Adding Vectors Geometrically — Question 1

Question 1

The resultant joins the first tail to the last head

A marker is displaced 8.0 m east and then 6.0 m north. These are exact straight legs in a fixed horizontal plane. North is up on every diagram. To add free displacement vectors graphically, translate the second arrow so its tail meets the first arrow’s head, preserving both its length and direction. Draw the resultant from the initial tail to the final head.

1. Use a scale of 2.0 m per paper centimetre to specify the lengths of the two leg arrows and the resultant arrow. Find the resultant magnitude and direction using right-triangle geometry.

2. Find the distance traveled. Explain why adding the two magnitudes does not give the resultant magnitude here.

3. A student draws the diagonal from the final head back to the starting tail. What physical displacement does this arrow represent?

4. Would moving the entire construction elsewhere on the page change the sum? Would rotating only the second arrow to point east preserve the sum? Explain both cases.

Question 1 - Solution

The resultant joins the first tail to the last head

(a) Preserve each arrow and connect the outer endpoints.

$$\ell_A=8/2=4\,\mathrm{cm},\ \ell_B=6/2=3\,\mathrm{cm},\ \boxed{R=\sqrt{8^2+6^2}=10\,\mathrm m},\ \theta=\tan^{-1}(6/8)=36.870^\circ$$ The resultant arrow is 5 cm long at the chosen scale and points from the initial tail to the final head. The right-triangle calculation determines its physical length and heading.

(b) Route length and directed separation are different quantities.

$$s=8+6=14\,\mathrm m>R,\qquad|\vec A+\vec B|\le A+B$$ Magnitudes add to the resultant only for arrows aligned in the same direction. Here the bent route is longer than the direct separation.

(c) Reversing the diagonal reverses the displacement.

$$\vec R_{\mathrm{return}}=-\vec R,\quad R_{\mathrm{return}}=10\,\mathrm m,\quad\theta_{\mathrm{return}}=216.870^\circ$$ The reversed diagonal represents the return displacement: $36.870^\circ$ south of west. It is a different physical vector.

(d) Translation and rotation do different things.

$$(\vec r_f+\vec a)-(\vec r_i+\vec a)=\vec R,\qquad\vec R_{\mathrm{both\ east}}=14\hat i\,\mathrm m$$ Translating the construction preserves directed separations. Rotating only the second leg east changes that leg and gives the new 14 m eastward sum.