Machines Class 10 ICSE — Selina Chapter 3

Machines Class 10 ICSE is chapter 3 in Selina Concise Physics. This ICSE chapter explains how machines multiply force, change direction, and gain speed. You will meet load, effort, mechanical advantage, velocity ratio, and efficiency. The chapter applies these ideas to levers and pulleys with formulas for numericals. For official syllabus details, see the CISCE board website.

Download Machines Class 10 ICSE Physics Chapter 3 PDF

Download the Machines Class 10 ICSE Physics chapter PDF from the Selina Concise textbook to revise offline or print for practice. It is the full chapter as printed in the book — 22 pages, 11 sections, 83 questions.

What is inside Chapter 3

22 pages · 11 sections · 2 worked examples · 83 exercise questions

• EXERCISE-3(A): 37 questions
• MULTIPLE CHOICE TYPE: 15 questions
• EXERCISE-3(B): 31 questions

3.1 MACHINES

A machine is a device that overcomes a large resistive force, called the load, by applying a smaller force, called the effort, at a convenient point. A machine can also give a gain in speed.

Machines are useful in four ways:

• As a force multiplier. A car jack lifts a heavy car with a small effort.
• To change the point of application of effort. A bicycle chain moves effort from the rear wheel to the pedal.
• To change the direction of effort. A fixed pulley lets you pull downward to raise a bucket from a well.
• To gain speed. Scissors blades move further than the handles move.

A wrench opening a nut shows the force multiplier idea. The longer the handle, the smaller the effort you need. A machine cannot act as a force multiplier and a speed multiplier at the same time. You must pick one function for a given design.

3.2 TECHNICAL TERMS RELATED TO A MACHINE

Load is the resistive force a machine must overcome. Effort is the force applied to the machine to overcome the load.

Mechanical advantage is the ratio of load to effort:

\text{M.A.} = \frac{L}{E} \quad \text{(3.1)}

When effort is less than load, M.A. is above 1 and the machine acts as a force multiplier. When effort exceeds load, M.A. is below 1 and the machine gives a gain in speed. M.A. has no unit because it is a ratio of two forces.

Velocity ratio is the ratio of the velocity of effort to the velocity of load. In the same time t, if the effort moves d_E and the load moves d_L:

\text{V.R.} = \frac{d_E}{d_L} \quad \text{(3.2)}

V.R. also has no unit. A V.R. above 1 means the load moves less than the effort, so the machine multiplies force. A V.R. below 1 means the load moves more, giving gain in speed.

Work input is the work done by the effort on the machine. Work output is the work done by the machine on the load. Efficiency is the ratio of work output to work input, usually shown as a percentage:

\eta = \frac{W_{\text{output}}}{W_{\text{input}}} \times 100\% \quad \text{(3.6)}

One note matters for numericals. Friction and the weight of moving parts do not change the velocity ratio of a machine. They reduce both mechanical advantage and efficiency.

3.3 PRINCIPLE OF A MACHINE

The effort point is where energy enters a machine. The load point is where the machine delivers useful energy. Input energy equals work done at the effort point. Output energy equals work done at the load point.

For an ideal machine:

\text{Output energy} = \text{Input energy} \quad \text{(3.7)}

No machine can have efficiency above 100 percent. This follows from conservation of energy. Output energy can never exceed input energy.

An ideal machine loses no energy, so its efficiency is 100 percent. An actual machine always loses some energy. The chapter gives three reasons for this loss. The moving parts are neither weightless nor smooth. The string, if the machine has one, is not perfectly elastic. The parts are not perfectly rigid.

If a machine is 80 percent efficient, then 80 percent of the input energy appears as useful output at the load point. The remaining 20 percent is lost, mostly to friction, and appears as heat in the parts.

The ideas here build on work, energy, and power. You can review those concepts in the chapter on work, energy, and power from Selina.

3.4 RELATIONSHIP BETWEEN EFFICIENCY (eta), MECHANICAL ADVANTAGE (MA) AND VELOCITY RATIO (VR)

This section derives the central relation of the chapter. Suppose a machine overcomes a load L with an effort E. The effort moves a distance d_E, and the load moves a distance d_L.

Work input is E \times d_E. Work output is L \times d_L. By definition, efficiency is work output over work input:

\eta = \frac{L \times d_L}{E \times d_E} = \frac{L}{E} \times \frac{1}{d_E/d_L}

But L/E is M.A., and d_E/d_L is V.R. Therefore:

\boxed{\text{M.A.} = \text{V.R.} \times \eta} \quad \text{(3.8)}

For an ideal machine, \eta = 1, so M.A. equals V.R. For a practical machine, \eta \lt 1, so M.A. is always less than V.R.

CISCE numericals on machines class 10 icse often test this relation. A typical question gives two of the three quantities and asks for the third.

Two mistakes trap students here. First, some think M.A. and V.R. are equal even when efficiency is below 100 percent. They are not. M.A. equals V.R. times efficiency. Second, some think a machine with M.A. below 1 is useless. It is not. Such a machine gives a gain in speed, which is a valid function.

