Inertia 1 Force 1 Next Slide Inertia Aristotles

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Inertia 1 Force 1 Next Slide Inertia • Aristotle’s old belief • Constant speed

Inertia 1 Force 1 Next Slide Inertia • Aristotle’s old belief • Constant speed motion requires constant force • Galileo’s law of inertia Photo • It is as natural for a moving object to keep moving with a constant speed along a straight line as for a stationary object to remain at rest.

Inertia 2 Force 1 Next Slide Inertia and Newton’s first law of motion •

Inertia 2 Force 1 Next Slide Inertia and Newton’s first law of motion • Galileo’s thought experiment Diagrams • Newton’s first law of motion: • Every object remains in a state of rest or uniform speed along a straight line (constant velocity) unless acted on by an unbalanced force. Photo

Inertia 3 Force 1 Next Slide Inertia and Newton’s first law of motion •

Inertia 3 Force 1 Next Slide Inertia and Newton’s first law of motion • Force : changes the state of rest or uniform motion of an object (vector! Why? ) • Inertia : the resistance of an object to a change in its state of rest or uniform motion in a straight line. • Mass : is a measure of inertia (scalar! Why? ) • Mass can be considered as a measure of inertia • Inertial balance Diagrams

Force 1 Next Slide Friction • Origin of wrong concept of Aristotle’s old belief

Force 1 Next Slide Friction • Origin of wrong concept of Aristotle’s old belief • Nature of friction Diagrams • Direction : opposite to the motion (velocity) • Examples of smooth surface Photo • Demonstration of Newton’s first law on smooth surfaces

Force 2 Force 1 Next Slide Unbalanced force (resultant force, net force) • Falling

Force 2 Force 1 Next Slide Unbalanced force (resultant force, net force) • Falling of object in liquid (terminal velocity) Photo • Its relation with Newton’s first law of motion Diagram • Friction-compensated inclined plane • Experiment to study resultant force • Results Calculation Photo Diagram

Force 3 Force 1 Next Slide Newton’s second law of motion • Acceleration is

Force 3 Force 1 Next Slide Newton’s second law of motion • Acceleration is not zero if the resultant force is not zero. • Deductions from the results 1. Acceleration force (mass is constant) 2. Acceleration 1/mass (force is constant) 3. F ma or F = constant ma

Force 4 Force 1 Next Slide Newton’s second law of motion • Definition: The

Force 4 Force 1 Next Slide Newton’s second law of motion • Definition: The acceleration of an object is directly proportional to, and in the same direction as, the unbalanced force acting on it, and inversely proportional to the mass of the object • Unit of force : Newton (N)

Force 5 Force 1 Next Slide Newton’s second law of motion • 1 newton

Force 5 Force 1 Next Slide Newton’s second law of motion • 1 newton of force will give a mass 1 kg an acceleration 1 • constant in F = constant ma becomes 1 • Mathematical form of Newton’s second law • Direction of resultant force = direction of acceleration

Weight & Mass 1 Force 1 Next Slide Weight and Mass • Definition: The

Weight & Mass 1 Force 1 Next Slide Weight and Mass • Definition: The force of gravity acting on the object is called the weight of the object and is measured in newton • Acceleration in free falling = 10 m s-2 1 kg of mass has a weight 10 N (downwards)

Weight & Mass 2 Force 1 Next Slide Weight and Mass • Instruments to

Weight & Mass 2 Force 1 Next Slide Weight and Mass • Instruments to measure weight and mass Photo • Weightlessness • Discussion about the restrictions of the above machine Calculation

Addition of Force 1 Next Slide Addition of Forces (vectors) • More than one

Addition of Force 1 Next Slide Addition of Forces (vectors) • More than one force acting on an object • Add them together to get ONE resultant force • F = ma can only be applied for resultant force • Tip-to-tail method (Revision) • Example 1 Calculation • Example 2 Calculation

Addition of Force 2 Force 1 Next Slide Addition of Forces (vectors) • Method

Addition of Force 2 Force 1 Next Slide Addition of Forces (vectors) • Method of resolving components • Adding forces or vectors without drawing diagrams • Example Calculation • Examples for components of forces Calculation

END of Force 1

END of Force 1

Back to Inertia 1 Force 1 Click Back to • Galileo Galilei (1564 -

Back to Inertia 1 Force 1 Click Back to • Galileo Galilei (1564 - 1642)

Inertia 2 Force 1 Next Slide • Small bearing is released from rest on

Inertia 2 Force 1 Next Slide • Small bearing is released from rest on a smooth track at A. A C D E F • A ball reaches point C which is of the same height • Same situation for D and E • If the track is infinite long, the ball will never stop.

