chapter 49 vectors - documents.routledge...

© 2014, John Bird

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CHAPTER 49 VECTORS

EXERCISE 206 Page 562

1. State the difference between scalar and vector quantities. A scalar quantity has magnitude only; a vector quantity has both magnitude and direction. In Problems 2 to 9, state whether the quantities given are scalar or vector. 2. A temperature of 70°C 3. 5 3m volume 4. A downward force of 20 N 5. 500 J of work 6. 30 2cm area 7. A southwesterly wind of 10 knots 8. 50 m distance 9. An acceleration of 15 2m/s at 60° to the horizontal 2. Scalar 3. Scalar 4. Vector 5. Scalar 6. Scalar 7. Vector 8. Scalar 9. Vector

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1. A force of 7 N is inclined at an angle of 50° to a second force of 12 N, both forces acting at a point. Calculate magnitude of the resultant of the two forces, and the direction of the resultant with respect to the 12 N force. The two forces are shown in the diagram

The horizontal component of the 12 N force is 12 cos 0° and the horizontal component of the 7 N force is 7 cos 50° The total horizontal component of the two forces, H = 12 cos 0° + 7 cos 50° = 12 + 4.4995 = 16.4995 The vertical component of the 12 N force is 12 sin 0° and the vertical component of the 7 N force is 7 sin 50° The total vertical component of the two forces, V = 12 sin 0° + 7 sin 50° = 0 + 5.3623 = 5.3623

From the diagram, magnitude of resultant vector = 2 2H V+ = 2 216.4995 5.3623+ = 17.35 N

The direction of the resultant vector, 1tan VH

θ − =

= 1 5.3623tan16.4995

−

= 1tan 0.3249977...− = 18.00° Thus, the resultant of the two forces is a single vector of 17.35 N at 18.00° to the 12 N vector

© 2014, John Bird

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2. Velocities of 5 m/s and 12 m/s act at a point at 90° to each other. Calculate the resultant velocity and its direction relative to the 12 m/s velocity. The velocity vectors are shown in the diagram below

The horizontal component of the 12 m/s velocity is 12 cos 0° and the horizontal component of the 5 m/s velocity is 5 cos 90° The total horizontal component of the two velocities, H =12 cos 0° + 5 cos 90° = 12 + 0 = 12 The vertical component of the 12 m/s velocity is 12 sin 0° and the vertical component of the 5 m/s velocity is 5 sin 90° The total vertical component of the two velocities, V =12 sin 0° + 5 sin 90° = 0 + 5 = 5

From the diagram, magnitude of resultant vector, R = 2 2H V+ = 2 212 5+ = 13 N


θ − =

= 1 5tan12

−

= 22.62°

Thus, the resultant of the two velocities is a single vector of 13 m/s at 22.62° to the 12 m/s vector There is an alternative method of calculating the resultant vector in this case Since a right-angled triangle results, we could use Pythagoras’s theorem without needing to go

through the procedure for horizontal and vertical components. In fact, the horizontal and vertical

components are 12 m/s and 5 m/s, respectively.

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3. Calculate the magnitude and direction of the resultant of the two force vectors shown below.

Using the triangle method results in the right-angled triangle shown:

Using Pythagoras, resultant, R = 2 213 10+ = 16.40 N

and 1 10tan 37.5713

θ − = = °

Thus, the resultant of the two forces is a single vector of 16.40 N at 37.57° to the 13 N force. 4. Calculate the magnitude and direction of the resultant of the two force vectors shown below.



and 1 22tan 50.7118

α − = = °

© 2014, John Bird

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Hence, θ = 180º – 50.71º = 129.29º Thus, the resultant of the two forces is a single vector of 28.43 N at 129.29° to the 18 N force 5. A displacement vector 1s is 30 m at 0°. A second displacement vector 2s is 12 m at 90°. Calculate magnitude and direction of the resultant vector 1s + 2s The displacement vectors are shown in the diagram below.



and 1 12tan 21.8030

θ − = = °

Thus, the resultant of the two displacements is a single vector of 32.31 N at 21.80° to the 30 N displacement 6. Three forces of 5 N, 8 N and 13 N act as shown below. Calculate the magnitude and direction of the resultant force.

