Physics JAMB

Simple Machines

Overview

Welcome to the course material on 'Simple Machines' in Physics. In this module, we delve into the fundamental concepts of machines that play a vital role in making our daily tasks easier by altering the magnitude or direction of the force applied. Let's begin by understanding the essence of simple machines.

Simple machines are basic mechanical devices that aid in performing work with the application of a single force. These machines form the building blocks for more complex mechanisms and are pivotal in various engineering applications. They operate on principles that involve the transmission or modification of forces to achieve desired outcomes.

There are various types of machines categorized based on their functions and structural designs. These include levers, pulleys, inclined planes, wedges, screws, and wheels and axles. Each type of simple machine serves a specific purpose and offers mechanical advantages that enable efficient work performance.

One key aspect of machines is the concept of mechanical advantage, which refers to the ratio of the output force to the input force. Mechanical advantage allows us to amplify the force applied through the use of machines, making tasks more manageable. Additionally, velocity ratio is another critical factor that determines the speed at which a machine operates relative to the input and output distances.

Efficiency in machines is a crucial metric that evaluates the effectiveness of a machine in converting input energy into useful work output. It is defined as the ratio of the output work to the input work and is usually expressed as a percentage. Understanding the efficiency of machines helps in optimizing their performance and minimizing energy wastage.

By the end of this course material, you will be able to identify different types of simple machines and solve problems involving simple machines. Through practical examples and problem-solving exercises, you will gain a deeper insight into the mechanics of these fundamental devices and their significance in various applications.

Objectives

Solve Problems Involving Simple Machines
Identify Different Types Of Simple Machines

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Lesson Evaluation

Congratulations on completing the lesson on Simple Machines. Now that youve explored the key concepts and ideas, its time to put your knowledge to the test. This section offers a variety of practice questions designed to reinforce your understanding and help you gauge your grasp of the material.

You will encounter a mix of question types, including multiple-choice questions, short answer questions, and essay questions. Each question is thoughtfully crafted to assess different aspects of your knowledge and critical thinking skills.

Use this evaluation section as an opportunity to reinforce your understanding of the topic and to identify any areas where you may need additional study. Don't be discouraged by any challenges you encounter; instead, view them as opportunities for growth and improvement.

Define a simple machine. A. A machine that has complex moving parts B. A machine that involves electrical components C. A machine that makes work easier by changing the direction or magnitude of a force D. A machine that operates on its own Answer: C. A machine that makes work easier by changing the direction or magnitude of a force
What are the types of simple machines? A. Lever, Wheel and Axle, Pulley, Inclined Plane, Wedge, Screw B. Waterwheel, Windmill, Gear system, Electrical motor C. Microscope, Telescope, Telephone, Microprocessor D. Bicycle, Car, Train, Airplane Answer: A. Lever, Wheel and Axle, Pulley, Inclined Plane, Wedge, Screw
What is the relationship between mechanical advantage, effort, and load? A. Mechanical advantage = Load / Effort B. Mechanical advantage = Effort / Load C. Mechanical advantage = Load * Effort D. Mechanical advantage = Load - Effort Answer: A. Mechanical advantage = Load / Effort
What does velocity ratio tell us about a machine? A. How fast the machine can operate B. The comparison of the distance the effort moves and the distance the load moves C. The weight of the load applied by the machine D. The rate of efficiency of the machine Answer: B. The comparison of the distance the effort moves and the distance the load moves
What is the efficiency of a machine? A. The speed of operation of the machine B. The comparison of work input to work output of the machine C. The amount of load the machine can lift D. The durability of the machine Answer: B. The comparison of work input to work output of the machine

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Recommended Books

Simple Machines: Forces in Action

Subtitle: Understanding the Basics of Simple Machines

Publisher: Penguin Books

Year: 2015

ISBN: 978-0147516061

Mechanical Advantage and Efficiency in Machines

Subtitle: Mastering the Calculations

Publisher: HarperCollins

Year: 2018

ISBN: 978-0064467644

Past Questions

Wondering what past questions for this topic looks like? Here are a number of questions about Simple Machines from previous years

NECO SSCE - Physics - 2009

Question 1 Report

An effort P applied at one end of a crowbar just overcomes the resistance W at the lid of a tin. The mechanical advantage of the crowbar is expressed as

W + P

P - W

\(\frac{w}{p}\)

\(\frac{p}{w}\)

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JAMB UTME - Physics - 2024

Question 1 Report

The velocity ratio of an inclined plane at 60º to the horizontal is

1.055

1.155

1.255

2.00

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WAEC SSCE - Physics - 2015

Question 1 Report

You are provided with a retort stand, boss head, clamp, stopwatch, slotted weights, hanger, grooved pulley, thread, measuring tape, and other necessary materials.

i. Measure and record the radius \(R\) of the pulley.

ii. Setup the apparatus as illustrated in the diagram above, such that the clamp is 1.5 m above the floor.

iii. Tie one end of the thread to the pulley.

iv. Tie the other end of the thread to the hanger.

