Using Newton's Laws effectively
Launching rockets
The lift-off of a space shuttle is an example of an unbalanced force in action. The space shuttle accelerates upwards from its launch pad. The thrust \(T\) from the rocket engines is greater than the weight \(W\) of the rocket system. Since force \(T\) is greater than the force \(W\), the effect of one force does not cancel that of the other. The forces acting are unbalanced.
The rocket launch is a very good example to look at when considering Newton's First and Second Laws.
When preparing for launch, the rocket is stationary.
If the rocket is stationary then the forces on the rocket are balanced.
Question
What forces are acting on the rocket while waiting for lift-off?
The weight of the rocket acting downwards and the reaction force of the launcher on the rocket keeping it upright and ready to go.
Remember:
\(Weight = mass \times gravitational\,field\,strength\)
\(W = m \times g\)
(where weight (\(W\)) is in Newtons (\(N\)), mass (\(m\))is in kilograms (\(kg\)) and gravitational field strength (\(g\)) is in Newtons per \(kg\) (\(Nkg^{-1}\))
When the engines are fired, an unbalanced force accelerates the rocket into the sky.
The upward force of the thrust from the rocket engines is greater than the downward weight of the rocket. This results in an unbalanced upward force, causing the rocket to accelerate upwards.
Example
Question
If the weight of the rocket is \(20,000N\) and the thrust of the engines is \(50,000N\), what is the unbalanced force accelerating the rocket?
Answer
Unbalanced force= \(50,000 - 20,000 = 30,000N\)(upward)
This unbalanced force accelerates the rocket.
This is where Newton's Second Law is applied.
Newton's Second Law states that
\(Unbalanced\,force = mass \times acceleration\)
\(F = m \times a\)
(where unbalanced force is in Newtons \(N\)), mass is in kilograms (\(kg\)) and acceleration is in metres per second per second (\(m\,s^{-2}\))
Example
Question
If the mass of the rocket above is \(2041\, kg\), what is the acceleration of the rocket at take off?
Remember, from the above question, the unbalanced force (accelerating force) is \(30,000N\).
Answer
\(F = m \times a\)
\(30000 = 2041 \times a\)
\(a \times 2041 = 30000\)
\(a = 30000 \div 2041\)
\(a = 14.7\,m\,s^{-2}\)
If the weight is known, then the mass can be calculated.
Look for the thrust (upward force) of the engines.
\(Unbalanced\,\,force=Thrust\,-\,Weight\)
Find the unbalanced force.
The acceleration can be calculated using \( \frac{f}{m}\)
