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New

ton's laws an

d application questions

FORCES AND NEWTON'S LAWS

QUANTITY SYMBOL UNIT VECTOR

force F N

mass m kg acceleration a m.s2

velocity m.s

time t S

static frictional coefficient

kinetic frictional coeficient

(normal) reaction force N N

gravitational acceleration 9,8 m.s2

universal gravitational constant G 6,7 x 10" N.m.kg

distance between centres m

EQUATIONS:

m(v, V AV F m

At Fnet ma

At

F (with F restricting force; N

H (with F = kinetic frictional force) N

Gm,m2 F

THEORY:

- An object persists in its state of rest or uniform motion, unless

a non-zero resultant force is exerted upon it. Newton

- If a non-zero resultant force is exerted on an object, the object will

experience a constant acceleration in the direction of the force.

The acceleration is directly proportional to the resultant force and

inversely proportional to the mass of the object. (a « F+m)

Newton I

- When object A exerts a force on object B, object B will exert an equal,

but opposite force on object A. Newton Ill

Forces

Force is a vector quantity - it has both magnitude and direction.

Forces are measured in Newtons - I Newton is the force acting when I kg accelerates at 1 m.s-2

Most forces are contact forces. There is direct contact between the force and the object on which it acts.

Sometimes a force is exerted through a rope or a solid rod. A force that acts through a rope is called a TENSION

Some forces act through space; for example the force between two magnets or the gravitational force between the earth and an object. The gravitational torce between the earth and an object is called WEIGHT.

The difference between Mass and Weight The mass of an object is the quantity of matter in the object. Mass is a scalar quantity and is measured in Kilograms. The mass of an object does not change.

The weight of an object is the gravitational force exerted by the earth on the object. and is measured in Newtons. Weight always acts down towards the centre of the earth.

When an object falls the earth causes it to accelerate at 1,4 m.s-2, To calculate the weight of an object we muitiply its mass by this acceleration.

Weight is a force

W mg ( where g = 13 m.s-2)

As a result of its weight, an object resting on a surface will exert a downwards force on the surface. In response to this the surface will exert an upwards force on the object. We call this upwards force the NORMAL REACTION (FN) of the surface (the force of the surface on the object).

15kg Normal Reaction

150N bOX

ground Force down on ground = 150N

Adding Forces

We can add forces to find their resultant. the individual forces.

The resultant has the same effect as the combination of

Forces in a straight line.

When forces are in a straight line we can use and to show direction.

+7

R +10 R +4

These forces are in the same direction. These forces are in the opposite direction.

The angle between them is 00. R will have the MAXIMUM value when the

The angle between them is 180° R will have the MINIMUM value when the

angle between two forces is 00. angle between two forces is 180,

When the forces acting on an object balance, the resultant of the forces must be zero. We say that such an object is in EQUILIBRIUM.

2

Newton's First Law

f all the forces on an object are balanced it will not change its momentum because its velocity will not change. This follows from Newton's First Law.

An object remains at rest or continues with constant velocity unless acted upon by an unbalanced force.

WHEN ALL FORCES BALANCE: if the object is at rest it stays at rest. If it is already moving it continues to move but does not speed up or slow down or change direction.

INERTIA: An object has inertia because it will not change its velocity unless an UNBALANCED force acts on it. Inertia is the resistance to change in velocity. The inertia of an object increases with increasing mass.

Do not confuse INERTIA with MOMENTUM. Inertia depends only on the mass of the body whereas momentum depends on the mass and velocity of the body. A stationary object has inertia but has no momentum.

Exercisel1 State Newton's First Law in full.

2. A small metal sphere hangs from a thread attached to the roof of a railway carriage which is moving to the right as shown in the diagram. What will happen to the sphere if the carriage 2.1 Speeds up? 2.2 Slows down? 2.3 Travels with constant velocity?

3. Use Newton's First law to explain the following: 3.1 3.2 3.3 3.4 3.5

When a car stops suddenly the passengers are "thrown" forwards. Seatbelts are necessary to stop passengers hitting the windscreen. When a car accelerates forwards the passengers are "jerked" backwards. When a car goes around the corner the parcels slide across the seat in the opposite direction. Space rockets can continue to move at constant velocity in space even though their engines are switched off.