v. Slot a mass \(m = 50\ \text{g}\) on the hanger.

vi. Wind the thread around the groove of the pulley until the base of the hanger is at a height \(h = 1.4\ \text{m}\) above the floor. Maintain this height \(h\) for every other value of \(m\) through out the experiment.

vii. Release the mass to unwind the thread.

viii. Determine and record the time \(t\) taken by the mass \(m\) to reach the floor.

ix. Evaluate \(t^{2}\)

x. Also evaluate
a = \(\frac{2h}{t^{2}}\), T = \(\frac{m}{1000}(10 - a)\) and \(\propto = \frac{a}{R}\)

xi. Repeat the procedure for four other values of \(m = 70\ \text{g}, 90\ \text{g}, 110\ \text{g}\) and \(130\ \text{g}\)

xii. Tabulate your readings.

xiii. Plot a graph with \(\propto\) on the vertical axis and T on the horizontal axis.

xiv. Determine the slope s, of the graph.

xv. Evaluate \(I = \frac{R}{s}\).

xvi. State two precautions taken to obtain accurate results.

(b)i. Define centripetal force

ii. An object drops to the ground from a height of 2.0 m. Calculate the speed with which it strikes the ground. [g=10 ms\(^{-2}\)]

Answer

(a) Determination of the moment of inertia of a pulley

A mass \(m\) on the hanger unwinds the thread from the pulley (radius \(R\)) and falls through a fixed height \(h = 1.4\ \text{m}\). The apparatus is set up as shown below, with the clamp \(1.5\ \text{m}\) above the floor.

Apparatus: falling mass unwinding a thread from a grooved pulley clamped 1.5 m above the floor; the hanger base starts at h = 1.4 m.

For each value of \(m\) the time of fall \(t\) is recorded twice and averaged, and the derived quantities are computed from

\[ a = \frac{2h}{t^{2}} = \frac{2(1.4)}{t^{2}} = \frac{2.8}{t^{2}},\qquad T = \frac{m}{1000}\,(10 - a),\qquad \alpha = \frac{a}{R},\quad R = 0.08\ \text{m}. \]

xii. Table of readings

S/N	\(m\)/g	\(t_1\)/s	\(t_2\)/s	\(t=\dfrac{t_1+t_2}{2}\)/s	\(t^{2}\)/s\(^2\)	\(a=\dfrac{2.8}{t^{2}}\)/m s\(^{-2}\)	\(T=\dfrac{m}{1000}(10-a)\)/N	\(\alpha=\dfrac{a}{R}\)/rad s\(^{-2}\)
1	50.0	5.00	5.00	5.00	25.000	0.110	0.490	1.380
2	70.0	4.80	4.80	4.80	23.040	0.120	0.690	1.500
3	90.0	4.60	4.60	4.60	21.160	0.130	0.890	1.630
4	110.0	4.40	4.40	4.40	19.360	0.140	1.080	1.750
5	130.0	4.20	4.20	4.20	17.640	0.150	1.280	1.880

where \(h = 1.4\ \text{m} = 140\ \text{cm}\) and \(R = 0.08\ \text{m} = 8\ \text{cm}\).

Worked check of row 1 (\(m = 50.0\) g, \(t = 5.00\) s):

\[ t^{2} = 5.00^{2} = 25.000\ \text{s}^2,\qquad a = \frac{2.8}{25.000} = 0.110\ \text{m s}^{-2}, \] \[ T = \frac{50}{1000}(10 - 0.110) = 0.050 \times 9.890 = 0.490\ \text{N},\qquad \alpha = \frac{0.110}{0.08} = 1.380\ \text{rad s}^{-2}. \]

xiii. Graph of \(\alpha\) against \(T\)

Plotting \(\alpha\) (vertical axis) against \(T\) (horizontal axis) gives a straight line:

xiv. Slope of the graph

Taking two well-separated points on the line of best fit, \((T_1,\alpha_1) = (0.50,\ 1.383)\) and \((T_2,\alpha_2) = (1.30,\ 1.891)\):

\[ s = \frac{\alpha_2 - \alpha_1}{T_2 - T_1} = \frac{1.891 - 1.383}{1.30 - 0.50} = \frac{0.508}{0.800} = 0.635\ \text{rad s}^{-2}\,\text{N}^{-1}. \]

xv. Moment of inertia

Since the driving torque \(TR = I\alpha\), we have \(\alpha = \dfrac{R}{I}\,T\), so the slope \(s = \dfrac{R}{I}\) and

\[ I = \frac{R}{s} = \frac{0.08}{0.635} = 0.126\ \text{kg m}^{2}. \]

xvi. Two precautions

I avoided parallax error by reading the metre rule and the height marks with the line of sight perpendicular to the scale.
I ensured the boss head and clamp were firmly tightened so the pulley did not slip or wobble during the fall.

