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Statics 1 H1/H2 Physics

JC1 H1/H2 Physics

STATICS Statics n. a branch of physics that studies objects at rest

Aims

To be able to

(a) draw the weight, tension, normal reaction and friction

(b) draw the free body diagram

(c) use vector triangle to represent forces in equilibrium

(d) calculate the moment of a force

(e) calculate the torque of a couple

(f) apply the principle of moments


1 About the topic

Statics is the branch of physics that studies objects at rest.

What keeps an object at rest?

If we place an object far out in the outer space, away from any planets, stars and galaxies, the object remains at rest because there is no force pulling it.

But if we release an object here on Earth, it cannot remain at rest and falls vertically downwards because the Earth is pulling it down with the gravitational force.

A force is something that has a tendancy to move an object at rest.

If there is no force acting on an object at rest, the object will remain at rest.

If a force acts on the object, the object cannot remain at rest and will start to move in the direction of the force.

When two or more forces act on the object, if the forces cancel out, then the object remains at rest, otherwise, it would start to move in the direction of the resultant force.

Condition for object to remain at rest

For an object at rest to remain at rest, either there is no force acting on it or the forces acting on it cancel out such that the resultant force is zero.

What’s in this topic?

In this topic, we will be dealing with situations where there are two or more forces acting on an object but the forces cancel out each other and the object remains at rest.

But first, we must know how to identify the forces acting on the object.

2 Common types of forces

Here are the different types of forces commonly found in our daily life:

Force Symbol

How to identify this force?

How to draw this force? Diagram

Weight W when object has a

mass

starting from centre of gravity draw vertically

downwards

Tension T when object is attached to a

string

starting from end of string draw

towards centre of string

Normal reaction

N when object is in

contact with a surface

starting from point of contact draw

normal and away from the surface

Friction f when contact

surface is rough

starting from point of contact draw

along the surface

Note:

• we draw an arrow to represent the force

• the arrow must have an arrow head to indicate the direction

• the arrow must be straight, not curved

• the length of the arrow indicates the magnitude of the force; the longer the arrow, the larger the force.

T

W

N

f

An object released in outer space remains at rest because no force acts on it.

An object released on Earth cannot remain at rest and falls vertically downwards because gravitational force is pulling it down.


QED 1

Which diagram shows the most proper way of drawing the weight?

QED 2

A box is hung from the ceiling by a thread.

Which of the above diagrams shows:

a. the tension acting on the box

b. the tension acting on the ceiling

QED 3 What is the proper way of drawing the normal reaction?

a. object resting on a horizontal surface

b. object resting on a slope

c. object resting against a wall

QED 4

What is the correct direction of friction?

Object resting on a rough slope

Object resting on rough floor and rough wall

QED 5

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D


The diagram shows a pail being swirled in a vertical circle by a rope and a box is inside the pail:

Draw

a. the weight acting on the box:

b. the normal reaction acting on the box:

c. the tension in the rope acing on the pail:

d. the weight acing on the pail:

QED 6

In the following diagrams, draw all the forces acting on the object.

c. ping pong ball being chopped by table-tennis bat

f. spiderman swinging

a. basketball after it was thrown

b. tennis ball after hit directly by tennis racket

direction of movement of

table-tennis bat

d. sprinter running e. high jumper clearing the bar


3 Free body diagram

A free body diagram is a sketch showing an object and all the forces acting on the object. Drawing such a diagram helps in solving problem.

For the free body diagram:

• do not draw forces exerted by the object,

• do not draw forces acting on other objects,

• draw only the forces acting on the object,

• draw all the forces acting on the object, don’t miss out any force.

Practice: Free body diagram

1. Box at rest on smooth table:

Free body diagram of the box:

2. Box at rest on rough slope:

Free body diagram of the box:

3. A box on a rough ground being pulled by a person:

Free body diagram of

(a) box (b) person

4 Two connected boxes (slope is smooth):

Free body diagram of

(a) box A: (b) box B:

B

A

B

A


Combined normal reactions

Sometimes there are more than one contact point with the surface. The normal reactions at the contact points can be combined together to form the total normal reaction exerted by the surface:

Similarly, frictional forces can also be combined together.

Practice: Combined normal reactions and combined frictions

5. Practice 3(b) cont’d: combine the normal reactions and the frictions.

Combined normal reaction and friction

A rough surface can exert the following forces on a box:

• normal reaction

• friction

The normal reaction and friction can be combined together as a force called the total reaction force R:

Practice: Combined normal reaction and friction

6. Practice 5 cont’d: combine normal reaction and friction together.

QED 7

A uniform ladder on a rough floor rests against a rough vertical wall. Draw the forces acting on the ladder, drawing only the total reaction force acting at each end of the ladder.

