Everything Science Grade 11

12.3 CHAPTER 12. FORCE,MOMENTUM AND IMPULSE

As the lift is now slowing down there is a resultant force downwards. This means that the force acting
downwards is greater than the force acting upwards. To find the magnitude of the force applied by
the cable (Fc) we can do the following calculation: Again wehave chosen upwards aspositive, which
means that the acceleration will be a negative number.

FR = ma

Fc+ Fg = ma

Fc+ (−4900) = (500)(−2)

Fc = 3900 N upwards

This makes sense as weneed a smaller force upwards to ensure that theresultant force is downward.
The force of gravity is now greater than the upward pull of the cable andthe lift will slow down.

Stage 5:
The cable snaps.
When the cable snaps,the force that used to beacting upwards is no longer present. The onlyforce
that is present would bethe force of gravity. Thelift will freefall and its acceleration can be calculated
as follows:

FR = ma

Fc+ Fg = ma

0 + (−4900) = (500)(a)

a =− 9 ,8m· s−^2

a = 9,8m· s−^2 downwards

Rockets

As with lifts, rockets are also examples of objects in vertical motion.The force of gravity pulls the
rocket down while the thrust of the engine pushes the rocket upwards. The force that the engine exerts
must overcome the force of gravity so that therocket can accelerate upwards. The worked example
below looks at the application of Newton’s Second Law in launching a rocket.

Example 11: Newton II - rocket

QUESTION

A rocket is launched vertically upwards into thesky at an acceleration of20 m·s−^2. If the mass

of the rocket is 5000 kg, calculate the magnitude and direction of thethrust of the rocket’s

engines.

SOLUTION

Step 1 : Analyse what is given and what is asked

We have the following:

m = 5000 kg

a = 20 m·s−^2

Fg= 5000 x 9,8 = 49000N

We are asked to find thethrust of the rocket engine F 1.