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Addition and resolution of forcesPowerPoint Presentation

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.

Addition and resolution of forces

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

A

B

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

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.

F2

F1

F1

+

F2

F2

F1

aGraphical method

'Tip-to-tail' method

resultant force

F2

F1

F1

+

F2

aGraphical method

Parallelogram of forces method

parallel

parallel

resultant force

aGraphical method

Video

Simulation

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

Addition of forces

Video

F1

F1

=

=

F1

F1

+

+

F2

F2

F2

F2

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

cAlgebraic method (2-demensions)

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

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

3 N

3

4

4 N

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.

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

3 N

60°

7 N

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

2

2

+

120

120

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

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

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

Fy

Fx

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

Fy

Fx

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

bAlgebraic method

Simulation

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.

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

mass = 1 kg

20

= 20

W

W

Component of weight along runway:

Weight (W ) = mg

= 10 N

Component of weight along runway

= 10 sin 20

= 3.42 N

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)

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

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

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

=

+

R

5252

76.62

N

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 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

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?

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

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

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.

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

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

F1

F1

F1

F2

F2

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

F1

F2

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