Addition of VectorsExample

Q. A pilot has selected a course of 100km/h, due West. Ifthere’s a wind blowing with a velocity of 40 km/h, due South,what is the resultant velocity of the plane?

A. Scale Drawing Method

Add vectors “nose” to “tail”.

Choose a scale.

NOTES p.4

scale : 1cm 10 km/h North

Startselected course (100 km/h)nose

tail

Resultant velocity : 10.9 cm (measured with a ruler) = 109 km/h

Direction is 248 o from North. (measure 22 o

with a protractor, then subtract from 270 o (due W)

22 o

+ wind(40 km/h)

SO…velocity = 109 km/h at a bearing of 248 o

Pythagoras/Trig. Method• Diagram does not need to be to scale BUT

must be a vector diagram!100 km/h

resultant velocity magnitude:

x2 = 1002+402

= 11600

x =108 km/h

40 km/h

direction : tan = opp = 40 adj 100 = 22o (Direction is 248 o from North)

x

SO…velocity = 108 km/h at a bearing of 248 o

Answer problems 16 – 22 in problems jotter (blue). These are the traditional Higher problems.Yellow Booklet

5.a) 80km b) 40 kmh-1 c) 20 km, N d) 10 kmh-

1, N

6.a) 70 m b) 50m at 0370 c) 70 s d) 0.71 ms-1 at 0370

Traditional Problems16.a) 6.8N at 160 from 3N (or 140 from 4N)

b) 11.3N at 450 above horizontal (or 450 from vertical) c) 6.4N at 390 below horizontal (or 510 from vertical)

17.900 kmh-1, N 18. 26 ms-1 at 0230.19. 804 kmh-1 at 3540. 20. --- (see notes) ---21. a) 11 kmh-1 b) 5 kmh-1 at 2330.22. 4.5 ms-1 at 0630.

Do “Tutorial 1 – Vectors in Physics” – see HW booklet.

Homework for Friday 24 August 2012

Physics: 1995 Q.1

1a. 1cm = 1N

20o

6cm = 6N

4cm = 4N

9.9cm = 9.9N

FR = 9.9N

m = 2kg

a = ?

a = FR / m

= 9.9 / 2

= 4.95 m/s2

Tutorial 1

Try Higher Physics 2014 Q. 21 and 22 in teams. Q.22 was a wee stinker!

Q.21a)(i) 0.172 m (ii) ± 0.007 m

b)s = ½ a t2 t2 = 2s / a

t = 0.187 s

Q.22 a) (i) The resultant of a number of

forces is the single equivalent force that produces the same effect as the original forces.

(ii)

20o

x

110o

900 N

1200 N

X = 1730N= 30o as shown

So angle from horizontal is 30o + 20o = 50o.20o

Q.22 b)The only forces remaining on the paraglider are the weight and the force of the parasail.As the vertical component of the force of the parasail is greater than the 900N weight, then there is an unbalanced force upwards so the paraglider will rise higher.