N

f

R

N1 N2

N

N1

N2

N3 N4

N

ladder


4 Forces in equilibrium

After drawing all the forces acting on an object, the next step is to analyse the forces.

If the object is at rest, the forces would cancel out each other. We say that the forces are in equilibrium.

Case 1: Two forces in equilibrium

If there are only two forces acting on an object and the forces are in equilibrium, the two forces must be equal and opposite in direction.

QED 8

A box rests on a horizontal table.

Which of the following is the correct free body diagram of the box?

If the weight of the box is 10 N, what is the normal reaction?

A 20 N B 10 N C 5 N D 0 N

QED 9

A box is suspended by a vertical string.

Which of the following correctly shows the forces acting on the box?

Case 2: Three forces in equilibrium

For three forces acting on an object, if the forces are in equilibrium and the forces are not parallel, then the forces will

• form a closed triangle in the vector diagram (so that the resultant force is zero), and

• intersect at a common point (so that the resultant moment is zero, why is this so – see Appendix 2)

QED 10

A uniform ladder on a rough floor rests against a smooth vertical wall. Draw the forces acting on the ladder, drawing only the total reaction force acting at each end of the ladder.

Draw the vector diagram of the forces.

The weight of the ladder is 40 N, the reaction force by the wall is 30 N. Find the reaction force by the floor.

A B C D

ladder

A B C D


5 Solve by vector diagram

A vector diagram is a sketch showing how the vectors are related to each other. For forces in equilibrium, the force vectors would form a closed triangle (3 vectors) or in general a closed polygon.

QED 11

A 20 N sphere is supported by two cords as shown in the diagram.

Problem: Find tension in cord 1.

Step 1 Draw the forces acting on the sphere.

Step 2 Draw the vector diagram of the forces acting on the sphere.

Step 3 Use the vector diagram to find the tension in cord 1:

6 Solve by resolving forces

Instead of using vector diagram, we can also solve the problems by resolving forces. The forces are resolved along two axes (usually vertically and horizontally) and the components are equated.

QED 12

A 20 N sphere is supported by two cords as shown in the diagram.

Problem: Find tension in cord 1.

Step 1 Draw the forces acting on the sphere.

Step 2 Resolve into horizontal and vertical components:

cord 2 tension horizontal component =

cord 2 tension vertical component =

Step 3 Equate horizontal and vertical components:

horizontally: →

vertically: ↓

Step 4 Use the equations to find the tension in cord 1:

50°

cord 1

cord 2 50°

cord 1

cord 2


7 Moment of a force

The diagram shows a steering wheel. When you exert a downward force on the wheel, the wheel turns. We say that the force has a turning effect; not only can it make things move, it can also make things turn.

The spanner is a tool that makes use of the turning effect of the force. The amount of turning effect it produces depends not only on the magnitude of the force employed but also the perpendicular distance between the axis of rotation and the line of action of the force:

Suppose a force F is applied at a certain point on the spanner such that the perpendicular distance between the axis of rotation and the line of action of the force is d. The amount of turning effect will depend on the product Fd. This product is known as the moment of the force. The bigger the moment, the greater the turning effect.

QED 10

A man is trying to lift a big stone.

Which of the following will require the least applied force F?

Force

force F

A B C

F F

F


When finding the moment of a force, there are basically two ways, either

• by first finding the perpendicular distance, or

• by first finding the perpendicular force

Then moment = force × perpendicular distance

or perpendicular force × distance

QED 13

Find the moment of the force and state whether the direction is clockwise or anticlockwise.

Method 1 Find perpendicular distance first

Perpendicular distance =

Hence moment =

Method 2 Find perpendicular force

For distance 0.30 m, perpendicular force =

Hence moment =

Direction of moment =

Method 1 Find perpendicular distance first

Perpendicular distance =

Hence moment =

Method 2 Find perpendicular force

For distance 0.30 m, perpendicular force =

Hence moment =


Moment =


0.30 m

25° 40 N

c.

0.30 m

50°

30 N

b.

20 N

0.30 m

60°

a.


8 Principle of moments

When a system is at rest, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point.

QED 14

A uniform metre rule, pivoted at the 50 cm mark, with a 4.0 N weight placed at the 20 cm mark, is held horizontal by a downward force F at the 100 cm mark.

