Adding Vectors Geometrically — Question 4

Section 3-2 · Physics pages 338–339

Question 4 Symmetry fixes a sum direction before a ruler is used Two displacement vectors each have magnitude 10 m. Drawn tail-to-tail, their directions are 60◦ apart and symmetric about east: one points 30◦north of east and the other 30◦south of east. 10 m 10 m east The two input arrows enclose 60 degrees; their common-tail diagonal bisects that angle. (a) Use reflection symmetry and the parallelogram construction to determine the sum direction. (b) Find the magnitude using the congruent right triangles formed by the rhombus diagonals. You may use cos 30◦= √ 3/2. (c) When one arrow is translated to form a head-to-tail triangle, what is the triangle’s interior angle at the join? Explain why it is not 60◦. (d) Keeping both magnitudes 10 m, let the tail-to-tail angle grow from 0◦toward 180◦, symmetrically about east. Describe how the sum changes, including the direction at the limiting zero result.
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Question 4 - Solution Symmetry fixes a sum direction before a ruler is used 10 m 10 m east (a) The sum diagonal lies on the symmetry line. Ry = 10 sin 30◦−10 sin 30◦= 0, Rx > 0 ⇒east Reflection about east exchanges the two inputs and leaves their sum unchanged. Their nonzero sum therefore lies on the eastward symmetry line. (b) Bisect the rhombus into right triangles. R = 2(10) cos 30◦= 10 √ 3 m = 17.321 m Bisection of the rhombus gives two congruent right triangles. Twice the adjacent leg is the sum diagonal, whereas 20 m is the two-leg route length. (c) At the join, one triangle side points backward along the first leg. ψ = 180◦−60◦= 120◦, R2 = 102 + 102 −2(10)2 cos 120◦= 300 m2 At the join, the triangle side points backward along the first arrow. Its angle with the second arrow is supplementary to the tail-to-tail angle. (d) Larger angular separation produces more cancellation. R(ϕ) = 20 cos(ϕ/2) m : R(0) = 20 m, R(180◦) = 0 For separations less than 180◦, the resultant points east and decreases continuously in magnitude. At 180◦it vanishes and its direction is undefined.

Question 4

Symmetry fixes a sum direction before a ruler is used

Two displacement vectors each have magnitude 10 m. Drawn tail-to-tail, their directions are 60∘60^\circ apart and symmetric about east: one points 30∘30^\circ north of east and the other 30∘30^\circ south of east.

Diagram shown in the original SVG worksheet.

  1. Use reflection symmetry and the parallelogram construction to determine the sum direction.

  2. Find the magnitude using the congruent right triangles formed by the rhombus diagonals. You may use cos⁡30∘=3/2\cos30^\circ=\sqrt3/2.

  3. When one arrow is translated to form a head-to-tail triangle, what is the triangle’s interior angle at the join? Explain why it is not 60∘60^\circ.

  4. Keeping both magnitudes 10 m, let the tail-to-tail angle grow from 0∘0^\circ toward 180∘180^\circ, symmetrically about east. Describe how the sum changes, including the direction at the limiting zero result.

Question 4 - Solution

Symmetry fixes a sum direction before a ruler is used

Diagram shown in the original SVG worksheet.

(a) The sum diagonal lies on the symmetry line.

Ry=10sin⁡30∘−10sin⁡30∘=0,Rx>0⇒eastR_y=10\sin30^\circ-10\sin30^\circ=0,\quad R_x>0\Rightarrow\text{east} Reflection about east exchanges the two inputs and leaves their sum unchanged. Their nonzero sum therefore lies on the eastward symmetry line.

(b) Bisect the rhombus into right triangles.

R=2(10)cos⁡30∘=103m=17.321m\boxed{R=2(10)\cos30^\circ=10\sqrt3\,\mathrm m=17.321\,\mathrm m} Bisection of the rhombus gives two congruent right triangles. Twice the adjacent leg is the sum diagonal, whereas 20 m is the two-leg route length.

(c) At the join, one triangle side points backward along the first leg.

ψ=180∘−60∘=120∘,R2=102+102−2(10)2cos⁡120∘=300m2\psi=180^\circ-60^\circ=120^\circ,\quad R^2=10^2+10^2-2(10)^2\cos120^\circ=300\,\mathrm{m^2} At the join, the triangle side points backward along the first arrow. Its angle with the second arrow is supplementary to the tail-to-tail angle.

(d) Larger angular separation produces more cancellation.

R(ϕ)=20cos⁡(ϕ/2)m:R(0)=20m,R(180∘)=0R(\phi)=20\cos(\phi/2)\,\mathrm m:\quad R(0)=20\,\mathrm m,\quad R(180^\circ)=0 For separations less than 180∘180^\circ, the resultant points east and decreases continuously in magnitude. At 180∘180^\circ it vanishes and its direction is undefined.