Everything Science Grade 11

CHAPTER 11. VECTORS 11.8

Notice how the two components acting togethergive the original vectoras

their resultant.

Step 3 : Determine the lengthsof the component vectors

Now we can use trigonometry to calculate the magnitudes of the components of

the original displacement:

xN = (250)(sin30◦)

= 125 km

and

xE = (250)(cos30◦)

= 216,5 km

Remember xNand xEare the magnitudes of the components – they are in

the directions north andeast respectively.

Extension: Block on an incline

As a further example ofcomponents let us consider a block of mass m placed on a frictionless

surface inclined at some angle θ to the horizontal. Theblock will obviously slide down the

incline, but what causesthis motion?

The forces acting on theblock are its weight mg and the normal force N exerted by the

surface on the object. These two forces are shown in the diagram below.

θ

θ

mg

N

Fg�

Fg⊥

Now the object’s weight can be resolved into components parallel andperpendicular to

the inclined surface. These components are shown as dashed arrowsin the diagram above

and are at right angles toeach other. The components have been drawn acting from the same

point. Applying the parallelogram method, thetwo components of theblock’s weight sum to

the weight vector.

To find the componentsin terms of the weight we can use trigonometry:

Fg� = mg sin θ

Fg⊥ = mg cos θ

The component of the weight perpendicular to the slope Fg⊥exactly balances the normal force

N exerted by the surface. The parallel component, however, Fg�is unbalanced and causes

the block to slide downthe slope.