Find F.

QED 15

A uniform metre rule of weight 12 N is supported by two vertical cords.

Find T. [Draw all forces first]

QED 16 The diagram shows a uniform platform of weight 200 N held in position by two rods X and Y with a lady of weight 500 N standing on the right-hand end.

Draw all the forces acting on the platform.

Find the force exerted by rod X.

Find the force exerted by rod Y.

T

40 cm

20 cm

100 cm 4.0 N

F

50 cm

6 m

2 m

500 N

X

Y


9 Torque of a couple

Suppose we place an object on a smooth table and a horizontal force is applied on it, the block would move in the direction of the force. The force is causing a linear motion:

If an equal and opposite force also acts on it and is in the same line of action of the first force, the forces cancel out and the object stays at rest:

If the lines of action of the forces do not coincide, the forces cancel out and do not produce linear motion but produce rotation instead:

The pair of equal and opposite forces that produce no linear motion but rotational motion is called the couple.

The amount of moment (called the torque of the couple, or moment of the couple) provided by the couple is given by the product of one of the forces and the perpendicular distance between the forces.

Case study

Consider the case of holding a racket:

Draw the net force exerted by the hand on the racket and explain how you deduce the direction of this net force exerted by the hand.

Explain why the hand also produces a couple that exerts a torque on the racket and deduce the direction of this torque.

torque (or moment) of the couple = Fd

d

F

F

smooth horizontal table

object undergoes rotation

lines of action of the forces do not coincide


object stays at rest

lines of action of the forces coincide



Appendix 1 More types of forces

Upthrust: when object is immersed in a fluid

Static friction: friction when object is not moving

When you push an object and the object does not move, the friction exerted on the object is called static friction.

Kinetic friction: friction when object is moving

Air resistance: when object is moving against air

Lift: caused by aeroplane wings, rotating helicopter blades

Thrust: generated by jet engines, car

Reaction force at hinge

Thrust

Thrust

Thrust

kinetic friction

box moving

Kinetic friction is less than static friction, so once an

object starts moving, it becomes easier to move.

static friction

box not moving

air resistance

water resistance

when moving in water, an object experiences water resistance

air and water resistance are collectively called viscous force

Lift

Lift Lift

hinge

hinge force

object being hinged

hinge force can be in any direction


Appendix 2 Equilibrium and stability

3 forces in equlibrium

If there are 3 non-parallel forces acting on an object and the object is in equilibrium, the three forces must intercept at a common point.

What if the three forces do not intercept at a common point?

What if there are 4 or more forces?

For 4 or more forces, when a system is equilibrium, it is not necessary that the lines of actions of the forces must meet at a common point.

Stability

If an object resting on a horizontal surface is tilted, under which situation will the object topple over?

When the object is tilted, the normal reaction acts at the edge as shown in Diagram 2. But if the tilting is small as in Diagram 2, the weight and normal reaction form a couple which produces an anticlockwise moment that will seek to restore the object back to its original equilibrium position in Diagram 1.

If the tilting is large as in Diagram 3 where the centre of gravity of the block has shot past the point of contact, the weight and normal reaction now form a couple that produces a clockwise moment which will topple the object over.

To increase the stability, the centre of gravity is to be as low as possible, so that a larger tilting angle is required to topple the object.

High CG: topples Low CG: does not topple

W

W N N

hinge

X

If the three forces do not intercept at a common point, there will be a non-zero net moment, this can be seen by taking moment about X, the moment by tension and weight are zero but not for the hinge reaction force.

T R

W

hinge

T R

W

Diagram 1 Diagram 2 Diagram 3

W

N

W W

N N


STATICS Tutorial

1 A ball is initially suspended by a thread but when a strong wind blows horizontally, the thread makes an angle of 30° with the vertical.

F is the force exerted by the wind on the ball.

Identify the name of the other two forces acting on the ball:

W = ..................................................

T = ..................................................

[Solving by the Vector Diagram Method – see QED 11 for help]

Draw a vector diagram of the forces, indicating the angle 30° and the right angle in the diagram.

Given W = 10 N, use the vector diagram to find T and F.

[Solving by the Force Resolution Method – see QED 12 for help]

Write down in terms of T the horizontal and vertical components of T:

horizontal component of T = .........................

vertical component of T = .........................

In the following, draw the horizontal and vertical components of T:

Write down the two equations that balance the horizontal and vertical components of the forces:

Using W = 10 N and the two equations, find T and F.