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The horizontal component of the 8 N force = 8 cos 70° = 2.7362 The horizontal component of the 5 N force is 5 cos 0° = 5 N The horizontal component of the 13 N force is 13 cos 300° = 6.5 N The total horizontal component of the three velocities H = 2.7362 + 5 + 6.5 = 14.2362 N The vertical component of the 8 N force = 8 sin 70° = 7.5175 N The vertical component of the 5 N force is 5 sin 0° = 0 N The vertical component of the 13 N force is 13 sin 300° = –11.2583 N The total vertical component of the three velocities, V = 7.5175 + 0 – 11.2583 = –3.7408 N

From the diagram, magnitude of resultant vector, R = 2 2H V+ = 2 214.2362 3.7408+ = 14.72 N


α − =

= 1 3.7408tan14.2362

−

= 14.72°

Thus, the resultant of the three forces is a single vector of 14.72 N at –14.72° to the 5 N force.

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7. If velocity 1v = 25 m/s at 60° and 2v = 15 m/s at –30°, calculate the magnitude and direction of 1v + 2v The velocities are shown in the diagram below.

The horizontal component of the 25 m/s velocity = 25 cos 60° = 12.5 m/s The horizontal component of the 15 m/s velocity is 15 cos(–30°) = 12.99 m/s The total horizontal component of the two velocities, H = 12.5 + 12.99 = 25.49 m/s The vertical component of the 25 m/s velocity = 25 sin 60° = 21.65 m/s The vertical component of the 15 m/s velocity is 15 sin(–30°) = – 7.5 m/s The total vertical component of the two velocities, V = 21.65 – 7.5 = 14.15 m/s

From the diagram, magnitude of resultant vector, R = 2 2H V+ = 2 225.49 14.15+ = 29.15 m/s


θ − =

= 1 14.15tan25.49

−

= 29.04°

Thus, the resultant of the two velocities is a single vector of 29.15 m/s at 29.04° to the horizontal

© 2014, John Bird

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8. Calculate the magnitude and direction of the resultant vector of the force system shown below.

Total horizontal component, H = 8 cos 30° + 4 cos 105° + 6 cos 300° = 8.8929 N Total vertical component, V = 8 sin 30° + 4 sin 105° + 6 sin 300° = 2.6676 N Magnitude of resultant vector, R = 2 2H V+ = 2 28.8929 2.6676+ = 9.28 N


α − =

= 1 2.6676tan8.8929

−

= 16.70°

Thus, the resultant of the three forces is a single vector of 9.28 N at 16.70° to the horizontal

9. Calculate the magnitude and direction of the resultant vector of the system shown below.

Total horizontal component, H = 2 cos 105° + 3.5 cos 135° + 4 cos 210° = –6.4566 m/s Total vertical component, V = 2 sin 105° + 3.5 sin 135° + 4 sin 210° = 2.4067 m/s

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From the diagram, magnitude of resultant vector, R = 2 2H V+ = 2 26.4566 2.4067+ = 6.89 m/s


α − =

= 1 2.4067tan6.4566

−

= 20.44°

from which, θ = 180° – 20.44° = 159.56° Thus, the resultant of the three velocities is a single vector of 6.89 m/s at 159.56° to the horizontal

10. An object is acted upon by two forces of magnitude 10 N and 8 N at an angle of 60° to each

other. Determine the resultant force on the object.

The two forces are shown in the diagram

The horizontal component of the 10 N force is 10 cos 0° The horizontal component of the 8 N force is 8 cos 60° The total horizontal component of the two forces, H = 10 cos 0° + 8 cos 60° = 10 + 4 = 14 N The vertical component of the 10 N force is 10 sin 0° The vertical component of the 8 N force is 8 sin 60° The total vertical component of the two forces, V = 10 sin 0° + 8 sin 60° = 0 + 5.3623 = 6.928 N

From the diagram, magnitude of resultant vector = 2 2H V+

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= 2 214 6.928+ = 15.62 N


θ − =

= 1 6.928tan14

−

= 26.33º

Thus, the resultant of the two forces is a single vector of 15.62 N at 26.33° to the 10 N force 11. A ship heads in a direction of E 20° S at a speed of 20 knots while the current is 4 knots in a

direction of N 30° E. Determine the speed and actual direction of the ship. The vector diagram is shown below.

