}$.

2. Technical Terms

MECHANICAL ADVANTAGE (MA)

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

$MA > 1$: Force Multiplier (e.g., Crowbar).
$MA < 1$: Speed Multiplier (e.g., Scissors).
$MA = 1$: Change in direction (e.g., See-saw).

VELOCITY RATIO (VR)

$$ VR = \frac{V_E}{V_L} = \frac{d_E}{d_L} $$

($d_E$ = displacement of effort, $d_L$ = displacement of load)

VR depends on the geometry/design of the machine and does not change with friction.

EFFICIENCY ($\eta$)

$$ \eta = \frac{\text{Work Output}}{\text{Work Input}} = \frac{MA}{VR} $$

For Ideal Machine: $\eta = 1$ (100%), so $MA = VR$.

For Practical Machine: $\eta < 1$, so $MA < VR$ (due to friction/weight of parts).

Practice Q1: Efficiency Logic

REASONING Can a machine have an efficiency of 100%? If not, why?

Ans: No practical machine is 100% efficient. Energy is always wasted in overcoming friction between moving parts and lifting the weight of the machine parts (like pulleys or lever arms). Thus, Output Work < Input Work.

Practice Q2: MA vs VR

CONCEPTUAL Explain why the Mechanical Advantage (MA) of a machine is always less than its Velocity Ratio (VR) in practice?

Ans: $\eta = MA/VR$. Since efficiency $\eta < 1$ due to friction and weight of moving parts, $MA < VR$.

3. Levers

A rigid bar capable of turning about a fixed point (Fulcrum).

[IMAGE PLACEHOLDER: CLASSES OF LEVERS]

Diagrams showing Class 1, 2, and 3 levers. For each, clearly mark F (Fulcrum), L (Load), and E (Effort). Include one real-life example sketch for each (e.g., scissors, nutcracker, tongs).

Principle of Levers: Load $\times$ Load Arm = Effort $\times$ Effort Arm.

Types of Levers

Class I (F in middle - L F E): Fulcrum between Load and Effort.

Examples: Scissors, Crowbar, See-saw, Pliers.
MA can be $>1$, $<1$, or $=1$.

Class II (L in middle - F L E): Load between Fulcrum and Effort.

Examples: Nutcracker, Wheelbarrow, Bottle Opener.
MA is always > 1 (Force Multiplier).

Examples: Sugar tongs, Forearm, Knife.
MA is always < 1 (Speed Multiplier).

Practice Q3: Lever Analysis

[IMAGE PLACEHOLDER: HUMAN FOREARM LEVER]

Diagram of human forearm holding a ball. Elbow is Fulcrum (F). Biceps muscle provides Effort (E) upward. Ball is Load (L) at hand. Show E is between F and L. (Class III). (Invert in dark mode).

APPLICATION Classify the human forearm as a lever. What is its mechanical advantage status?

Ans: Class III Lever (Effort is between Fulcrum and Load). Its MA is always less than 1, meaning it acts as a speed multiplier (hands move faster than biceps contraction).

Practice Q4: Crowbar (Class I)

NUMERICAL A crowbar of length 120 cm has its fulcrum situated at a distance of 20 cm from the load. Calculate mechanical advantage.

Solution: Length=120cm. Load Arm ($L_{arm}$)=20cm. Effort Arm ($E_{arm}$)=$120-20=100cm$.

$MA = E_{arm}/L_{arm} = 100/20 = 5$.

Practice Q5: Fire Tongs (Class III)

THINKING Why do we use fire tongs if they have a mechanical advantage less than 1?

Ans: Safety. They extend reach to handle hot objects. Even though effort required is more, the safety and convenience are the advantages.

4. Pulleys

A. Single Fixed Pulley

Fixed to a rigid support.
$L = T$ and $E = T$.
MA = 1, VR = 1, Efficiency $\approx$ 100%.
Use: To change direction of effort (e.g., drawing water from well).

B. Single Movable Pulley

Pulley moves with the load.
$L = 2T$ and $E = T$.
MA = 2, VR = 2.
Use: Force Multiplier.

C. Combination of Pulleys (Block and Tackle)

[IMAGE PLACEHOLDER: BLOCK AND TACKLE]

Diagram of a Block and Tackle system with 5 pulleys (3 fixed, 2 movable). Show rope winding. Clearly label Tension vectors 'T', Load 'L' downwards, and Effort 'E' downwards at the free end.

System with two blocks: Upper dynamic (fixed) and Lower movable.