Inertia 2 Force 1 Next Slide • Galileo’ pin-and-pendulum experiment • Consider the swing

Inertia 2 Force 1 Next Slide • Galileo’ pin-and-pendulum experiment • Consider the swing of a simple pendulum

Back to Inertia 2 Force 1 Click Back to • The bob rises to

Back to Inertia 2 Force 1 Click Back to • The bob rises to the same height as before • Even we have a pin, the bob rises to the same height

Back to Inertia 2 Force 1 Click Back to • Isaac Newton (1642 -1727)

Back to Inertia 2 Force 1 Click Back to • Isaac Newton (1642 -1727)

Back to Inertia 3 Force 1 Click Back to • We set the platform

Back to Inertia 3 Force 1 Click Back to • We set the platform into vibration and record the period. • Fix load on the platform and repeat the vibration, we find that a longer period can be found. • The larger the load, the longer the period.

Back to Force 1 Click Back to • Friction is caused by the interlocking

Back to Force 1 Click Back to • Friction is caused by the interlocking of surface irregularities.

Force 1 Next Slide • A mass placed on a thin layer of polystyrene

Force 1 Next Slide • A mass placed on a thin layer of polystyrene beads on a glass plate • A balloon is blown up and attached to a short pipe

Back to Force 1 Click Back to • Air-layer Ball • Motion on a

Back to Force 1 Click Back to • Air-layer Ball • Motion on a air track

Back to Force 2 Force 1 Click Back to • An object is falling

Back to Force 2 Force 1 Click Back to • An object is falling inside liquid.

Back to Force 2 Force 1 Click Back to • The object is falling

Back to Force 2 Force 1 Click Back to • The object is falling downwards with constant velocity. Do you know why? • Liquid resistance is equal to the weight • No unbalanced force liquid resistance weight

Back to Force 2 Force 1 Click Back to • An inclined plane is

Back to Force 2 Force 1 Click Back to • An inclined plane is prepared so that when we give the trolley a hard push, it moves down with constant velocity. It is called to be friction-compensated. • Careful adjustment for the plane is needed to achieve this situation.

Force 2 Force 1 Next Slide • Identical elastic strings are used to pull

Force 2 Force 1 Next Slide • Identical elastic strings are used to pull the trolley. • At first, we use one string and then two, and three. • We always maintain the same length for all the strings so that each string produces the same force. • The accelerations in each case are recorded. trolley Friction-compensated inclined plane elastic string

Back to Force 2 Force 1 Click Back to • One elastic string is

Back to Force 2 Force 1 Click Back to • One elastic string is used to pull several trolleys. • At first, we use one trolley and then two, and three. • We always maintain the same length for the string so that the string produces the same force in each case. • The accelerations in each case are recorded. elastic string friction-compensated inclined plane

Force 2 Force 1 Next Slide • Different tape charts for different no. of

Force 2 Force 1 Next Slide • Different tape charts for different no. of strings with one trolley are shown. 1 string 2 strings a = 2 m s-2 a = 4 m s-2 3 strings a = 6 m s-2 • We find that the acceleration is directly proportional to the no. of strings used (Force) when the mass of trolley is kept constant.

Back to Force 2 Force 1 Click Back to • Different tape charts for

Back to Force 2 Force 1 Click Back to • Different tape charts for different no. of trolleys with one string are shown. 1 trolley a=2 2 trolleys a=1 3 trolleys a = 0. 67 • We find that the acceleration is inversely proportional to the no. of trolleys used (mass) when 1 string is used (constant force).

