11.3 Lifts and rockets  (Page 78/2)

Lifts and rockets

So far we have looked at objects being pulled or pushed across a surface, in other words motion parallel to the surface the object rests on. Here we only considered forces parallel to the surface, but we can also lift objects up or let them fall. This is vertical motion where only vertical forces are being considered.

Let us consider a 500 kg lift, with no passengers, hanging on a cable. The purpose of the cable is to pull the lift upwards so that it can reach the next floor or to let go a little so that it can move downwards to the floor below. We will look at five possible stages during the motion of the lift.

Stage 1:

The 500 kg lift is stationary at the second floor of a tall building.

Because the lift is stationary (not moving) there is no resultant force acting on the lift. This means that the upward forces must be balanced by the downward forces. The only force acting down is the force of gravity which is equal to (500 x 9,8 = 4900 N) in this case. The cable must therefore pull upwards with a force of 4900 N to keep the lift stationary at this point.

Stage 2:

The lift moves upwards at an acceleration of 1 m $·$ s ${}^{-2}$ .

If the lift is accelerating, it means that there is a resultant force in the direction of the motion. This means that the force acting upwards is now greater than the force of gravity F ${}_g$ (down). To find the magnitude of the force applied by the cable (F ${}_c$ ) we can do the following calculation: (Remember to choose a direction as positive. We have chosen upwards as positive.)

The answer makes sense as we need a bigger force upwards to cancel the effect of gravity as well as make the lift go faster.

Stage 3:

The lift moves at a constant velocity.

When the lift moves at a constant velocity, it means that all the forces are balanced and that there is no resultant force. The acceleration is zero, therefore F ${}_R$ = 0. The force acting upwards is equal to the force acting downwards, therefore F ${}_c$ = 4900 N.

Stage 4:

The lift slows down at a rate of 2m $·$ s ${}^{-2}$ .

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 (F ${}_c$ ) we can do the following calculation: Again we have chosen upwards as positive, which means that the acceleration will be a negative number.

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

Stage 5:

The cable snaps.

When the cable snaps, the force that used to be acting upwards is no longer present. The only force that is present would be the force of gravity. The lift will freefall and its acceleration can be calculated as follows:

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 the rocket can accelerate upwards. The worked example below looks at the application of Newton's Second Law in launching a rocket.