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Physics / Chapter 3: Vector / Section 3-5

Adding Vectors by Components — Question 3

Section 3-5 · Physics pages 396–397

Question 3 Find the missing closing leg A survey follows three legs:⃗A = 6ˆı m, a 4 m leg B at 60◦north of west, and⃗C = −3ˆȷ m. A final straight leg D returns to the start. A B C D The dashed final arrow must close the polygon. (a) Resolve B and find the position after the first three legs relative to the start. (b) Find the required closing vector D and its magnitude. (c) What endpoint error results if the surveyor follows −⃗D instead?
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Question 3 - Solution Find the missing closing leg A B C D (a) Convert the bearing before adding.⃗ B = 4(cos 120◦, sin 120◦) = (−2, 2 √ 3) m,⃗ S = (4, 2 √ 3 −3) m The standard angle of B is 120◦, so⃗B = (−2ˆı + 2 √ 3ˆȷ) m. The sum of the first three legs is⃗ S = [4ˆı + (2 √ 3 −3)ˆȷ] m, approximately (4, 0.4641) m. (b) Set the total vector to zero.⃗ D = −⃗S = (−4, 3 −2 √ 3) m , D = √ 37 −12 √ 3 m = 4.0268 m⃗ A +⃗B +⃗C +⃗D =⃗0 requires⃗D = −⃗S = [−4ˆı + (3 −2 √ 3)ˆȷ] m. Its magnitude is √ 37 −12 √ 3 ≈4.0268 m, directed 6.62◦south of west. (c) Add the mistaken leg to the actual endpoint.⃗ e =⃗S −⃗D = 2⃗S = (8, 4 √ 3 −6) m, e = 8.0537 m The wrong leg is −⃗D =⃗S, so the final position is 2⃗S ≈(8, 0.9282) m. The error magnitude is 2|⃗S| ≈8.0537 m. A correct closing length is not enough; its sense must point back toward the start.

Question 3

Find the missing closing leg

A survey follows three legs: A→=6ı̂\vec A=6\hat\imath m, a 4 m leg B at 60∘60^\circ north of west, and C→=−3ȷ̂\vec C=-3\hat\jmath m. A final straight leg D returns to the start.

Diagram shown in the original SVG worksheet.

  1. Resolve B and find the position after the first three legs relative to the start.

  2. Find the required closing vector D and its magnitude.

  3. What endpoint error results if the surveyor follows −D→-\vec D instead?

Question 3 - Solution

Find the missing closing leg

Diagram shown in the original SVG worksheet.

(a) Convert the bearing before adding.

B→=4(cos⁡120∘,sin⁡120∘)=(−2,23)m,S→=(4,23−3)m\vec B=4(\cos120^\circ,\sin120^\circ)=(-2,2\sqrt3)\,\mathrm m,\quad\vec S=(4,2\sqrt3-3)\,\mathrm m The standard angle of B is 120∘120^\circ, so B→=(−2ı̂+23ȷ̂)\vec B=(-2\hat\imath+2\sqrt3\hat\jmath) m. The sum of the first three legs is S→=[4ı̂+(23−3)ȷ̂]\vec S=[4\hat\imath+(2\sqrt3-3)\hat\jmath] m, approximately (4,0.4641)(4,0.4641) m.

(b) Set the total vector to zero.

D→=−S→=(−4,3−23)m,D=37−123m=4.0268m\boxed{\vec D=-\vec S=(-4,3-2\sqrt3)\,\mathrm m},\quad D=\sqrt{37-12\sqrt3}\,\mathrm m=4.0268\,\mathrm m A→+B→+C→+D→=0→\vec A+\vec B+\vec C+\vec D=\vec0 requires D→=−S→=[−4ı̂+(3−23)ȷ̂]\vec D=-\vec S=[-4\hat\imath+(3-2\sqrt3)\hat\jmath] m. Its magnitude is 37−123≈4.0268\sqrt{37-12\sqrt3}\approx4.0268 m, directed 6.62∘6.62^\circ south of west.

(c) Add the mistaken leg to the actual endpoint.

e→=S→−D→=2S→=(8,43−6)m,e=8.0537m\vec e=\vec S-\vec D=2\vec S=(8,4\sqrt3-6)\,\mathrm m,\quad e=8.0537\,\mathrm m The wrong leg is −D→=S→-\vec D=\vec S, so the final position is 2S→≈(8,0.9282)2\vec S\approx(8,0.9282) m. The error magnitude is 2|S→|≈8.05372|\vec S|\approx8.0537 m. A correct closing length is not enough; its sense must point back toward the start.

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