Addition and resolution of forces
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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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Slide2 l.jpg

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

A

B


Slide3 l.jpg

2 A 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

a Graphical 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

a Graphical method

'Tip-to-tail' method

resultant force


Adding forces6 l.jpg

F2

F1

F1

+

F2

Adding forces

a Graphical method

Parallelogram of forces method

parallel

parallel

resultant force


Adding forces7 l.jpg
Adding forces

a Graphical 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



1 adding forces l.jpg

F1

F1

=

=

F1

F1

+

+

F2

F2

F2

F2

1 Adding forces

b Algebraic 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
1 Adding forces

c Algebraic 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
Q1 If 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

Q2 Find 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

Q3 2 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
2 Resolving 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

a Graphical 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

b Algebraic 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

b Algebraic 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
Q1 A 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
Q1 A 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
Q2 Find 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
Q3 What happens to the trolleys...

What happens to the trolleys when they are released ?

smooth pulley

3 kg

2 kg

20

30

A The 3-kg trolley moves downwards.

B Both trolleys remain at rest.

C The 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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