Addition of Vectors Example Q. A pilot has selected a course of 100km/h, due West. If there’s a wind blowing with a velocity of 40 km/h, due South, what.

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Addition of Vectors Example Q. A pilot has selected a course of 100km/h, due West. If there’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 Start selected 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: x 2 = = x =108 km/h  40 km/h direction : tan  = opp = 40 adj 100  = 22 o ( 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 c) 70 s d) 0.71 ms -1 at Traditional Problems 16.a) 6.8N at 16 0 from 3N (or 14 0 from 4N) b) 11.3N at 45 0 above horizontal (or 45 0 from vertical) c) 6.4N at 39 0 below horizontal (or 51 0 from vertical) kmh -1, N ms -1 at kmh -1 at (see notes) a) 11 kmh -1 b) 5 kmh -1 at ms -1 at

Do “Tutorial 1 – Vectors in Physics” – see HW booklet. Homework for Friday 24 August 2012 Physics: 1995 Q.1

1a. 1cm = 1N 20 o 6cm = 6N 4cm = 4N 9.9cm = 9.9N F R = 9.9N m = 2kg a = ? a = F R / m = 9.9 / 2 = 4.95 m/s 2 Tutorial 1

Try Higher Physics 2014 Q. 21 and 22 in teams. Q.22 was a wee stinker! Q.21 a)(i)0.172 m (ii) ± m b)s = ½ a t 2 t 2 = 2s / a t = 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) 20 o x  110 o 900 N 1200 N X = 1730N  = 30 o as shown So angle from horizontal is 30 o + 20 o = 50 o. 20 o

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.