3.6 A passenger who jumps from a moving train has to run forwards to stop himself from falling Over.

Ben is riding on his skateboard when his friend Arnie offers to tow him with his bicycle. What happens when Arnie suddenly has to brake?

4.1

Why is it necessary for caravans and trailers to have their own braking systems?

4.2

In a movie an open truck which is moving forwards brakes suddenly and a body rolls out of the back. Use concepts in physics and explain why this is unlikely to happen.

5.

Place a coin on a piece of card which is on top of a glass. Flick the piece of card away. The coin falls into the glass.

5.

TA Try to analyse the forces acting and use the concept of inertia to explain what happens.

Newton's Third Law

Newton realized that forces act in pairs. Consider two magnets floating on a dish of water with opposite poles facing. The magnets move towards each other because they BOTH experinece the same force of attraction. We express this as Newton's Third Law:/

If body A exerts a force on body B, then B will exerta force on A. The forces are equal in

magnitude but opposite in direction.

These forces are called an ACTiON-REACTION pair. Action and reaction CANNOT ADD UP TO ZERO becauseTHEY ACT ON DIFFERENT OBJECTS BALANCED FORCES can add up to zero because THEY ACT ON THE SAME OBJECT.

Example 1 Force 1 earth pulls BOOK down (Weight). Force 2- book pulls EARTH up. Action-Reaction pair (different objects). Force 3-book pushes TABLE down. Force 4 table pushes BOOK up. Action- Reaction pair (different objects). The book does not move because Force 1 (earth pulls BOOK down) = Force 4 (table pushes BOOK up). Balanced forces (SAME OBJECT).

Example 2 Man pushes CAR ; car pushes MAN Action Reaction pair (different objects). The car will not move if: push on CAR = friction on CAR (Balanced forces SAME OBJECT). .

friction friction The car will accelerate if:

push on CAR > friction on CAR (there is a resultant force). The man will accelerate backwards if: push on man > friction on man's feet (there is a resultant force).

Think Tank| 1

This exercise gives several examples of the interaction between two bodies. For each example. write down the reaction to the named force (the action). i) on the diagram draw an arrow to indicate the direction and magnitude of the reaction force. say whether forces acting on the same object are balanced.

1.1 The earth pulls the box down with a force of 40N. Earth pulls box down

1.2 Why does the box fall down while the earth does not appear to move?

2.1 Magnet M pulls magnet P to the left with a force of 4N.

M pulls P to the left.

Pis on a rough surface and does not move.

Mark in the size and direction of the force that keeps

P from moving.

2.2

N N

magnet M magnet P 2.3 Mis on a frictionless surface.

Explain what happens to M.

3.1 The boy on the skateboard pushes against the wall with a force of

50N. 5ON

There is a friction force of 50N between the wheels and the ground.

Draw in this force. Why does the boy not move? 3.2

3.3 Explain what happens when the boy pushes against the wall witha

force of 100N.

4.1 The athlete's foot pushes backwards on the ground with a force of 240N.

4.2 Why does the athlete accelerate?

4.3 Why does the earth not appear to move?

5.1 The jet engines push thee exhaust gases backwards with a force of 30 000 N.

30 000 N

5.2 The jet is flying at a constant velocity of 300 km.h1. Suggest a reason why the jet does not accelerate.

5.3 On the diagram mark in the magnitude and direction of the force which prevents the jet from

accelerating.

6. In diagram (i) a learner pulls on the spring balance with a force of 20 N. In diagram (ii) each

learner pulls on the spring balance with a force of 20 N.

diagram () diagram (i)

6.1 Mark and name the action /reaction forces at points P and Q.

6.2 What does the spring balance read in diagram(i)/ in diagram (i)?

5

Exercise 2

1. The sketch shows a book being pressed against the wall. The book is stationary and a number of forces act on it.

1.1 Draw a diagram to show the forces acting and give the direction of each. 1.2 1 Why does the book not move?

2. An orange rests on a table. 2.1 Draw and name two forces acting on the orange.

Which bodies exert these forces? On what bodies does the orange exert forces? In which directions do these forces act?

2.2 2.3

3. The diagram shows two boxes which are pushed by a force of 140N exerted on box M. Box M exerts a force of 4ON on box N.