Back to Weight and Mass 2 Force 1 Click Back to • Beam balance

Back to Weight and Mass 2 Force 1 Click Back to • Beam balance (measure mass) • Spring balance (measure weight)

Back to Weight and Mass 2 Force 1 Click Back to • Can we

Back to Weight and Mass 2 Force 1 Click Back to • Can we use the beam balance or spring balance on Moon to get correct readings of mass and weight of an object with 1 kg mass? • The acceleration due to gravity on Moon is only about 1. 8 m s-2. • 1 kg slot-mass is still needed to balance the object. • The reading from the spring balance = 1 1. 8 = 1. 8 N! • Mass is the same anywhere while weight depends on position and is not a constant even for the same object.

Back to Addition of Force 1 Click Back to • Two forces 3 N

Back to Addition of Force 1 Click Back to • Two forces 3 N and 4 N are acting on an object (2 kg) as shown below. What are the resultant force and acceleration? 3 N 2 kg 4 N N • Use a scale of 1 cm to 1 N to draw the forces in the form of arrows. The direction of the force is indicated by the arrow. 3 N (3 cm in length) 4 N (4 cm in length)

Back to Addition of Force 1 Click Back to • Attach the tip of

Back to Addition of Force 1 Click Back to • Attach the tip of an arrow to the end of another arrow. (Tip-to-tail method) • Draw an arrow from the starting point to the end 5 N (5 cm) point. It is the net force. 3 N (3 cm 53. 1° in length) • Length of the arrow : 5 cm • Direction : N 53. 1°E 4 N (4 cm in length) • Direction of net force : N 53. 1°E • Magnitude of net force : 5 N (Why? ) • Direction of acceleration : N 53. 1°E (Why? ) • Magnitude of acceleration :

Addition of Force 2 Force 1 Next Slide • By using the concept of

Addition of Force 2 Force 1 Next Slide • By using the concept of tip-to-tail method, one force can also be separated into two different forces, for example, • Use a scale of 1 cm to 1 N 10 sin 60°N (10 sin 60°cm) 10 N (10 cm) 60° 10 cos 60° N (10 cos 60° cm)

Addition of Force 2 Force 1 Next Slide • We want to add the

Addition of Force 2 Force 1 Next Slide • We want to add the following two forces using the method of resolving components. Scale : 1 cm to 1 N 5 N 60° 6 N 30°

Addition of Force 2 Force 1 Next Slide y 6 N 5 N 60°

Addition of Force 2 Force 1 Next Slide y 6 N 5 N 60° 30° • We place them together on xy-coordinate plane with both the nails at the origin. x y 5 sin 60°N 6 sin 30° N • Each force can be represented by a force (component) along x -axis and a component along yaxis. 30° 5 cos 60°N 6 cos 30°N x

Addition of Force 2 Force 1 Next Slide y • Add the components along

Addition of Force 2 Force 1 Next Slide y • Add the components along x-axis together. Then add the components along y-axis. 6 sin 30° N + 5 sin 60° N 6 cos 30° N - 5 cos 60° N x • They are of the same direction and we can add them like scalars.

Addition of Force 2 Force 1 Next Slide • Combine the components of force

Addition of Force 2 Force 1 Next Slide • Combine the components of force along each axis to form the net force vector. y 6 cos 30° N - 5 cos 60° N 6 sin 30° N + 5 sin 60° N x

Back to Addition of Force 2 Force 1 Click Back to • Magnitude of

Back to Addition of Force 2 Force 1 Click Back to • Magnitude of the net force: • Direction of force ( ):

Addition of Force 2 Force 1 Next Slide • HK Convention & Exhibition Centre

Addition of Force 2 Force 1 Next Slide • HK Convention & Exhibition Centre • By resolving components, the roof will not fall even no support is directly below the roof

Addition of Force 2 Force 1 Next Slide • Hong Kong Space Museum •

Addition of Force 2 Force 1 Next Slide • Hong Kong Space Museum • By resolving components, the roof will not fall even no support is directly below the roof

Back to Addition of Force 2 Force 1 Click Back to • The top

Back to Addition of Force 2 Force 1 Click Back to • The top of a tunnel