Vector Resolution

Angled vectors are in 2 dimensions, and have two components, a horizontal and vertical.

Two Methods of Vector Resolution Two methods of vector resolution are: Parallelogram method Trigonometric method

The Parallelogram Method Select a scale and accurately draw the vector to scale in the indicated direction. Sketch a parallelogram around the vector: beginning at the tail of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the head of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram). Draw the components of the vector. The components are the sides of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right). Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc. Measure the length of the sides of the parallelogram and use the scale to determine the magnitude of the components in real units. Label the magnitude on the diagram. The Parallelogram Method

The method of employing trigonometric functions to determine the components of a vector are as follows: Construct a rough sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal. Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. Beginning at the tail of the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the head of the vector. The sketched lines will meet to form a rectangle. Draw the components of the vector. The components are the sides of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right). Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward force velocity component might be labeled vx; etc. To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse. Use some algebra to solve the equation for the length of the side opposite the indicated angle. Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

Adding 3 or more right angle vectors A student drives his car 6.0 km, North before making a right hand turn and driving 6.0 km to the East. Finally, the student makes a left hand turn and travels another 2.0 km to the north. What is the magnitude of the overall displacement of the student?

Vectors can be rearrange into a right triangle R2 = (8.0 km)2 + (6.0 km)2 R2 = 64.0 km2+ 36.0 km2 R2 = 100.0 km2 R = SQRT (100.0 km2) R = 10.0 km

Next find direction Tangent(Θ) = Opposite/Adjacent Tangent(Θ) = 6.0/8.0 Tangent(Θ) = 0.75 Θ = tan-1 (0.75) Θ = 36.869 …° Θ =37° α = 90°- 37°= 53°

Mac and Tosh are doing the Vector Walk Lab Mac and Tosh are doing the Vector Walk Lab. Starting at the door of their physics classroom, they walk 2.0 meters, south. They make a right hand turn and walk 16.0 meters, west. They turn right again and walk 24.0 meters, north. They then turn left and walk 36.0 meters, west. What is the magnitude of their overall displacement?

Rearrange vectors into a right triangle

R2 = (22.0 m)2 + (52.0 m)2 R2 = 484.0 m2 + 2704.0 m2 R2 = 3188.0 m2 R = SQRT (3188.0 m22) R = 56.5 m

Next find direction Tangent(Θ) = Opposite/Adjacent Tangent(Θ) = 52.0/22.0 Tangent(Θ) = 2.3636 … Θ = tan-1 (2.3636 …) Θ = 67.067 …° Θ =67.1° α = 90° + 157°

Addition of Non-Perpendicular Vectors V = Vx + Vy

Max plays middle linebacker for South's football team Max plays middle linebacker for South's football team. During one play in last Friday night's game against New Greer Academy, he made the following movements after the ball was snapped on third down. First, he back-pedaled in the southern direction for 2.6 meters. He then shuffled to his left (west) for a distance of 2.2 meters. Finally, he made a half-turn and ran downfield a distance of 4.8 meters in a direction of 240° counter-clockwise from east (30° W of S) before finally knocking the wind out of New Greer's wide receiver. Determine the magnitude and direction of Max's overall displacement.

Resolve angled vector into components

tangent(Θ) = (6.756… m)/(4.6 m) = 1.46889… R2 = (6.756… m)2 + (4.6 m)2 R2 = 45.655… m2 + 21.16 m2 R2 = 66.815… m2 R = SQRT(66.815… m2 ) R = 8.174 … m R = ~8.2 m tangent(Θ) = (6.756… m)/(4.6 m) = 1.46889… Θ = tan-1 (1.46889…) = 55.7536… ° Θ = ~56° α = 180 ° + 56° = 236°

Examples –Vector Addition worksheet

A: 2.65 km, 140° CCW B: 4.77 km, 252° CCW C: 3.18 km, 332° CCW Cameron Per (his friends call him Cam) and Baxter Nature are on a hike. Starting from home base, they make the following movements. A: 2.65 km, 140° CCW B: 4.77 km, 252° CCW C: 3.18 km, 332° CCW Determine the magnitude and direction of their overall displacement.

Resolve each vector into components

North-South Component Vector East-West Component North-South Component A 2.65 km 140° CCW (2.65 km)•cos(40°) = 2.030… km, West (2.65 km)•sin(40°) = 1.703… km, North B 4.77 km 252° CCW (4.77 km)•sin(18°) = 1.474… km, West (4.77 km)•cos(18°) = 4.536… km, South C 3.18 km 332° CCW (3.18 km)•cos(28°) = 2.808… km, East (3.18 km)•sin(28°) = 1.493… km, South Sum of A + B + C 0.696 km, West 4.326 km, South

R2 = (0. 696 km)2 + (4. 326 km)2 R2 = 0. 484 km2 + 18. 714 km2 R2 = 19 R2 = (0.696 km)2 + (4.326 km)2 R2 = 0.484 km2 + 18.714 km2 R2 = 19.199 km2 R = SQRT(19.199 km2) R = ~4.38 km Tangent(Θ) = opposite/adjacent Tangent(Θ) = (4.326 km)/(0.696 km) Tangent(Θ) = 6.216 Θ = tan-1(6.216) Θ = 80.9° α = 180° + 80.9° = 261°