3. Forces and Pressure

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Mastery Practice 3.1 pg 81

1. (a) minimum pressure = Force

Area maximum

=2 2 2

64

(20 10 )(10 10 )

N

x x m

= 3200 N m-2

(b) Maximum pressure = ForceArea minimum

= 2 2 264

(10 10 )(8 10 )

N

x x m

= 8000 N m-2

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2. Pressure = Force

Area

=2

36000

0.08

N

m2 2 2

36000

2(400)(10 )

N

m

=2

36000

0.08

N

m

= 450 000 Pa

The platform can withstand the pressure because it is

less than 700 000 Pa.

Challenge Yourself pg 81

1. Pave the soft ground with wooden planks of large

surface area.

This will increase the area of contact.

When the area of contact increases, pressure

decreases.

Hence pressure on the ground is reduced while the


1 (a) The tea level is high above the tap. The tea exert

a high pressure at the tap.

(b) When there is very little tea, the pressure exerted

on the tap by the tea is very small. If the

container is tilted, the tea level above the tap

increases and hence the liquid pressure

increases.

2 (a) Pressure = h g

= 8000 x 1025 x 10

= 8.2 x 107 Pa

(b) Force = Pressure x area

= 8.2 x 107 x 0.16

= 1.312 x 107 N

Challenge Yourself pg 85

1.

(i) Water flows in from the top.

(ii) The water column above the sprinkler exerts

pressure to the sprinkles.

(iii) The pressure forces the water sprinkling

out .

(iv) The higher the water column, the bigger the

pressure.

3.1 UNDERSTANDING PRESSURE

3.1

3.1

3.2

3.2

3.2 UNDERSTANDING PRESSURE

IN LIQUID

Water from tap

sprinkler

3.2


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Oil will flow back to the oil reservoir and

platform will be lowered down.


1. (a) Buoyant force = 15 10

= 5 N

(b) Weight of water displaced = buoyant force

= 5 N

(c) Volume of water displaced = m = 0.5

1000= 0.0005 m3

Hence, volume of the block = 0.0005 m3

Density of the block = 1.5

Objective Questions pg 104

3.5 APPLYING

ARCHIMEDES PRINCIPLE

3.5 Chap

3

Chapter 3 PRACTISE YOUR

SKILL


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0.0005

= 3000 kg m3

2. Weight of wood = weight of water displaced

100 = (0.8 V)(1000)(10)

Volume of wood, V = 0.125 m3

Weight of wood + weight of stone = buoyant force100 + W = 0.125 (1000)(10)

= 125

W = 25 N


1. Velocity of air flowing through A is faster than that

through B. Hence pressure of air at A is lower than

that at B.

2. (a)

(b) The ball will curve off to one side.

(c)

Structured Questions pg 105

1(a) Because the bicycle in Figure 3.49(b) sinks lessinto the sand.

(b) (i) Comparison of the features :

Tyres of bicycle in Figure 3.49(a) are narrower

than that of Figure 3.49(b).

Wheels of bicycle in Figure 3.49(a) has a

bigger diameter than that of Figure 3.49(b)

Two physical quantities :

Area of contact and the pressure

(ii) When the area of contact increases, pressure

decreases

3(a)(i) The pointer moves upwards

(ii) Upper region of the aerofoil is flatter, hence

air flows slower and has higher pressure than

the lower region.

Aerofoil moves downward and the pointer

moves upward.

This follows Bernoullis Principle

(b) The pointer shows a bigger movement

upwards.

(c) To stabilize the racing car at high speed.

(d)

3.6 UNDERSTANDING

BERNOULLIS PRINCIPLE

3.6

Chap

3


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(c) Let off some air from the tyres . Flatten tyres have

bigger areas of contact .

2. (a) When air is pumped out, a partial vacuum is

created in the space between the cup and the

glass panel.

The surrounding atmospheric pressure forces

the cups tightly against the glass panel.

(b) To ensure the contact is airtight.

(c) Air rushes into the cup. Hence air pressure

inside and outside the cup are equal.

(d) The same method cannot be used because the

air in the moon is very thin. The atmosphericpressure is not high enough to press the cup

against the glass panel.

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When moving through the air, the air flows

faster over the top and creates a region of low

pressure. The higher pressure below the wing

causes a loft which helps the plane to float in

air.

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3. Forces and Pressure