Everything Science Grade 11

CHAPTER 12. FORCE,MOMENTUM AND IMPULSE 12.3

and the normal force present in this situation.

SOLUTION

Step 1 : Draw a force diagram

Draw a force diagram and fill in all the detailson the diagram. This makes it

easier to understand theproblem.

50 kg

N Ff

Fx= 490 sin 30◦= 245 N

Fg= 50 x 9,8 = 490 N Fy= 490 cos 30◦= 224 N

30 ◦

30 ◦

Figure 12.9: Friction andthe normal forces on a slope

Step 2 : Calculate the normal force

The normal force acts perpendicular to the surface (and not vertically upwards).

It’s magnitude is equal tothe component of the weight perpendicular to theslope.

Therefore:

N = Fgcos 30 ◦

N = 490 cos 30 ◦

N = 224N perpendicular to the surface

Step 3 : Calculate the frictionalforce

The frictional force actsparallel to the sloped surface. It’s magnitude is equal to

the component of the weight parallel to the slope. Therefore:

Ff = Fgsin 30 ◦

Ff = 490 sin 30 ◦

Ff = 245N up the slope

We often think about friction in a negative waybut very often friction isuseful without us realising it.
If there was no frictionand you tried to prop aladder up against a wall, it would simply slideto the
ground. Rock climbers use friction to maintain their grip on cliffs. The brakes of cars would be useless
if it wasn’t for friction!