Worked examples · Sheet 3

Vector components and sums of displacements

Theory used: Cartesian components Cartesian components and polar coordinates Operations in components Sum of vectors

Exercise 1

A ball has a velocity of magnitude $20 m/s$ directed at $15 0^{\circ}$ from the positive $x$ -semiaxis, measured counterclockwise. Compute the components of the velocity. Then determine the magnitude and the angle of the displacement $s = (- 6.0 \overset{}{^} - 8.0 \overset{}{^}) m$ .

v_{x} = v cos θ = 20 \cdot cos 15 0^{\circ} m/s = - 20 \cdot \frac{3}{2} m/s \approx - 17.3 m/s

v_{y} = v sin θ = 20 \cdot sin 15 0^{\circ} m/s = 20 \cdot \frac{1}{2} m/s = 10.0 m/s

The displacement

s = (- 6.0, - 8.0) m

lies in the third quadrant

s = s_{x}^{2} + s_{y}^{2} = (- 6.0)^{2} + (- 8.0)^{2} m = 100 m = 10 m

arctan \frac{s _{y}}{s _{x}} = arctan \frac{- 8.0}{- 6.0} \approx 53. 1^{\circ}, θ = 53. 1^{\circ} + 18 0^{\circ} = 233. 1^{\circ}

Given the magnitude and the angle, the Cartesian components are the magnitude times the cosine and the magnitude times the sine of the angle, and the formulas hold for any angle. The angle of $15 0^{\circ}$ lies in the second quadrant, where the cosine is negative and the sine positive; hence $v_{x}$ is negative and $v_{y}$ positive, and the velocity is $v = (- 17.3 \overset{}{^} + 10.0 \overset{}{^}) m/s$ .

For the inverse problem the magnitude is obtained with the Pythagorean theorem and equals $10 m$ . The two components of the displacement are negative, so the vector lies in the third quadrant, as the figure shows. The arctangent returns an angle between $- 9 0^{\circ}$ and $9 0^{\circ}$ , namely the angle of the opposite vector $(6.0, 8.0) m$ ; therefore the angle of the displacement is obtained by adding $18 0^{\circ}$ and equals $233. 1^{\circ}$ counterclockwise, that is, $126. 9^{\circ}$ clockwise.

Exercise 2

A hiker walks $3.0 km$ east and then $4.0 km$ northeast. Determine the magnitude and direction of the total displacement, taking the $x$ -axis toward east and the $y$ -axis toward north.

The total displacement

R

goes from the tail of

d_{1}

to the tip of

d_{2}

(lengths in

km

)

R = d_{1} + d_{2}

d_{1} = (3.0, 0) km, d_{2} = (4.0 cos 4 5^{\circ}, 4.0 sin 4 5^{\circ}) km = (2.83, 2.83) km

R = (3.0 + 2.83, 0 + 2.83) km = (5.83, 2.83) km

R = 5.8 3^{2} + 2.8 3^{2} km \approx 6.5 km, θ = arctan \frac{2.83}{5.83} \approx 25. 9^{\circ}

The diagram shows the two displacements arranged with the tip-to-tail rule: the second starts where the first ends, and the total displacement is their vector sum. The northeast direction forms $4 5^{\circ}$ with east, so the components of the second displacement are equal and are $4.0 \cdot \frac{2}{2} km \approx 2.83 km$ . The operations in components reduce the sum to two additions of numbers, one per axis. The magnitude of the sum is obtained with the Pythagorean theorem; since $R_{x} > 0$ , the arctangent gives the angle directly. The hiker is about $6.5 km$ from the starting point, in the direction $25. 9^{\circ}$ north of east. As a check, $6.5 km$ is less than the distance walked, $7.0 km$ , as it must be for two non-aligned displacements.

Exercise 3

A robot in a warehouse makes three successive displacements: $5.0 m$ at $0^{\circ}$ , $8.0 m$ at $12 0^{\circ}$ and $4.0 m$ at $21 0^{\circ}$ , with the angles measured counterclockwise from the positive $x$ -semiaxis. Determine the total displacement and the displacement that brings the robot back to the starting point.

The three displacements arranged tip to tail and the total displacement

R

(lengths in meters)

R = d_{1} + d_{2} + d_{3}

R_{x} = 5.0 cos 0^{\circ} + 8.0 cos 12 0^{\circ} + 4.0 cos 21 0^{\circ} = (5.0 - 4.00 - 3.46) m = - 2.46 m

R_{y} = 5.0 sin 0^{\circ} + 8.0 sin 12 0^{\circ} + 4.0 sin 21 0^{\circ} = (0 + 6.93 - 2.00) m = 4.93 m

R = (- 2.46)^{2} + 4.9 3^{2} m \approx 5.5 m, θ = arctan \frac{4.93}{- 2.46} + 18 0^{\circ} \approx 116. 6^{\circ}

- R = (2.46 \overset{}{^} - 4.93 \overset{}{^}) m, θ^{'} = 116. 6^{\circ} + 18 0^{\circ} = 296. 6^{\circ}

Each displacement is resolved with the formulas $d_{x} = d cos θ$ and $d_{y} = d sin θ$ , and the $x$ component of the sum is the sum of the $x$ components, and likewise for $y$ . The cosines of $12 0^{\circ}$ and $21 0^{\circ}$ are negative, so the second and third displacements move the robot back toward the left, and the component $R_{x}$ turns out negative. The vector $R$ has $R_{x} < 0$ and $R_{y} > 0$ , therefore it lies in the second quadrant, and $18 0^{\circ}$ is added to the angle given by the arctangent, $- 63. 4^{\circ}$ . The displacement that brings the robot back to the start cancels $R$ , so it is its opposite: it has the same magnitude, $5.5 m$ , and direction $296. 6^{\circ}$ , that is, $63. 4^{\circ}$ below the positive $x$ -semiaxis.