Total horizontal component = 4 cos 60° + 20 cos 340° = 20.794 knots

Total vertical component = 4 sin 60° + 20 sin 340° = –3.376 knots

From the diagram above, R = ( )2 220.794 3.376+ = 21.07 knots

and 1 3.376tan 9.2220.794

θ − = = °

Hence, the speed of the ship is 21.07 knots and its actual direction is E 9.22° S

© 2014, John Bird

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1. Forces of F1 = 40 N at 45° and F 2 = 30 N at 125° act at a point. Determine by drawing and by calculation: (a) F1 + F 2 (b) F1 – F 2 (a) Total horizontal component of 1F + 2F = 40 cos 45° + 30 cos 125° = 11.077 N

Total vertical component of 1F + 2F = 40 sin 45° + 30 sin 125° = 52.859 N

From the diagram below, R = ( )2 211.077 52.859+ = 54.0 N,

and 1 52.859tan 78.1611.077

θ − = = °

i.e. the resultant of 1F + 2F is 54.0 N at an angle of 78.16° to the horizontal (b) Total horizontal component of 1F – 2F = 40 cos 45° – 30 cos 125° = 45.492 N

Total vertical component of 1F – 2F = 40 sin 45° – 30 sin 125° = 3.710 N

From the diagram below, R = ( )2 245.492 3.710+ = 45.64 N,

and 1 3.710tan 4.6645.492

θ − = = °

i.e. the resultant of 1F – 2F is 45.64 N at an angle of 4.66° to the horizontal 2. Calculate the resultant of (a) v1 + v 2 – v 3 (b) v 3 – v 2 + v1 when v1 = 15 m/s at 85°, v 2 = 25 m/s at 175° and v 3 = 12 m/s at 235° (a) Total horizontal component of 1v + 2v – 3v = 15 cos 85° + 25 cos 175° – 12 cos 235° = –16.715

Total vertical component of 1v + 2v – 3v = 15 sin 85° + 25 sin 175° – 12 sin 235° = 26.952

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From the diagram below, R = ( )2 216.715 26.952+ = 31.71 m/s,

1 26.952tan 58.1916.715

α − = = °

and thus θ = 180° –58.19° = 121.81°

i.e. the resultant of 1v + 2v – 3v is 31.71 m/s at angle of 121.81° (b) Total horizontal component of 3v – 2v + 1v = 12 cos 235° – 25 cos 175° + 15 cos 85° = 19.329

Total vertical component of 3v – 2v + 1v = 12 sin 235° – 25 sin 175° + 15 sin 85° = 2.934

From the diagram below, R = ( )2 219.329 2.934+ = 19.55 m/s

and 1 2.934tan 8.6319.329

θ − = = °

i.e. the resultant of 3v – 2v + 1v is 19.55 m/s at angle of 8.63°

© 2014, John Bird

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1. A car is moving along a straight horizontal road at 79.2 km/h and rain is falling vertically downwards at 26.4 km/h. Find the velocity of the rain relative to the driver of the car. The space diagram is shown in diagram (a). The velocity diagram is shown in diagram (b) and the velocity of the rain relative to the driver is given by vector rc where rc = re + ec

rc = ( )2 279.2 26.4+ = 83.5 km/h and 1 79.2tan 71.626.4

θ − = = °

(a) (b)

i.e. the velocity of the rain relative to the driver is 83.5 km/h at 71.6° to the vertical 2. Calculate the time needed to swim across a river 142 m wide when the swimmer can swim at 2 km/h in still water and the river is flowing at 1 km/h. At what angle to the bank should the swimmer swim? The swimmer swims at 2 km/h relative to the water, and as he swims the movement of the water

carries him downstream. He must therefore aim against the flow of the water at an angle θ shown in

the triangle of velocities below, where v is the swimmer’s true speed

v = 2 22 1 3− = km/h = 1000360

m/min = 28.87 m/min

Hence, if the width of the river is 142 m, the swimmer will take 14228.87

= 4.919 minutes

© 2014, John Bird

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= 4 min 55 s

In the above diagram, sin θ = 12

from which, θ = 30°

Hence, the swimmer needs to swim at an angle of 60° to the bank (shown as angle α in the diagram.

