Thrust and Pressure
Thrust:
Force applied to an area in the perpendicular direction.
F (Thrust)
SI unit of thrust is N. It is a vector quantity.
Pressure:
Thrust per unit area.[ It is a scalar quantity]
SI unit:
SI unit of thrust is ‘N’
SI unit of area is ‘m2’
SI unit of pressure =Nm-2 or Pascal (Pa)
1Pa= 1N/1m2
Definition of 1Pa:
If 1N thrust is applied to an area of 1m2, then the pressure is 1Pa.
Pressure (P) =F/A
Important conversion formula: 1cm2=10-4m2[1cm=10-2m]
Give reason:
• The edge of a knife is sharp
• The snowshoes are wider
• The tyres of truck are wider in comparison with a car
Pressure exerted by a fluid (Liquid and gas at rest):
The cylinder is filled with a liquid of density ‘ρ’
Then the pressure at a depth ‘h’ on the walls of the container is
P=h. ρ. g [g = Acceleration due to gravity]
P∝h
P∝ 𝜌
The pressure exerted by a fluid (at rest) is independent of the
• Mass of the liquid
• Area or the shape of the container
• Volume of liquid.
Derivation of fluid (liquid and gas) pressure:
Volume of the liquid= h x A [h – height of the liquid; A – area of the
container]
Let the density of the liquid be ‘ρ’
Mass of the liquid (M) = Volume x density
=hxAxρ
Weight (F) = h x A x ρ x g{F=Mg} [g= Acceleration due to gravity]
Pressure (P) = F/A= (h x A x ρ x g)/A= h x ρ x g
Give reason:
The water dams are wider at the base than the top.
For deep see diving we need specially designed dress.
Numerical:
1. The thrust applied on an area is 10N. The magnitude of the area is
1cm2. Calculate the pressure exerted on the area.
Answer:
Thrust (F) = 10N
Area (A) =1cm2= (1 x 10-4) m2
Pressure (P) = F/A=10/ (1 x 10-4)
P=105Pa
2. The length, breadth and height of a brick are 10cm, 5cm and 8cm
respectively. The mass of the brick is 5kg. Calculate the maximum
pressure exerted by the brick if it is kept on sand. [g=10N/kg]
Answer:
L=10cm
B=5cm
H=8cm
Mass (M) =5kg
Thrust (F) = mg= (5 x 10) =50N
The minimum area is possible when we keep the brick by its breadth and
height
Amin=(B x H)=(40 x 10-4)m2[B x H=40cm2]
Pmax= F/Amin
Pmax=50/(40 x 10-4)
Pmax= (1.25 x 104) Pa
3. Calculate the pressure exerted by water at depth 10m below the water
surface. [g=10N/kg]
Answer:
H=10m
g=10N/kg; Density of water (ρ) =1000kg.m-3- This is SI unit
CGS unit – density of water (ρ) =1gm.cm-3
Pressure (P) = h x ρ x g = (10 x 1000 x 10) =105Pa
Note:
1gcm-3
1g= (1/1000)kg=10-3kg
1cm= (1/100)m
1cm3= (10-2)3 m3=10-6m3
1gcm-3=10-3kg/10-6m3
1gcm-3= (10-3+6)kgm-3
1gcm-3=1000 kgm-3
Buoyancy and Archimedes’s Principle
Buoyancy:
It is a property of any substance due to which a force is applied on it, if
it is kept in fluid. The force on the substance is governed by density of
liquid and volume of displaced liquid.
If an object is kept in a fluid, why force will act on the object?
Answer:
Weight
Buoyant force
The force is acting on the object due to its weight. This force is
called the buoyant force. The buoyant is equal to the weight of the
displaced fluid.
Weight of the object (F1) = mg
= V1 x ρ1 x g
V1- Volume of the object
‘ρ1’- Density of the object
Weight of the displaced fluid (F2) = V2 x ρ2 x g [V2- Volume of
displaced fluid]- buoyant force. ‘ρ2’- density of the fluid.