3.1 Draw separate diagrams for M and N and show ALL the forces acting on each box. Give the magnitude and direction of the resultant

140N M 50kg N

3.2 20kg force on each box.

3.3 Name one action reaction pair in this example.

4. Car X is towing car Y. the tow rope is 800N. There is a friction force of 200N acting on each car.

The engine of car X exerts a forward force of 1800N and the tension in

1800N

80ON

4.1 Draw separate force diagrams for each car showing the horizontal forces acting. 4.2 4.3

Give the magnitude and direction of the resultant horizontal force on each car. Name one action reaction pair in this example.

5. Two men wish to break a length of rope that has a reported breaking strain of 900 N. They try two methods. they pull on opposite ends of the rope exerting 500N each

they attach one end to the wall and both pull on the other end exerting 500 N each. Will either of these methods work? Explain your answer.

A girl pulls a box across rough ground at constant velocity as shown in the diagram. forces acting on the box?

Which of the following diagrams correctly shows all the

B C

A parachutist of mass 70 kg is descending at a constant velocity of 2 m.s. The force exerted by the parachute (in N) is.

A. 70 B. 700 C. smaller than his weight D. greater than his weight

8. How can Newton's Third Law be represented? A. C. If F = 0 then V = Constant

D. W = mg

p mv

B. FAB-FBA

9. An athlete starts a race by pushing hard against the starting blocks. If the force which the athlete exerts on the starting blocks has a magnitude X then the force with which the starting blocks push back on the athlete is...

A. equal to X B. greater than X

smaller than X D. equal to the athlete's weight

6

10. A man of mass 90 kg collides head on with a boy of mass 45 kg on a skating rink. It follows that:

the magnitude of the force exerted by the man on the boy is double that of the force exerted by A.

the boy on the man.

the magnitude of the force exerted by the man on the boy equals that of the force exerted by the

boy on the man.

the magnitude of the force exerted by the man on the boy is half that of the force exerted by the

boy on the man. the boy exerts no force on the man because the man pushes him.

B.

C D.

Finding the components of forces

The opposite of adding two forces together is finding the effect of a force in a certain direction. We

call this resolving the forceinto its components. We normally resolve into components that are at 90

to each other. we know the value of the force R and the angle it makes with the horizontal (e) we

can find its vertical component Rsin 6 and its horizontal component Rcos .

Y = R sin e

X = R cos

Think Tank 2

1.1 Ris a force of 55 N which acts on a bearing of 600. of 10mm 10 N). Mark in the components in the north and east directions then measure them.

Draw R to scale on the axis provided (Scale

N

**************************************************e***********

1.2 Check your answer to 1.1 by calculation.

2. Draw a sketch of a force of 0,75 N acting at 400 to the horizontal and then find its vertical an horizontal components.

200N 3. The rope is pulling the box of mass 15kg but the box does not

move. The rope exerts a force of 200N and acts at an angle of 25° to the horizontal.

3.1 Calculate the horizontal and vertical components of the force exerted by the rope.

25

3.2 How much force does the box exert on the ground

3.3 Mark and name all the forces acting on the box.

Newton's Second Law

If there is an unbalanced force acting on an object, the object will accelerate in the direction of

the force. The acceleration is directly proportional to the force and inversely proportional to

the mass of the object.

Accelerate can mean speed up or slow down or change direction.

A CONSTANT force causes a CONSTANT acceleration. This means that the object will speed up or

slow down by the same amount in each second.

CONSTANT VELOCITY means there is NO resultant force. CONSTANT ACCELERATION means there is CONSTANT resultant force.

If the mass stays the same the acoeleration depends directly on the force If the force stays the same the acceleration depends inversely on the mass.

a « F

a c 1/m

We put this together in the equation Fres ma

V-u We can also use the tact that FAt = mv- mu and a=

At

FV-u) ma

To get At

To calculate the acceleration of an object you need to add all the forces on it and find the RESULTANT force.

