This chapter discusses pressure and its relationship to force and area. Pressure is defined as force per unit area, with the formula P=F/A. Several examples are provided to demonstrate calculating pressure given the force and contact area. Quizzes throughout assess understanding of key concepts like how elephants and ladies exert different pressures due to differences in their contact areas.

1. Chapter 7
Pressure

2. At the end of this chapter,
you should be able to
 Define pressure in terms of force
and area and recall the relationship:
Pressure = Force/Area
 Apply the relationship between
pressure, force and area to new
situations or to solve related
problems.

3. Quiz Time
 Why do camels have large flat feet?
 Why do eskimos wear snow shoes?

4. 6.1 Pressure
 Pressure is defined as force acting
per unit area.
 Formula P =
F
A
where P = Pressure
F = Force (N)
A = Area where the force
is acting on (m2)

5. 6.1 Pressure
 The S.I. unit of pressure is
Newton per square metre (N m-2)
Pascal (Pa).
or

6. Quiz Time
 An elephant weighs 40 000 N
while a lady wearing high heels
weighs 400 N. Who will exert a
greater pressure on the ground if
both were to stand on one leg?
Given:
Area of a elephant foot = 0.1 m2
Area of a heel = 0.0001 m2

7. Quiz Time
Pressure exerted by elephant
= Force / Area
= 40 000 / 0.1
= 400 000 Pa
Pressure exerted by lady
= Force / Area
= 400 / 0.0001
= 4 000 000 Pa

8. Quiz Time
Pressure exerted by elephant
= 400 000 Pa
So who hurts more
when they step on you?
Pressure exerted by lady
The elephant’s foot
= 4 000 000 Pa
or the lady’s heel?
The lady’s heel exerts a higher
pressure than the elephant foot.

9. Quiz
 When a balloon is pressed against a
single nail, it will burst. However, when
the balloon is pressed against a bed of
nails with the same amount of force, it
will NOT. WHY??

10. Quiz
 Applying the same amount of force, a
single nail having a smaller area of
contact will exert a higher pressure to the
balloon as compared to the pressure
exerted by a bed of nails.
 On the bed of nails, the force is
distributed over all the nails, hence the
area of contact is larger which result in a
lower pressure acting on the balloon.
Hence the balloon does not pop easily on
it.

11. 6.1 Pressure
 In conclusion,
To increase pressure, we can either
 Increase the force applied.
 Decrease the area of contact.

12. Example 1
 The atmospheric pressure is 100 kPa.
What is the force exerted by the
atmosphere on a rectangular surface
that measures 0.5 m by 0.4 m?
Pressure P = F
A
F=P×A
= (100×103) ×(0.5×0.4)
= 20 000 N

13. Example 2
 The tyres of a car are in contact
with the ground over a total area of
3.0 × 10-2 m2. The total weight of the
car is 6300 N. Calculate the
pressure exerted by the tyres on the
ground. = F
Pressure
A
= 6300
3.0×10-2
= 2.1 × 105 Pa

14. Example 3
 A standing man has a total contact
area between his feet and the
ground of 400 cm2. He is 60 kg.
Calculate the pressure exerted on
the ground by the man.
Weight of the man W = mg
= 60 × 10
= 600 N
Area A = 400 cm2
= 400 × 10-4
= 4 × 10-2 m2

15. Example 3
Pressure = W
A
= __600___
4 × 10-2
= 15 000 Pa