Chapter 3, Sections 1 Chapter 3, Sections 1 & 2& 2

Pg. 94 - 97Pg. 94 - 97

Inclined & Non-Perpendicular

Vectors

Vectors on an InclineVectors on an InclineRose is sliding down an ice covered hill inclined at an angle of 15° with the horizontal. If Rose and the sled have a combined mass of 54.0 kg, what is the force pulling them down the hill? (neglect friction)

mg

Fn

Fapp

θ = 15°

mgFn

Fapp

15°

m = 54 kgg = 9.81 m/s²

θ= 15°

Sin θ = Fx/mgFapp = mg sin θFapp = (54 kg)(9.81 m/s²) sin 15°

Fapp = 137 N

Non-Perpendicular Non-Perpendicular VectorsVectors

When addressing situations When addressing situations where vectors that are not where vectors that are not perpendicular to each other, you perpendicular to each other, you can break up the problem by can break up the problem by individual vectors and solve it individual vectors and solve it piecewise (algebraically).piecewise (algebraically).

Sample Problem Sample Problem (pg. 95)(pg. 95) A hiker walks 25.5 km from her A hiker walks 25.5 km from her

base camp at 35° south of east. base camp at 35° south of east. On the second day she walks 41.0 On the second day she walks 41.0 km in a direction 65° north of km in a direction 65° north of east, at which point she east, at which point she discovers a forest ranger’s tower. discovers a forest ranger’s tower. Determine the magnitude and Determine the magnitude and direction of her resultant direction of her resultant displacement between the base displacement between the base camp and the ranger’s tower.camp and the ranger’s tower.

θ1 = 35°

θ2 = 65°

???

Ok………lets break up the Ok………lets break up the problem!!problem!!

25.5 km41.0

km

θ1 = 35°25.5 km

Sin 35 = y / Sin 35 = y / 25.525.5

y

x

25.5 Sin 35 25.5 Sin 35 = y= y14.6 km 14.6 km

= y= y

Cos 35 = x / Cos 35 = x / 25.525.525.5 Cos 35 25.5 Cos 35 = x= x20.9 km = x20.9 km = x

θ2 = 65°

41.0

km

y

x

Sin 65 = y / Sin 65 = y / 41.041.041.0 Sin 65 41.0 Sin 65 = y= y37.2 km 37.2 km = y= y

Cos 65 = x / Cos 65 = x / 41.041.041.0 Cos 65 41.0 Cos 65 = x= x17.3 km = x17.3 km = x

θ1 = 35°

θ2 = 65°

???

25.5 km41.0

km

y2

x2

y1

x1

yy3 3 = y= y22 – – yy11yy3 3 = 37.2 km – = 37.2 km – 14.6 km14.6 km

yy33

= 22.6 = 22.6 kmkm

xx3 3 = x= x22 + + xx11xx3 3 = 17.3 km + = 17.3 km + 20.9 km20.9 km

= 38.2 = 38.2 kmkm

C

= ??? 22.6 22.6 kmkm

38.2 38.2 kmkmc² = (38.2 km)² + c² = (38.2 km)² +

(22.6 km)²(22.6 km)²c² = 1459.24 km² + c² = 1459.24 km² + 510.76 km²510.76 km²c = 44.4 c = 44.4 kmkm

Tan Tan θθ = 22.6 / = 22.6 / 38.238.2θθ = Tan = Tan--¹ (22.6 / ¹ (22.6 / 38.2)38.2)θθ = 30.6° north = 30.6° north of eastof east

AssignmentAssignment

Pg. 97 Practice # 2Pg. 97 Practice # 2