Everything Science Grade 11

CHAPTER 12. FORCE,MOMENTUM AND IMPULSE 12.3

involve using the valuefor P (for in case you made a mistake in calculating P).

tan 55◦ =

R

T

tan 55◦ =

R

312

R = tan 55◦× 312

R = 445,6 N

R = 446 N

Note that the question asks that the answers begiven to 3 significant figures. We

therefore round 445,6 Nup to 446 N.

Lifts and rockets

So far we have looked at objects being pulled or pushed across a surface, in other words motion par-
allel to the surface the object rests on. Here weonly considered forces parallel to the surface, but we
can also lift objects upor let them fall. This isvertical 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 down-
wards to the floor below. We will look at five possible stages during themotion 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 isno resultant force actingon the lift. This means that
the upward forces mustbe balanced by the downward forces. The onlyforce acting down is theforce
of gravity which is equal to (500 x 9,8 = 4900 N) in this case. The cablemust therefore pull upwards
with a force of 4900 N to keep the lift stationaryat this point.

Stage 2:
The lift moves upwardsat an acceleration of 1 m·s−^2.
If the lift is accelerating, it means that there isa resultant force in thedirection of the motion. This
means that the force acting upwards is now greater than the force of gravity Fg(down). To find the
magnitude of the forceapplied by the cable (Fc) we can do the following calculation: (Remember to
choose a direction as positive. We have chosenupwards as positive.)

FR = ma

Fc+ Fg = ma

Fc+ (−4900) = (500)(1)

Fc = 5400 N upwards

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

Stage 3:
The lift moves at a constant velocity.
When the lift moves ata constant velocity, it means that all the forces are balanced and that there is
no resultant force. The acceleration is zero, therefore FR= 0. The force acting upwards is equal to the
force acting downwards, therefore Fc= 4900 N.

Stage 4:
The lift slows down at arate of 2m·s−^2.