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Addition and resolution of forces. 1 Which of the following schoolbags would you feel heavier to carry if the same number of books are put in them?. A . B. 2 A light rope is stretched tightly between two poles. . A T-shirt of weight 8 N is hung at the midpoint of the rope.

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Addition and resolution of forces

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Addition and resolution of forces l.jpg

Addition and resolution of forces


Slide2 l.jpg

1Which of the following schoolbagswould you feel heavier to carry if thesame number of books are put inthem?

A

B


Slide3 l.jpg

2A light rope is stretched tightlybetween two poles.

A T-shirt of weight8 N is hung at the midpoint of therope.

What is the tension T in the string?

AT < 4 N

BT = 4 N

CT > 4 N

8 N


Adding forces l.jpg

Adding forces

aGraphical method

Force is a vector quantity.

It has both magnitude and direction.

Like displacements, forces can also be added graphically using the 'tip-to-tail' method.

Lines representing forces should be drawn to scale.


Adding forces5 l.jpg

F2

F1

F1

+

F2

F2

F1

Adding forces

aGraphical method

'Tip-to-tail' method

resultant force


Adding forces6 l.jpg

F2

F1

F1

+

F2

Adding forces

aGraphical method

Parallelogram of forces method

parallel

parallel

resultant force


Adding forces7 l.jpg

Adding forces

aGraphical method

Video

Simulation


Slide8 l.jpg

F1

F2

Addition of forces

Stretch a rubber band with a spring balance.

F

B

O

A

Stretch the rubber band by thesame amount using 2 springbalances.

Note the magnitude & direction of the F1 & F2.

Check if resultant = F(in magnitude & direction) by

–using 'tip-to-tail' method

–using parallelogram of forces method


Slide9 l.jpg

Addition of forces

Video


1 adding forces l.jpg

F1

F1

=

=

F1

F1

+

+

F2

F2

F2

F2

1Adding forces

bAlgebraic method (1-dimension)

The magnitude of the resultant force isthe algebraic sum of the forces.

Adding forces in the samedirection & along the same line.

Adding forces in the oppositedirection & along the same line.

F

F

direction is the same as the forces

direction is the same as the larger force


1 adding forces11 l.jpg

1Adding forces

cAlgebraic method (2-demensions)

Forces in 2 dimensions can also be added algebraically. See this example:


Example 8 l.jpg

Example 8

C

Use 1 cm to represent 25 000 N

Length of AC = 6.9 cm

resultant

Resultant

D

B

= 6.9  25 000 N

= 173 000 N

100 000 N

100 000 N

(with an  of 30 to either force)

60

A

oil rig


Example 9 l.jpg

3 N

3

4

4 N

Example 9

2 forces, 3 N and 4 N, act at right angles to each other. Find the magnitude & direction of the resultant force.

R

= 5 N

tan  =

5 N

 =36.9

= 36.9

Resultant force is 5 N with 36.9 to the 4-N force.


Q1 if janice is pushing with l.jpg

Q1If Janice is pushing with...

If Janice is pushing with a force of 60 N &

Tommy ispushing with 80 N,

what are the maximum and the minimum magnitudes of their combined force?

Maximum force Minimum force

A 100 N 80 N

B120 N60 N

C 140 N20 N

D 160 N 40 N


Q2 find the resultant of the l.jpg

3 N

60°

7 N

Q2Find the resultant of the...

Find the resultant of the forces below with graphical method.

(1 cm represents 1.5 N)

The resultant force is ________ Nmaking an angle__________with the 3-N force.

8.9

43


Q3 2 ropes are used to pull l.jpg

2

2

+

120

120

Q32 ropes are used to pull...

2 ropes are used to pull a tree as shown.

Find themagnitude of the resultant force acting on the tree.

120 N

Magnitude of resultant

= ____________ (Pythagoras’ theorem)

= ________ N

170

120 N


2 resolving forces into components l.jpg

2Resolving forces into components

We 've learnt how to combine 2 forces into 1.

Let's see how to spilt (resolve) 1 force into 2.


Resolving forces into components l.jpg

Fy

Fx

Resolving forces into components

aGraphical method

Suppose a force F is represented by OC :

y

component

C

Magnitudes of Fx & Fy can be measured directly.

F

component

x

O


Resolving forces into components19 l.jpg

Fy

Fx

Resolving forces into components

bAlgebraic method

Magnitudes of Fx & Fy can be also be found by

algebraic method:

y

C

F 2= Fx2 + Fy2

(Pythagoras’ theorem)

Fx = F cos 

F

Fy = F sin 

Fy

tan =

Fx

O

x


Resolving forces into components20 l.jpg

Resolving forces into components

bAlgebraic method

Simulation


Example 821 l.jpg

Example 8

2 tug boats pull an oil rig, each exerting a force of 100 000 N, they are at 60 to each other.