3.5 LEVERS

A lever is a rigid, straight or bent bar that turns about a fixed axis. The axis passes through a point called the fulcrum, marked F.

A lever works on the principle of moments. In equilibrium, the moment of the load about the fulcrum equals the moment of the effort about the fulcrum. The two moments act in opposite directions.

L \times \text{FB} = E \times \text{FA} \quad \text{(3.9)}

Here FA is the effort arm and FB is the load arm. From this balance:

\text{M.A.} = \frac{\text{effort arm}}{\text{load arm}} \quad \text{(3.10)}

This is the law of levers.

Three cases follow from equation (3.10). If effort arm equals load arm, M.A. = 1. If effort arm exceeds load arm, M.A. is above 1. If effort arm is shorter than load arm, M.A. is below 1.

You can increase M.A. by lengthening the effort arm or shortening the load arm.

Worked pattern: A class II lever has an effort arm of 40 cm and a load arm of 8 cm. The ideal M.A. is 40/8 = 5. If the machine has 70 percent efficiency, actual M.A. becomes 5 \times 0.7 = 3.5, while V.R. stays at 5.

3.6 KINDS OF LEVERS

Levers come in three classes, based on the positions of fulcrum, load, and effort.

In a class I lever, the fulcrum sits between the effort and the load. M.A. can be above 1, equal to 1, or below 1. Examples include a seesaw, crowbar, pair of scissors, and pliers.

In a class II lever, the load sits between the effort and the fulcrum. The effort arm is always longer than the load arm. Therefore M.A. is always above 1. Examples include a nut cracker, bottle opener, wheel barrow, and door.

In a class III lever, the effort sits between the fulcrum and the load. The effort arm is always shorter than the load arm. Therefore M.A. is always below 1. Examples include sugar tongs, fire tongs, knife, and fishing rod.

To remember the classes, use the order F-L-E: Fulcrum, Load, Effort. Class I puts the Fulcrum in the middle. Class II puts the Load in the middle. Class III puts the Effort in the middle.

A class I lever with a longer effort arm acts as a force multiplier. One with equal arms has M.A. of 1, like a physical balance. One with a shorter effort arm gives gain in speed, like scissors cutting cloth. These three classes appear across the Class 10 ICSE syllabus.

3.7 EXAMPLES OF EACH CLASS OF LEVERS AS FOUND IN THE HUMAN BODY

The human body contains examples of all three lever classes. Muscles exert force by contraction.

Nodding the head is a class I lever. The spine acts as the fulcrum. The load is at the front of the head. The effort, from neck muscles, acts at the rear.

Raising the body on toes is a class II lever. The toes are the fulcrum at one end. The body weight is the load in the middle. The calf muscles provide the effort at the other end.

Raising a load on the forearm is a class III lever. The elbow is the fulcrum at one end. The biceps exert effort in the middle. The load on the palm sits at the other end.

Worked pattern for a class I lever: The head weighs about 50 N at a load arm of 4 cm from the spine. The muscle effort acts at 8 cm on the opposite side. By moments:

E \times 8 = 50 \times 4

E = 25 \ \text{N}

Worked pattern for a class II lever: A person weighs 600 N. The toes are the fulcrum. The body weight acts at 5 cm from the toes. The calf muscle effort acts at 15 cm. By moments:

E \times 15 = 600 \times 5

E = 200 \ \text{N}

M.A. = 600/200 = 3. These worked patterns follow the same method as the textbook examples. The textbook works through crowbars, seesaws, fire tongs, and nut crackers with its own values.

3.8 PULLEY

A pulley is a metallic or wooden disc with a grooved rim. A string or rope passes around the groove. The disc rotates about an axle through its centre. The axle is fixed to a frame.

A block is a single pulley or a combination of pulleys fixed in a frame. A tackle is the string, rope, or chain that winds around the pulleys in different blocks.

A single pulley can be used in two ways. As a fixed pulley, the axis of rotation stays fixed in position. As a movable pulley, the axis moves with the frame.

The distinction between fixed and movable drives the rest of the chapter. A fixed pulley changes direction. A movable pulley multiplies force. A pulley system gives no gain in energy. The work done by the effort equals the work done on the load in an ideal system.

3.9 SINGLE FIXED PULLEY

A fixed pulley has its axis fixed in position. The axle is attached to a rigid support. A string of negligible mass passes around the grooved rim. One end connects to the load L. The effort E is applied at the other end.

If we neglect the mass of the string and friction at the axle, the tension T is the same throughout. The load equals the tension, and the effort also equals the tension:

E = L \quad \text{(3.11)}

Therefore:

\text{M.A.} = \frac{L}{E} = 1 \quad \text{(3.12)}

If the effort moves a distance d, the load also moves the same distance d. So:

\text{V.R.} = \frac{d_E}{d_L} = 1 \quad \text{(3.13)}

\eta = \frac{\text{M.A.}}{\text{V.R.}} = 1 \ \text{or} \ 100\% \quad \text{(3.14)}

In practice, friction makes the effort slightly larger than the load. A load of 100 N might need an effort of 105 N. M.A. becomes 0.95, but V.R. stays at 1.

