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Boat across the river

Done by : Mohamed S aeed A lsayyah Hamdan Abdullah Alremaithi Jasem Ahmed Alsuwaidi Rashed Mohammed A lhebsi. Boat across the river. Task 1. Students will search the web to find real life examples on working with vectors, e.g. adding forces,

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Boat across the river

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  1. Done by : Mohamed Saeed Alsayyah Hamdan Abdullah Alremaithi Jasem Ahmed Alsuwaidi Rashed Mohammed Alhebsi Boat across the river

  2. Task 1 • Students will search the web to find real life examples on working with vectors, e.g. adding forces, • Projectile, components, vectors in 3D, etc.… • Vector application: • Sports bull riding • Lighting system of a theatre • Projecting a rocket (will have magnitude and direction) • 3D films (length and direction of objects will give the affect) • Pulling a wagon

  3. Task 2 If the boat is heading due north as it crosses a river with velocity of 10 km/h relative to water, the river has a uniform velocity of 5 km/h due to east. Graph the vectors in the chart below. Then determine the magnitude and direction of the boat’s velocity. Scale: 2 boxes= 1km/h 5 km/h 10 km/h

  4. Magnitude: (Pythagoras Theorem): C2 = a2 + b2 R2 = (10)2 + (5)2 R2 = 100 + 25 R2 = 125 R = 125 R= 11.18 km/h Direction: Tan θ = oppositeadjacent Θ = tan-1oppadjacent Θ = tan-1105 Θ = 63.43°

  5. Task 3 Two horses are pulling the boat along a canal, one horse exerts a force of 300 N at an angle20°,and the other exerts 500 N at angle 30°.Find the magnitude of the resultant force. y- component= 500 sin 30 x- component = 500 cos 30 30° x- component= -300 cos 20 y-component: -300 sin 20

  6. Magnitude: Force in x-component= 500 cos30 + (-300 cos-20) Force in x-component= 714.92 N Force in Y-component= 500 sin30 + (-300 sin-20) Force in Y-component= 352.6 N C2= a2 + b2 C2= (714.92)2 + (352.6)2 C2= 635437.3664 C= 635437.3664 C= 797.14 N

  7. Task 4

  8. Graph

  9. Task 5 Suppose that the force acting on the boat can be expressed by vector 45,24,110 where each measure in the ordered triple represents the force in Newton. What is the magnitude of the force? r→= x2+y2+z2 r→= 452+242+ 1102 r→= 14701 r→= 121.24 N

  10. 200 N Task 6 60° 6 feet Workers are using a force of 200 Newton to push a cart up a ramp. The ramp is 6 feet long and is at 30°angle with the horizontal. How much work are the workers doing in the vertical direction? You may use sine ratio and formula W =F.xd W= f x d W= 200 x 6 W= 1200 Joules With vertical: W= 1200 sin60 Work = 1039.23 Joules with the vertical

  11. Resources • Wikipedia.com • Youtube.com

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