(b)(i) Centripetal force

Centripetal force is the resultant inward force, directed towards the centre of the circular path, that keeps a body moving with constant speed in a circle. Its magnitude is

\[ F = \frac{m v^{2}}{r}. \]

(b)(ii) Speed of a body dropped from a height

An object drops to the ground from a height \(h = 2.0\ \text{m}\). Taking \(g = 10\ \text{m s}^{-2}\) and equating potential energy to kinetic energy:

\[ \tfrac{1}{2} m v^{2} = m g h \;\Rightarrow\; v^{2} = 2 g h = 2 \times 10 \times 2.0 = 40\ \text{m}^2\text{s}^{-2}, \] \[ v = \sqrt{40} = 6.32\ \text{m s}^{-1}. \]

The object strikes the ground with a speed of \(6.32\ \text{m s}^{-1}\).

Answer Details

(a) Determination of the moment of inertia of a pulley

For each value of \(m\) the time of fall \(t\) is recorded twice and averaged, and the derived quantities are computed from

\[ a = \frac{2h}{t^{2}} = \frac{2(1.4)}{t^{2}} = \frac{2.8}{t^{2}},\qquad T = \frac{m}{1000}\,(10 - a),\qquad \alpha = \frac{a}{R},\quad R = 0.08\ \text{m}. \]

xii. Table of readings

S/N	\(m\)/g	\(t_1\)/s	\(t_2\)/s	\(t=\dfrac{t_1+t_2}{2}\)/s	\(t^{2}\)/s\(^2\)	\(a=\dfrac{2.8}{t^{2}}\)/m s\(^{-2}\)	\(T=\dfrac{m}{1000}(10-a)\)/N	\(\alpha=\dfrac{a}{R}\)/rad s\(^{-2}\)
1	50.0	5.00	5.00	5.00	25.000	0.110	0.490	1.380
2	70.0	4.80	4.80	4.80	23.040	0.120	0.690	1.500
3	90.0	4.60	4.60	4.60	21.160	0.130	0.890	1.630
4	110.0	4.40	4.40	4.40	19.360	0.140	1.080	1.750
5	130.0	4.20	4.20	4.20	17.640	0.150	1.280	1.880

where \(h = 1.4\ \text{m} = 140\ \text{cm}\) and \(R = 0.08\ \text{m} = 8\ \text{cm}\).

Worked check of row 1 (\(m = 50.0\) g, \(t = 5.00\) s):

xiii. Graph of \(\alpha\) against \(T\)

Plotting \(\alpha\) (vertical axis) against \(T\) (horizontal axis) gives a straight line:

xiv. Slope of the graph

Taking two well-separated points on the line of best fit, \((T_1,\alpha_1) = (0.50,\ 1.383)\) and \((T_2,\alpha_2) = (1.30,\ 1.891)\):

\[ s = \frac{\alpha_2 - \alpha_1}{T_2 - T_1} = \frac{1.891 - 1.383}{1.30 - 0.50} = \frac{0.508}{0.800} = 0.635\ \text{rad s}^{-2}\,\text{N}^{-1}. \]

xv. Moment of inertia

Since the driving torque \(TR = I\alpha\), we have \(\alpha = \dfrac{R}{I}\,T\), so the slope \(s = \dfrac{R}{I}\) and

\[ I = \frac{R}{s} = \frac{0.08}{0.635} = 0.126\ \text{kg m}^{2}. \]

xvi. Two precautions

I avoided parallax error by reading the metre rule and the height marks with the line of sight perpendicular to the scale.
I ensured the boss head and clamp were firmly tightened so the pulley did not slip or wobble during the fall.

(b)(i) Centripetal force

Centripetal force is the resultant inward force, directed towards the centre of the circular path, that keeps a body moving with constant speed in a circle. Its magnitude is

\[ F = \frac{m v^{2}}{r}. \]

(b)(ii) Speed of a body dropped from a height

An object drops to the ground from a height \(h = 2.0\ \text{m}\). Taking \(g = 10\ \text{m s}^{-2}\) and equating potential energy to kinetic energy:

\[ \tfrac{1}{2} m v^{2} = m g h \;\Rightarrow\; v^{2} = 2 g h = 2 \times 10 \times 2.0 = 40\ \text{m}^2\text{s}^{-2}, \] \[ v = \sqrt{40} = 6.32\ \text{m s}^{-1}. \]

The object strikes the ground with a speed of \(6.32\ \text{m s}^{-1}\).

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Simple Machines

Overview

Objectives

Lesson Note

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Lesson Evaluation

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Recommended Books

Past Questions

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(a) Determination of the moment of inertia of a pulley

xii. Table of readings

xiii. Graph of \(\alpha\) against \(T\)

xiv. Slope of the graph

xv. Moment of inertia

xvi. Two precautions

(b)(i) Centripetal force

(b)(ii) Speed of a body dropped from a height

(a) Determination of the moment of inertia of a pulley

xii. Table of readings

xiii. Graph of \(\alpha\) against \(T\)

xiv. Slope of the graph

xv. Moment of inertia

xvi. Two precautions

(b)(i) Centripetal force

(b)(ii) Speed of a body dropped from a height

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