Why use a fixed pulley if M.A. is only 1? It changes the direction of effort. You pull downward to lift the load upward. This lets you use your own body weight as effort.

3.10 A SINGLE MOVABLE PULLEY

A movable pulley has a movable axis. The load L hangs from the axle. One end of the string is tied to a rigid support. The effort E is applied at the free end.

Neglect friction and the weight of the pulley. The load is balanced by the tension in two segments of string. The effort balances the tension at the free end:

L = 2T, \quad E = T

E = \frac{L}{2} \quad \text{(3.15)}

Therefore:

\text{M.A.} = \frac{L}{E} = 2 \quad \text{(3.16)}

When the effort end moves up by 2d, the load rises by d. Each segment of string shortens by d. Therefore:

\text{V.R.} = \frac{2d}{d} = 2 \quad \text{(3.17)}

\eta = \frac{\text{M.A.}}{\text{V.R.}} = 1 \ \text{or} \ 100\% \quad \text{(3.18)}

In practice, friction and the pulley’s own weight make the effort more than half the load. M.A. drops below 2. V.R. stays at 2.

A movable pulley is a force multiplier. To apply effort downward, add a fixed pulley above it. This does not change M.A. or V.R.

3.11 COMBINATION OF PULLEYS

A single movable pulley gives M.A. of 2. Heavy loads need more. So we combine pulleys in two ways.

First arrangement: one fixed pulley with n movable pulleys. Each movable pulley has its own string. The tensions differ across strings. For n movable pulleys:

\text{M.A.} = 2^n \quad \text{(3.19)}

\text{V.R.} = 2^n \quad \text{(3.20)}

With 3 movable pulleys, M.A. = 8 and V.R. = 8.

Second arrangement: the block and tackle system. Two blocks hold several pulleys. The upper block is fixed. The lower block is movable. The load attaches to the lower block.

The tension is the same along the whole string. If n segments support the load:

E = \frac{L}{n} \quad \text{(3.22)}

\text{M.A.} = n \quad \text{(3.23)}

\text{V.R.} = n \quad \text{(3.24)}

Here n is the total number of pulleys in both blocks.

Now consider the weight w of the lower block. The load plus the lower block weight is balanced by n tensions:

\text{M.A.} = n - \frac{w}{E} \quad \text{(3.26)}

The velocity ratio stays at n. Efficiency becomes:

\eta = 1 - \frac{w}{nE} \quad \text{(3.28)}

A heavier lower block means lower efficiency. Use light pulleys and lubricate the bearings.

Worked pattern: A block and tackle has 3 pulleys. Load = 150 kgf. Ideal effort = 150/3 = 50 kgf. If the lower block weighs 6 kgf, effort = (150 + 6)/3 = 52 kgf. M.A. = 150/52 \approx 2.88. V.R. = 3. Efficiency \approx 2.88/3 = 0.96 or 96 percent.

A pulley system gives no gain in energy. The work done by the effort equals the work done on the load in an ideal case.

What to practise from Machines Class 10 ICSE

This chapter has three question blocks. Work through them in order.

• EXERCISE-3(A) — 37 questions on machines, technical terms, and levers
• MULTIPLE CHOICE TYPE — 15 questions covering concepts from both parts of the chapter
• EXERCISE-3(B) — 31 questions on pulleys and pulley systems

The numericals in both exercises test M.A., V.R., efficiency, and lever arms. For more physics practice material, visit the Class 10 Physics books hub.

Machines Class 10 ICSE frequently asked questions

What is the difference between mechanical advantage and velocity ratio in machines class 10 icse?

M.A. is the ratio of load to effort. V.R. is the ratio of displacement of effort to displacement of load. For an ideal machine, they are equal. For a real machine, M.A. = V.R. × efficiency, so M.A. is always less than V.R.

Why can no machine in this ICSE chapter have an efficiency of more than 100 percent?

Output energy can never exceed input energy. That would violate conservation of energy. The efficiency of any real machine is always below 100 percent.

Which class of lever always has a mechanical advantage greater than 1 in machines class 10 icse?

Class II levers. The load sits between the fulcrum and the effort. The effort arm is always longer than the load arm, so M.A. is always above 1.

What is the velocity ratio of a single movable pulley in this ICSE chapter?

The V.R. of a single movable pulley is 2. The effort moves twice as far as the load. M.A. is also 2 in ideal conditions.

Why is a single fixed pulley used when its mechanical advantage is only 1?

A fixed pulley changes the direction of effort. You pull downward instead of lifting upward. This lets you use your own body weight as part of the effort.

Reference: ICSE Class 10 Physics textbook material, chapter 3: Machines — CISCE-aligned.

More from this book

• Chapter 1: Force
• Chapter 2: Work Energy and Power
• Chapter 4: Refraction of Light At Plane Surfaces
• Chapter 5: Refraction Through a Lens
• Chapter 6: Spectrum
• Chapter 7: Sound
• Chapter 8: Current Electricity
• Chapter 9: Household Circuits
• Chapter 10: Electro Magnetism
• Chapter 11: Calorimetry
• Chapter 12: Radioactivity