Horizontal Motion

Forces in a straight line Forces at an angle

15 N 13 N 55 N

60 11 kg

15 kg

20 N friction

Fres 55-(20+ 13 ) = 22 N Fres ma

22 11a

Fres = 15 cos 60= ma

7,5 N 15a a = 0,5 m.s to the right

a = 2 m.s to the right

. Vertical motion

Accelerating upwards Accelerating downwards

F 117 N F 72 N

Fres= 117- 90 = 27 N

27 ma = 9a

res 90- 72 = 18N

9 kg 9 kg 18 ma = 9a

a = 3 m.sZ upwards a 2 m.s downwards W = mg

90 N W mg

90 N L

The Laws of Friction FRICTION is a force that acts to oppose motion. Friction is caused by one surface dragging over another suface. A smooth surface such as ice will be nearly frictionlesSs. The force of friction is increased by pressing surfaces together. This means that heavier objects will

experience more friction.

Force to overcome friction

The diagram shows a wooden block which is pulled by a

spring balance. Initially the block does not move but when the applied force is large enough the block starts to move and the reading on the spring balance now gives the value of the force of limiting (static) friction. When the block is moving, the reading on the spring balance decreases slightly, showing that the force of kinetic (dynamic or sliding) friction is silightly less than the force of limiting friction.

Experiments show the following to be true: The friction force between two surfaces opposes their motion. The friction force depends on the type of material and the roughness of the surtace. The friction force does not depend on the area of contact between the surfaces. When the surfaces are at rest the limiting friction force f, is directly proportional to the normal

reaction FN. When motion occurs the kinetic friction force fe is directly proportional to the normal reaction F and is independent of the velocity. fk a FN

fs aFN

The coefficients of limiting and kinetic friction are denoted by us and u, respectively and are defined by the following equations.

Hs Ts(max) FN and Hk=Tk/ Fu where k is usually less than us

Think Tank 3

1 What type of motion can an object have if 1.1 all the forces on the object are balanced? 1.2 there is a constant unbalanced force acting on the object?

1.3 there is a varying unbalanced force acting on the object?

What is the weight in Newtons of a mass of 2.1 45 kg?

2.2 459?

3 For each of the following diagrams decide, by calculation, if an unbalanced force acts. Describe the motion of the object using the phrases: at rest; moving with constant velocity; moving with constant positive or with constant negative acceleration.

3.1 Initial velocity = 0m.s 3.2 initial velocity = 0 m.s1

2 kg N 3N 2 kg 6N

frictionless surface friction

3.3 Initial velocity = 0 m.s 3.4 Initial velocity = 5 m.s to the right

12N 8N 60° 60°

sassesnse

2 kg 2kg 4N

frictionless surface fiction

3.6 Initial velocity = 5 m.s*l to the right

Initial velocity 3.5 5 m.s' to the right

8N 8N

60

60 *************

SN 2 kg 2N 2 kg

friction friction

A vertical force of 30 N on a 2 kg box.

initial velocity = 5 m.s upwards 3.8

3.7 initial velocity = 5 m.sl upwards

20 N 30 N

2 kg 2 kg

Vertical force of 10 N on a 2 kg box. Vertical force of 10 N on a 2 kg box. initial velocity = 5 m.sl upwards

3.9 3.10 initial velocity = 0 m.s1

10N 10N

2 kg 2kg

Think Tank 4

1. The diagram shows a car of 900 kg that is

travelling along a road at 20m.s1 1.1 Mark in the magnitude and direction of the

weight of the car. 1.2 The arrows show forces that act on the car.

Label the forces using the following labels: accelerating force of engine; weight; friction of road on tyres; reaction force of the road.

- 3000N

2250N

750N 750N

1.3 Give the magnitude and direction of the resultant force acting on the car.

1.4 Calculate the constant of kinetic friction for this car.

2. The diagram shows two ring magnets inside a glass tube. Magnet B is on the bottom of the tube and magnet A is suspended above it with the north poles of the magnets facing each other.

2.1 Mark and name the forces which are acting on magnet A. 2.2 Calculate the force between the two magnets if the mass of

magnet A is 100g.

Magnet A

Magnet B

10

Exercise 3

1. In which of the following cases is the resultant force on the object zero? An object which.. moves with constant speed around a corner.

B

moves with a constant acceleration. is in free fall.

A C

undergoes equal displacements in equal time intervals. D

2. lfa constant resultant force of 1N acts on a stationary piano the piano will.. A stay at rest since the force is too small to overcome the inertia of the piano.

stay at rest since the frictional force between piano and floor must be greater than 1N. move with constant velocity.

