Question

Here is my question :-
Q1. Describe thrust. How can it be calculated ? Give examples.

Answer 1

HEY DEAR HERE IS YOUR ANSWER ✌✌❤❤

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➡ Thrust means to push something forcibly which is normally acted upon the surface .
Its S.I. Unit is :- newton ( N )

IT CAN BE CALCULATED LIKE THIS :-

T = π/4 D²pv∆v

HERE ,

T = thrust [ N ]
D = propeller diameter [ m ]
v = velocity of air at the propeller [ m/s ]
∆v = velocity of air accelerated by propeller [ m/s ]
p = density of air . ( 1.225 kg/m3 )

EXAMPLE :-

A boat with a motor makes thrust or reverse thrust when the propellers are turned to push water backwards (or forwards). The thrust that this makes pushes the boat in the opposite direction than the water is being pushed.

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HOPE IT HELPS U.✌✌
@ RiyaSharma01 .❤✌
THANKS.❤❤

Answer 2

Thrust has mainly two different (but still somewhat similar) meanings.

We will see the more relevant one first

1) Thrust is simply the force acting perpendicularly on a surface.

This means that if some force is acting on an object, then the component of force perpendicular to the surface is called Thrust.

Let us see an example. Suppose a hammer hits a nail perpendiculary with a force of 50 N. Here, all of the 50 N force is perpendicular to the surface of the nail. So, the Thrust is 50 N.

For those who know Trigonometry and some basic ideas of Vector Components, here's a slightly more detailed explanation. Others can skip to Point 2.

Suppose a Force F is acting on a surface at an angle $\theta$ with the normal. [A Figure is Attached]

Then The component of Force perpendicular to the surface is Thrust. So, we have:

$Thrust = T = F \cos \theta$

For an example, suppose a force 100 N is acting on a surface at an angle of $60^{\circ}$ with the normal, then the thrust will become:

$T=F\cos\theta \\ \\ \implies T=100 \cos 60^{\circ} \\ \\ \implies T=50 \, \, N$

2) When a system throws some of its mass in one direction, then a Newton's Third Law Reaction Force acts in the opposite direction. This force is called Thrust.

For example, consider a rocket. A rocket ejects gases at high speeds from the nozzle in the downward direction. As a result, a force acts on the rocket in the upward direction. This force is nothing but thrust.

Now, some things are worth noting. A body accelerates when an external force acts on it. But for a rocket, there's no external force that can push it upwards. So, it uses the Newton's Third Law Mechanism. It ejects some of its own mass (in form of gases) in downward direction, so it can propel in the upward direction.

This force is the Thrust Force.

A diagram is attached.

For those who know some Basic Concepts of Differentiation and Vectors, here are some more details.

Newton's Second Law states that the time rate of change of momentum is force. So, we write:

$\displaystyle \vec{F} = \frac{d\vec{p}}{dt}$

We can write momentum as $p=mv$. And so, by Product Rule of Differentiation, we get the following result:

$\displaystyle \vec{F} = \frac{d\vec{p}}{dt} \\ \\ \\ \implies \vec{F} = \frac{d}{dt} \, (m\vec{v}) \\ \\ \\ \implies \vec{F} = m\,\frac{d\vec{v}}{dt}+\vec{v}\,\frac{dm}{dt} \\ \\ \\ \implies \vec{F} = m\vec{a}+\vec{v}\,\frac{dm}{dt}$

Here, $\vec{a}$ is the acceleration vector, and $m\vec{a}$ is the normal expression that we have been studying since lower standards. However, the interesting thing to note is the second term:

$\vec{v} \, \frac{dm}{dt}$

Here, $\frac{dm}{dt}$ is the rate of change of mass of the system, and $\vec{v}$ is the velocity by which the mass is added or removed.

This term is what we call thrust. So, thrust is:

$\boxed{\vec{T}=\vec{v} \, \frac{dm}{dt}}$

This represents the simple statement we put on earlier.

When a system throws out some mass with some speed in one direction, it gets a reaction force in the opposite direction, which is nothing but Thrust.

We see that Thrust is then just a Force.

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In whatever interpretation we take Thrust as, the fact remains common that Thrust is a Force.

And, so its unit is newton (N).