3. A ship is heading in a direction N 60° E at a speed which in still water would be 20 km/h. It is carried off course by a current of 8 km/h in a direction of E 50° S. Calculate the ship’s actual speed and direction. In the triangle of velocities shown below (triangle 0AB), 0A represents the velocity of the ship in

still water, AB represents the velocity of the water relative to the Earth, and 0B is the velocity of the

ship relative to the Earth.

Total horizontal component of v = 20 cos 30° + 8 cos 310° = 22.46

Total vertical component of v = 20 sin 30° + 8 sin 310° = 3.87

Hence, v = ( )2 222.46 3.87+ = 22.79 km/h,

and 1 3.87tan 9.7822.46

θ − = = °

Hence, the ship’s actual speed is 22.79 km/h in a direction E 9.78° N

© 2014, John Bird

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1. Given that q = –i + j + 4k evaluate and simplify the following vector in i, j, k form: –q –q = –(–i + j + 4k ) = i – j – 4k 2. Given that p = 2i + 0.5j – 3k evaluate and simplify the following vector in i, j, k form: 2p 2p = 2(2i + 0.5j – 3k) = 4i + j – 6k 3. Given that q = –i + j + 4k and r = 6j – 5k, evaluate and simplify the following vector in i, j, k form: q + r q + r = (–i + j + 4k ) + (6j – 5k) = –i + 7j – k 4. Given that p = 2i + 0.5j – 3k and q = –i + j + 4k, evaluate and simplify the following vector in i, j, k form: –q + 2p –q + 2p = –(–i + j + 4k) + 2(2i + 0.5j – 3k) = i – j – 4k + 4i + j – 6k = 5i – 10k 5. Given that q = –i + j + 4k and r = 6j – 5k, evaluate and simplify the following vector in i, j, k form: 3q + 4r 3q + 4r = 3(–i + j + 4k) + 4(6j – 5k) = –3i + 3j + 12k + 24j – 20k = –3i + 27j – 8k 6. Given that p = 2i + 0.5j – 3k and q = –i + j + 4k, evaluate and simplify the following vectors in i, j, k form: q – 2p q – 2p = (–i + j + 4k) – 2(2i + 0.5j – 3k) = (–i + j + 4k) – (4i + j – 6k)

© 2014, John Bird

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= –i + j + 4k – 4i – j + 6k = –5i + 0j + 10k = –5i + 10k 7. Given that p = 2i + 0.5j – 3k, q = –i + j + 4k and r = 6j – 5k, evaluate and simplify the following vectors in i, j, k form: p + q + r p + q + r = (2i + 0.5j – 3k) + (–i + j + 4k ) + (6j – 5k) = 2i + 0.5j – 3k – i + j + 4k + 6j – 5k = i + 7.5j – 4k 8. Given that p = 2i + 0.5j – 3k, q = –i + j + 4k and r = 6j – 5k, evaluate and simplify the following vectors in i, j, k form: p + 2q + 3r p + 2q + 3r = (2i + 0.5j – 3k) + 2(–i + j + 4k) + 3(6j – 5k) = 2i + 0.5j – 3k – 2i + 2j + 8k + 18j – 15k = 20.5j – 10k 9. Given that p = 2i + 0.5j – 3k, q = –i + j + 4k and r = 6j – 5k, evaluate and simplify the following vector in i, j, k form: 2p + 0.4q + 0.5r 2p + 0.4q + 0.5r = 2(2i + 0.5j – 3k) + 0.4(–i + j + 4k) + 0.5(6j – 5k) = 4i + j – 6k – 0.4i + 0.4j + 1.6k + 3j – 2.5k = 3.6i + 4.4j – 6.9k 10. Given that q = –i + j + 4k and r = 6j – 5k, evaluate and simplify the following vector in i, j, k

form: 7r – 2q

7r – 2q = 7(6j – 5k) – 2(– i + j + 4k ) = (42j – 35k) – (–2i + 2j + 8k ) = 42j – 35k + 2i – 2j – 8k = 2i + 40j – 43k

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