Number of Pulleys ($n$): Total pulleys in system.
MA = $n$ (Ideal).
VR = $n$ (Always equal to number of pulleys/strands supporting load).
Effort required: $E = L/n$.

Practice Q6: Block and Tackle (VR 4)

[IMAGE PLACEHOLDER: BLOCK AND TACKLE 4 PULLEYS]

A Block and Tackle system with 4 pulleys (2 in fixed block, 2 in movable block). Show string winding starting from *hook of fixed block* (since n is even). Mark Tension 'T' on all 4 strands support load + 1 effort strand. Equation: L=4T, E=T. (Invert in dark mode).

DIAGRAM & NUMERICAL A woman draws water from a well using a block and tackle system of velocity ratio 4.

(i) Draw a labelled diagram of the system.

(ii) If she exerts a pull of 200 N, what is the maximum load she can raise? (Assume $\eta = 100\%$).

Solution:

(ii) $MA = VR = 4$. $MA = Load/Effort$

$\implies 4 = L / 200 \implies L = 800 \text{ N}$.

Practice Q7: System with VR 5

NUMERICAL A block and tackle system has 5 pulleys. An effort of 1000 N is needed in the downward direction to raise a load of 4500 N. Calculate (a) MA, (b) VR, (c) Efficiency.

Solution: $n=5 \implies VR=5$. $E=1000, L=4500$.

MA = $4500/1000 = 4.5$.

$\eta = MA/VR = 4.5/5 = 0.9 = 90\%$.

Effect of Rope Weight & Friction

In actual systems, Efficiency < 100%.

MA decreases (Effective load decreases due to pulley weight).
VR remains same (depends on strands).
$\eta = \frac{MA}{VR} \times 100\%$.

NUMERICAL ADVICE

For Block & Tackle systems: simply count the number of string segments supporting the lower movable block. That number is your VR.
If efficiency is given (e.g., 80%), use $MA = \eta \times VR$ to find the actual $MA$, then use $MA = L/E$ to find Effort or Load.
Don't confuse Class II and Class III. Remember FLE (1-2-3 rule: Fulcrum middle=1, Load middle=2, Effort middle=3).
Weight of the movable block reduces the MA of the system but VR remains unchanged.

Classroom Practice Questions

Question 1: Levers

CONCEPTUAL A crowbar of length 120 cm has its fulcrum situated at a distance of 20 cm from the load. Calculate mechanical advantage.

Solution:

Length of crowbar = 120 cm. Fulcrum is at 20 cm from load.

Load Arm ($L_{arm}$) = 20 cm.

Effort Arm ($E_{arm}$) = Total length - Load Arm (Assuming load at end) = $120 - 20 = 100 \text{ cm}$. (Note: Crowbar is Class I lever).

$MA = \frac{\text{Effort Arm}}{\text{Load Arm}} = \frac{100}{20} = 5$.

Ans: MA is 5.

Question 2: Block and Tackle

NUMERICAL A block and tackle system has 5 pulleys. An effort of 1000 N is needed in the downward direction to raise a load of 4500 N. Calculate (a) MA, (b) VR, (c) Efficiency.

Solution:

Given: $n = 5$ (Total pulleys), $E = 1000 \text{ N}$, $L = 4500 \text{ N}$.

(a) $MA = \frac{L}{E} = \frac{4500}{1000} = 4.5$

(b) $VR = n = 5$ (Number of pulleys)

$= 0.9 \times 100\% = 90\%$

Ans: MA = 4.5, VR = 5, Efficiency = 90%.

Question 3: Fire Tongs (Class III)

THINKING Why do we use fire tongs if they have a mechanical advantage less than 1?

Solution:

Fire tongs are Class III levers where effort is in the middle. We use them not to multiply force, but for safety (handling hot objects from a distance) and convenience. Even though MA < 1, they provide a safe extension of our hands.

Practice Q8: Single Movable Pulley

DERIVATION For a single movable pulley, if the weight of the pulley is $w$, derive the expression for MA.

Solution:

In equilibrium, Upward Forces = Downward Forces.

$L + w = 2T \implies T = \frac{L+w}{2}$. And Effort $E = T$.

So, $E = \frac{L+w}{2} \implies 2E = L + w$

$\implies L = 2E - w$.

$MA = \frac{L}{E} = \frac{2E - w}{E} = 2 - \frac{w}{E}$.

Result: MA is less than 2 due to weight of pulley.

Practice Q9: Single Fixed Pulley

REASONING Why is a single fixed pulley used if its MA is only 1?

Ans: To change the direction of effort. It allows effort to be applied downwards (using body weight) rather than lifting upwards.