Condition for flotation:
V1 x ρ1 x g[Weight of the object (F1)]= Weight of the displaced
fluid (F2)]V2 x ρ2 x g
V1 x ρ1 x g= V2 x ρ2 x g F2
V1 x ρ1 = V2 x ρ2
M1=M2
M1= Mass of the object F1
M2= Mass of the displaced fluid. F2-F1=ma
The apparent weight of a floating object is ZERO.[F2-F1=0]
ρ1=ρ2 ρ1< ρ2
Normal reaction
Mg
Conditions for flotation according to the density:
ρ1> ρ2- The object will sink.
ρ1=ρ2- The object will completely submerge and float.
ρ1< ρ2- The object will partially submerge and float.
RD (relative density)
It is defined as the ratio of density of a substance to the density
of pure water at 4oC.
At 4oC the density of water is the maximum(1gcm-3)
ρ1/ ρ2- is greater than 1 – the object will sink.
ρ1/ ρ2- is less than 1, the object will float
ρ1/ ρ2- is equal to 1, the object will completely submerge and
float.
“An object will sink if the weight of the object is greater than
the buoyant force.”
V1 x ρ1 = V2 x ρ2
V2- Volume of the displaced fluid. It is the volume of the object
below the liquid surface.
Iceberg Situation:
V1 x ρ1 (Weight of the iceberg)x g = V2 x ρ2 x g
ρ1(9/10) x V1 =V2 x ρ2(=1)
V2=9/10V1
F2 F1
F1-F2=ma(F1> F2) F1-F2=0[F1= F2]
Pascal law:
The fluid (liquid +gas) must be enclosed (kept in a closed container)
The pressure applied to any enclosed fluid transmitted in all direction in
an undiminished form. The force applied on the walls of the container
are in perpendicular direction (Thrust).
This pressure applied is independent of the nature of the fluid and also
the nature of the container in which the fluid is kept.
The pressure applied on the left piston is P 1= F1/A1
The pressure applied on the right piston is F1/A1. It is because of
Pascal’s law.
The thrust applied on right piston is F2= P1 x A2
A2= Area of the 2nd piston
F2= P1 x A2
F2= (F1/A1) x A2 [because P1= F1/A1]
F2= (A2/A1) x F1
As A2>A1
Therefore F2>F1
PRACTICE QUESTIONS
Q1. What is the apparent weight of a floating body? Give reason for
your answer.
Q2. An object when kept in water, half of its volume remains below
water. If it is kept in a liquid of relative density 0.8, what portion of the
object will remain below the liquid?
Answer:
Let ‘V1’ be the volume of the object; V 2 is the volume of displaced
water. According to the question V2=V1/2
Density of water is 1gm/cc
V1 x ρ1 = V2 x ρ2 (1gm/cc)
ρ1=0.5gm/cc
Density of the new liquid ρ3=0.8gm/cc
Relative density= density of a substance/ density of water
Density of a substance= Relative density x density of water
Let , ‘V3’ is the volume of the displaced liquid
V1 x ρ1 = V3 x ρ3
V3 x 0.8= V1 x 0.5
V3=5/8 V1
Q3. The density of ice is 9/10th of water. What portion of ice will remain
below water?
Answer:
V1 x ρ1 = V2 x ρ2
Q4. The mass of an object is 100gm of volume 20cm3. Will it sink or
float in water. What is the maximum buoyant force that will act on the
object if it is placed in water? What is the maximum apparent weight
loss of the object?[g=10N/kg]
Answer:
Mass (m) =100gm
Volume (V) =20cm3
Density (ρ1) =100/20=5gm.cm-3
Density of H20=1gmcm-3
Relative density=5
The maximum buoyant force that will act on the object when it is
completely below water
The buoyant force= [(20x1)/1000] x 10=0.2N
Weight of the object= [100/1000] x 10=1N
Q5. The length, breadth and height of a brick are 10cm, 5cm and 8cm
respectively. The mass of the brick is 5kg. Calculate the maximum
pressure exerted by the brick if it is kept on sand. What is the ration of
the maximum to minimum pressure exerted by the body on the sand?
[g=10N/kg]