D move with constantly increasing velocity

A horizontal force of 12N is applied to a body and gives it an acceleration of magnitude 2 m.s along a horizontal surface. If the frictional force acting on the body is 8 N what is its mass?

A.

3.

6 kg B. 4 kg C. 2 kg D. 0,5 kg

The acceleration of an object of mass M is x when a force F acts on it. Another object with twice the mass experiences an acceleration of 3x. The force on the second object is..

A.

4.

6F B. 1/6 F C. 2/3 F D. 3/2 F

3N 9N The box shown in the diagram has a weight of 30 N. What is its acceleration?

5.

A. 0,2 m.s-2 B. 0,3 m.s2 C. 2 m.s2 D. 10 m.s2

6. A workman standing on scaffolding lowers a bag of weight 500N by means speed. lf the weight of the rope is negligible, the force that the workman exerts is..

a rope, at constant

A constant and equal to 500 N. constant and less than 500 N. B constant and greater than 500 N. greater than 500 N and decreasing.

7.The sit, vt and at graphs desoribe the motion of three diferent objects. In which case/s MUST the forces acting be balanced?

| III V a

A. lonly B. Il only C. and II D. Il and Ill

F The mass M shown in the diagram will move with constant velocity if.. X

A X = F BX = F sin X = F cos 6

GauroYeasdoese**

0 M

Exercise 4/

1. A Cupboard of mass.150kg is on a horizontal floor. A horizontal force of 400 N can just shift it. Find the coefficient of friction between the cupboard and the floor.

2. A body of mass 70 kg lies on a horizontal floor where the coefficient of friction is 0,4. Find the least horizontal force which can shift the body.

The diagram shows a body of mass 2kg which is pressed against a rough vertical wall by a force of 40 N. If the body is about to slide find the coefticient of limiting friction.

3.

40 N Kg

4. A horizontal force presses a mass of 5kg against a vertical wal where the coefficient of friction H is 0,5. Find the least value of the force which can prevent the body from slipping.

11

Tanya and Mitsie do an experiment to investigate the laws of friction. They measure the limiting force of friction (F) and the kinetic force of friction (F) for varying values of the normal reaction force (N). They use their resuts to draw the graphs shown in the

diagram. What relationship between F and N is indicated by the graphs? What is given by the gradient of the graph? Which graph shows the results for the limiting force of friction? Give a reason for your answer.

Calculate the force of limiting friction and kinetic friction for an

object of mass 600g.

5. A

A B

2,5 1,5 S.1

1,25 s.3

2,5 5 N (Newtons) S.4

Find the The diagrams show two forces of 5 N and 12 N that act at the same point.

esultant in each case.

6.1

5N 12N 5N 12N

12N

5N

6.2 6.3

When will two forces have (i) a maximum value (i) a minimum value? For each of the diagrams shown above give the magnitude of the force needed to balance the resultant.

An elephant is pulling a log of mass 100 kg , with a force of 500 N making an angle of 200 witk 1.

the horizontal.

7.1 Calculate the horizontal and vertical components of the force. Show that the log will not lift off the ground. If the log is just about to move,calculate the coefficient of friction.

1.2 7.3

A gardener is pushing a roller of mass 150 kg along the ground. He exerts a force of 300 N along the handle which makes an angle of 300 with the horizontal.

8.1 Calculate the vertical and horizontal components of the force exerted by the gardener. 8.2 What is the total force exerted on the ground? 8.3 Why is it easier to pull the roller than to push it? 8.4 Would there be more or less force exerted on the ground if the angle between the roller and the horizontal were 400 instead of 30°?

Objects on a slope

When an object is on a horizontal surface its weight acts at right angles to the surface. The surface exerts an equal and opposite force on the object. This is called the reaction force.

Reaction Force

When an object is against a vertical surface the weight of the object acts parallel to the surface. The object will slide along the surface because its weight is not supported.

Weight mg Weight mg

12

When an object is on a slope, part of the weight is supported by the siope but part of the weight acts parallel to the slope and makes the object slide. The slope divides the weight into two components. One component is perpendicular to the slope (W). This is equal and opposite to the reaction

force The other component is parallel to the slope (Wa). This has to be balanced by a force such as

friction.