F

W

F

30° T

W


2 A box is resting on a rough slope.

Identify the name of the forces acting on the box:

W = ..................................................

N = ..................................................

f = ..................................................

[This is Maths question] Find the angle θ.

[Solving by the Vector Diagram Method – see QED 11 for help]

Draw a vector diagram indicating the angle 20° and the right angle.

Given W = 20 N, use the vector diagram to find N and f.

[Solving by the Force Resolution Method – see QED 12 for help]

Write down in terms of W:

component of W along the slope = .........................

component of W perpendicular to slope = .........................

Given: W = 20 N

By balancing the components perpendicular to slope, find N.

By balancing the components along the slope, find f.

3 A weight is suspended by threads as shown:

If W = 30 N, find T1 and T2.

4 The diagram shows a stationary cable car supported by cable:

Given that W = 80 kN, find T1 and T2.

20°

N

f

W θ

20°

N

f

W

40° T1

T2

W

T1

T2

W

26°

32°


Steps in solving statics problems

Step 1 Draw all forces acting on the object

Step 2 EITHER draw a vector diagram OR resolve the forces

Step 3 Calculate from vector diagram OR from the components

5 A small ball of weight 50 N is suspended by a light thread. When a strong wind blows horizontally, exerting a constant force F on the ball, the thread makes an angle 30° to the vertical. Find F.

6 A picture weighing 10 N hangs freely by a cord as shown. Find T.

7 A 2.0 kg beam is held horizontal against a smooth wall by a cable.

Find the tension in the cable.

Steps in solving problems involving the Principle of Moments

Step 1 Draw all forces acting on the object

Step 2 Take moment about the point where unknown force acts

[See QED 14 for help]

8 The diagram shows a uniform metre rule pivoted at the 50 cm mark.

Where should the 3.0 kg mass be placed to keep the rule horizontal?

9 A uniform metre rule of mass 0.7 kg is pivoted at the 60 cm mark:

Find the mass m.

[See QED 15 for help]

10 A heavy uniform beam of length L is supported by two cords.

Find the ratio 2

1

TT

of the tensions in the cords.

beam

cable 55°

20 cm 100 cm

2.0 kg m

20 cm 100 cm

2.0 kg 50 cm

3.0 kg ?

T2

0.3 L 0.7 L

T1

F

30°

50° 50° T T


11 A uniform rod of weight 10 N is freely hinged to a wall. It is held horizontal by a string at an angle 20° to the rod.

Find the tension in the string.

[Hint: let the length of the rod be L; the weight acts at midpoint]

12 A uniform ladder of weight 50 N is resting on a rough floor and leaning against a smooth wall. The contact force exerted on the ladder by the wall and floor are R1 and R2 respectively.

Use the Principle of Moments to find R1.

[Hint: let the length of the ladder be L; the weight acts at midpoint]

Draw a vector diagram of the forces and use it to find R2.

13 [System of bodies] The diagram shows box A of weight 20 N resting on box B of weight 30 N, which is on the floor.

Draw the free body diagram of box A:

Draw the free body diagram of box B:

Draw the free body diagram of box A and B as a single system:

Find the force exerted by

(a) B on A, (b) A on B, (c) the force exerted by ground on B.

Answers

1. 11.5 N, 5.8 N; 2. 18.8 N, 6.8 N; 3. 46.7 N, 35.8 N; 4. 136 kN, 80 kN

5. 28.9 N; 6. 6.5 N; 7. 34.2 N; 8. 70 cm mark; 9. 2.9 kg; 10. 0.4

11. 14.6 N; 12. 2.1 N, 5.4 N

AB System

B

A

R2

50°

R1

smooth wall

rough floor

20°

hinge

A

B


ASSIGNMENT

Name: ______________________________ Class: __________

A picture is suspended from a hook on a wall by a cord. The picture is found to be too low and the cord is shortened to hang it higher.

However, the cord breaks immediately. Explain why this happens.

2 When a person bends, much stress is added to his back spine.

The spine can be considered as pivoted at its base. The various

muscles of the back are equivalent to a single muscle producing a force T as shown. W is the force that the upper part of the body exerts on the spine.

(a) Explain why, for equilibrium, the value of T is large (typically several times W).

(b) For equilibrium, a force P at the pivot is necessary. Draw a trian-gle of forces to show the equilibrium of the spine under the action of forces T, W and P. Comment on the size of P relative to W.

hook

pivot

W

T

hook 1

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