Find the resultant force.


Example 10 l.jpg

Example 10

A 1-kg trolley runs down a friction-compensated runway at a constant speed.

Find the component of weight of the trolley along runway,

Hence find friction between trolley & runway.

mass = 1 kg

 = 20


Example 1023 l.jpg

mass = 1 kg

20

 = 20

W

W

Example 10

Component of weight along runway:

Weight (W ) = mg

= 10 N

Component of weight along runway

= 10 sin 20

= 3.42 N


Example 1024 l.jpg

Example 10

Friction between trolley & runway:

constant v

friction

component of weight along runway = 3.42 N

 = 20

constant speed

 net force on trolley = 0

friction = 3.42 N (up the runway)


Example 11 l.jpg

Example 11

180 sin 12 N

180 N

12

180 cos 12 N

Resolve tension along each rope into 2 components

 along line of travel

 normal to line of travel

Components in each direction are added


Example 1126 l.jpg

Example 11

Total force along line of travel:

180 cos 12

+ 170 cos 10

+ 200 cos (10+15)

= 525 N

A

180 N

12

10

B

15

170 N

200 N

C


Example 1127 l.jpg

Example 11

Total force normal to line of travel:

180 sin 12

 170 cos 10

 200 cos (10+15)

= 76.6N

+ve

A

180 N

12

10

B

15

170 N

200 N

C


Example 1128 l.jpg

=

+

R

5252

76.62

N

Example 11

525 N

R

76.6 N

76.6

tan  =

525

= 531N

 =8.30

Resultant force exerted by the skiers is 531 N at an angle of 8.30°to the line of travel.


Forces in equilibrium l.jpg

Forces in equilibrium

Forces acting on 1-kg mass:

weight

2 forces from 2 springs

But it is at rest,

1-kg mass

 net force = 0

Implication:

in equilibrium

Resultant of any 2 forces

equal & opposite to the 3rd force


Q1 a 1 kg block on rest on a l.jpg

Q1A 1-kg block on rest on a...

A 1-kg block rests on a wedge.

Use graphicalmethod to find the weight components of the block along & normal to the wedge.

O

C

B

30

A

What is the frictionacting on the block?


Q1 a 1 kg block on rest on a31 l.jpg

Q1A 1-kg block on rest on a...

5

Use a scale of 1 cm to represent _____ N.

at rest

O

Length of OB = _____ cm

1.75

C

Weight componentalong the wedge = ______ N

8.75

B

Length of OC = _____ cm

1

30

A

Weight componentnormal to the wedge = ______ N

5

Since the block is at rest,

friction = weight component ____________ (along/normal to) the wedge = ________ N.

along

5


Q2 find the weight components l.jpg

Q2Find the weight components...

Find the weight components of the block by algebraic method.

at rest

O

C

Weight componentalong the wedge

= 10×_______

= ______ N

B

30

sin 30

A

5

Weight componentnormal to the wedge

= 10×_______

= _______ N

cos 30

8.66


Q3 what happens to the trolleys l.jpg

Q3What happens to the trolleys...

What happens to the trolleys when they are released ?

smooth pulley

3 kg

2 kg

20

30

AThe 3-kg trolley moves downwards.

BBoth trolleys remain at rest.

CThe 2-kg trolley moves downwards.


Example 1134 l.jpg

Example 11

Find the magnitude & direction of the resultant force the skiers exert on the speedboat?

180 N

A

12

10

170 N

15

B

200 N

C


Example 12 l.jpg

Example 12

2 spring balances are used to support an 1-kg mass.

What are the readings of the 2 balances?

F1

F2

30

1-kg mass


Example 1236 l.jpg

F1

F1

F1

F2

F2

Example 12

30

30

30

10 N

The mass is in equilibrium.

 resultant force of F2 & weight is equal & opposite to F1

10

10

 F1 =

= 20 N

= sin 30

sin 30

F1

10

10

 F2 =

= 17.3 N

= tan 30

tan 30

F2


Example 1237 l.jpg

F1

F2

Example 12

Alternative method by resolution of forces

30

The mass is in equilibrium.

 net force = 0

10 N

horizontal resultant = 0

vertical resultant = 0

horizontal: F1 cos 30 = F2

 F1= 20 N;

vertical: F1 sin 30 = 10

F2= 17.3 N


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