W The weight is divided into two

components W1 mg cose W2 = mg sine

W2 mg W

W mg

N

Three forces keep the object in equilibrium N W1= mg cos F W2 = mg sin

mg

mg

Think Tank|5

1 The mass of the block in the diagram is 5 kg and the angle of the siope - 150. Calculate.

W, the component of the weight perpendicular to the surface. 1.1

s Wa the component of the weight parallel to the surface.

mg 1.2

13 What is the magnitude of the friction force F that will just stop the block sliding?

What is the magnitude of the reaction force N? 1.4

1.5 Calculate the coefficient of limiting friction for the block on the slope. 1.6 Show that for an object on a slopeu = /N = tan 8.

2 A car of mass 1600 kg is at rest on a slope inclined at an angle ot 150 to the horizonta Calculate.. 2.1 the force exerted on the ground by the car. 2.2 The friction force which acts to keep the car stationary. 2.3 The coefficient of static friction for the slope.

A 3. Two boys pull a stone of mass 50 kg along a plank which slopes at 35 to the horizontal. At this point the stone is at rest on the plank. plank can withstand a force of 450 N perpandicular to its surtace before it breaks.

3.1 Calculate the force exerted by the two boys to stop the stone from sliding down the plank.

3.2 Do a calculation and decide whether or not the plank will break.

The

4. A frictional force of 540N keeps a box stationary on a sleping surtace which has a coefficient of friction u = 0,6. Calculate.,

4.1 The force exerted by the box on the surface. 4.2 The angle of the slope. 4.3 The mass of the box.

HOMEWORK FORCES AND NEWTON'S LAWS GRADE 11

MARK: NAME:

Ax (v, + v,)At Ax = V,At + sadt v-+ 2aAx; V V,+ aAt; F ma

Two blocks, masses 10 kg and 4 kg respectively, that are

in contact with each other on a frictionless, horizontal

surface, move at 20 m.s', as in the diagram. A braking force of 28 N is then exerted horizontally on the 10 kg block. Calculate ..

the magnitude of the acceleration of the system.

1. 20 m.s

A kg 10 kg 1.1

3) 1.2 the magnitude of the resultant horizontal force exerted on the 10 kg block.

(2) 1.3 the force exerted on the 10 kg block by the 4 kg block.

(2)

1.4 how long the 28 N force must be exerted in order to attain a velocity of 10 m.s' in the opposite

direction.

(4)

2 Two trolleys with masses of 2 kg and 10 kg respectively, are accelerated by a force of 36 N to the right. Friction of 6 N is

exerted on each trolley. Calculate .. 2.1 the acceleration of the system.

T 36N A

10kg

O 6N 6N

(3)

2.2 the tension (T) in the string between the trolleys.

(3)

Three trolleys, A with mass 10 kg, B with mass 20 kg and C with mass 4 kg, are connected to

each other on a table surface. These trolleys are then connected across a pulley to a hanging block D with mass 6 kg. Consider the table and the pulley to be frictionless. Calculate..

3.

Ti 2 C

A B 10 kg 20 kg 4 kg

6 kg

3.1 the tension T, between B and C.

(6) 3.2 the tension T, between A and B.

(3)

A helicopter lifts a motorcar with mass of 900 kg and a motorbike with mass of 600 kg as shown.

4.

0 Cable 1 4.1 Write an equation in symbols for

900 kg 4.1.1 the forces on the motorcar.

(2) Cable 2

4.1.2 the forces on the motorbike.600 kg

(2)

Draw separate force diagrams to indicate the forces acting on.

4.2.2 the motorbike. 4.2 4.2.1 the motorcar.

(3) (2)

4.3 The maximum tension in cable 1 is 18 000 N. Calculate the maximum acceleration of the

helicopter if the cable may not break.

(5)

4.4 Calculate the tension in cable 2 when the helicopter reaches this maximum acceleration.

3) If cable 1 breaks and the motorcar and motorbike fall together, what will the tension be in

2) 4.5

cable 2?

TOTAL: [45]

4

Calculate the force which the block will experience as a result of its weight, parallel to the slope. Calculate the dynamic frictional force which the block will experience while it is accelerated.

Calculate the magnitude of forceF.

17.3

17.4

17.5

Two crates (m, = 80 kg and m, = 120 kg) are

in contact and are pushed across the floor of

18.

M2 a warehouse. A horizontal force of 2 000 N is

m F 2 000 N exerted on the 80 kg crate. All forces of

friction can be ignored. Draw a free body diagram for each crate. 80 kg 120 kg 18.1

18.2 Calculate the acceleration of the crates.

Calculate the magnitude of the force which crate m, exerts on crate ma. Prove that the force crate m, exerts on crate m,, is equal to the force crate m, exerts on crate m

18.3

18.4

Susan exerts a horizontal force of 180 N on a rope, causing two blocks with masses of 20 kg and 40 kg to accelerate to the right, as shown in the diagram. All frictional forces, as well as the masses of the ropes can

19

be ignored.

F Fs 180 N B A

40 kg 20 kg Susan

19.1 Draw a free body diagram for each block. 19.2 Calculate the acceleration of the blocks. 19.3 Calculate the tension F in the rope between the blocks.

Bob then exerts a force of equal magnitude on the same blocks, but in the opposite direction.

F FB 180 N

B A

40 kg 20 kg

How will the tension F in the rope between the blocks compare with that in question 19.3? Use principles in physics and explain your answer to question 19.4.

19.4 19.5

A boy connects two blocks, with mass 1 kg and 3 kg, by a light string. The boy pulls the 3 kg-block to the right with a force F and both blocks accelerate at 0,5 m.s. The frictional force between the floor and the 1 kg-block is 2,5 N, while the frictional force between the floor and

20.

0,5 m-s2 FT

1 kg 3 kg

the 3 kg-block is 3,5 N. The tension in the string is FT Draw a free body diagram for each block.

Apply Newton II to the 1 kg-block and calculate the magnitude of the tension FT Now calculate the magnitude of force F.

20.1

20.2

20.3

A and B are two identical trolleys, each with mass 60 kg, attached to each other with a rope. They are drawn across a rough surface by a force of 600 N acting on trolley B at an angle of 60° to the horizontal. The tension (force) in the rope between A and B, is 160 N. The trolleys acelerate to the right at 2 m.s2

21

600 N A 160 N B

60 kg o O

60 60 KY

Three trolleys, A with mass 10 kg, B with mass 20 kg and C with mass 4 kg, are connected to eacr cre on the surface of a table. These trolleys are then attached to a hanging block D with mass 6 kg cver

a pulley

25.

Ti. T 10 kg 20 kg

Consider the table and the pulley to be frictionless. Calculate the tension T, between B and C.

Calculate the tension T, between A and B. Which rope in the system needs to be the strongest?

25.1

25.2 25.3

Three blocks, masses 5 kg and 6 kg, are attached to each other by two ropes of negligible mass, as shown in the diagram. The 15 kg mass-piece hangs over

the side of the rough table with the rope over a frictionless pulley. The blocks' surfaces differ and each experiences a kinetic frictional force of 20 N. The system accelerates to the right (clockwise).

26.

5 kg 6 kg Z7777777

15 kg

Draw a free body diagram for eacj of the blocks.

Calculate the acceleration of the system. Calculate T, and T, as in the diagram. Calculate H, for each block.

26.1

26.2

26.3

26.4

Two blocks, with mass 2 kg and 5 kg respectively, are connected by one rope over a frictionless pulley. The mass of the rope is negligibly smal. Draw free body diagrams for the two masses.

Calculate the acceleration of the system. Calculate the tension, T, in the rope.

27.

27.1 2 kg 27.2

5 kg 27.3

A system of objects are attached to each

other by strings (with negligible masses) and

set up as shown in the diagram. T, is the

tension in the string between the objects with masses 2 kg (Q) and 5 kg (R). Tz is the

tension in the string between the objects with masses 2 kg (Q) and 3 kg (P). The surface and the pulleys are frictionless.

28.

T2 2 kg T, Q

kg P 5 kg R

28.1 Calculate the tension in both strings. 28.2 Calculate the acceleration of the system. 28.3 How will the acceleration be influenced if the masses of the strings are taken into account?

29 A block, mass 4 kg, resting on an incline, is connected to

mass piece, mass 3 kg, by a light string over a frictionless pulley. The block experiences an acceleration of 0,75 m.s?

upwards, parallel to the surface. The incline is at an angle of 25 with the horizontal.

a = 0,75m.s

4 kg

250 3 kg

29.1 Calculate the weight of the mass piece.

Calculate the component of the weight of the block parallel to the incline Draw separate free body diagrams of all the forces acting on the block and the mass piece. Calculate the frictional force exerted by the surface on the block.

29.2

29.3 29.4

Two people exert a pushing force, A= 120 N, and a pulling force, B 250 N, respectively on a heavy crate, mass 60 kg, as shown in the accompanying diagram. The crate moves at

a constant velocity across a rough floor surface.

30.

120 N B 250 N

60

60 kg

Draw a labelled free body diagram of all the forces acting on the crate.. Calculate the resultant horizontal and vertical components of the forces A and B.

Calculate the magnitude of the normal force on the crate.

Calculate the magnitude of the frictional force on the crate.

Determine the magnitude of the pulling force B if it were to cause an acceleration of 2 m.s" on the crate.

30.1

30.2

30.3

30.4

30.5

Two objects of mass 6 kg and 3 kg respectively, are

connected by a light inelastic string. They are pulled up a frictionless inclined plane which forms an angle of 30 with the horizontal, by a force of magnitude F. The mass of the string may be ignored. Calculate..

31.

ka

6 kg 31.1 the tension in the string if the system accelerates

up the inclined plane at 4 m.s the magnitude of F if the system moves up the inclined plane at CONSTANT VELOCITY.

30.

31.2

32 Two objects of mass 6 kg and 3 kg respectively, are connected by a light inelastic string. They are pulled up an inclined plane which forms an angle of 30 with the horizontal, by a force of magnitude F. The mass of the string may be ignored. The coefficient of kinetic friction for the 3 kg object is 0,1 and for the 6 kg object it is 0,2. Calculate..

F

3kg

6 kg

...0

32.1 the tension in the string if the system accelerates up the inclined plane at 4 m.s2. the magnitude of F if the system moves up the inclined plane at CONSTANT VELOCITY.

32.2

13

A learner constructs a push toy using two blocks with masses 1,5 ka and 3 kg respectively. The blocks are connected by a massless, inextensible cord. The learner then a applies a force of 25 N at an angle or d to the 1,5 kg block by means of a light rigid rod, causing the toy to move across a flat, rougn, noriO surface, as shown. The coefficient of kinetic friction (u.) between the surface and each block is 0,15.

44.

25 N

30°

3 kg 1,5 kg

44.1 State Newton's Second Law of Motion in words. (2) (4,41 N)(

(5) 44.2 Calculate the magnitude of the kinetic frictional force acting on the 3 kg block.

Draw a labelled free-body diagram showing ALL the forces acting on the 1,5 kg block.

Calculate the magnitude of the.. 44.3

44.4

(4,08 N(3) 44.4.1 kinetic frictional force acting on the 1,5 kg block.

44.4.2 tension in the cord connecting the two blocks. (13,19 N(5)

An 8 kg block, P, is being pulled by constant force F up a rough inclined plane at an angle of 30° to the

horizontal, at CONSTANT SPEED. Force Fis parallel to the inclined plane, as shown. The kinetic frictional force between the block and the surface of the inclined plane is 20,37 N.

45

F

8 8 kg

30°

(2) (4)

45.1 State Newton's First Law in words.

Draw a labelled free-body diagram for block P.

Calculate the magnitude of force F.

Force F is now removed and the block ACCELERATES down the plane. The kinetic frictional force remains

20,37 N. Calculate the magnitude of the acceleration of the block.

45.2

45.3 (59,57 N5) 45.4

(2,35 m.s 4)

46. Two blocks A and B of equal mass 0,2 kg slide down an

inclined plane as shown. The coefficient of kinetic friction between block A and the inclined plane is = ,01 and

the coefficient of kinetic friction between block B and the

inclined plane is 1,00.

45

46.1 Draw a labelled free-body diagram showing ALL the

forces acting on block A as it slides down the incline. (4) The two blocks accelerate at 3,43 m.s. Calculate the magnitude of the force exerted by block B on 46.2

block A. (0,69 N)(3) The system is placed on the Moon where the gravitational acceleration is 1,62 m.s2 How will it affect the acceleration of the blocks? Write INCREASES, DECREASES or REMAINS THE SAME. Explain

46.3

your answer. (3)