Adding Vectors by Components — Question 10

Question 10

Find a fixed-length correction

An existing displacement is $\vec A=(6,2)$ m. A correction B must have length 5 m, and the combined displacement $\vec R=\vec A+\vec B$ must have horizontal component 3 m.

1. Find every possible correction vector B.

2. Find the corresponding resultant vectors, lengths and directions.

3. If R must also point above the horizontal, which correction remains? Explain why one scalar condition did not initially give a unique answer.

Question 10 - Solution

Find a fixed-length correction

(a) Fix one projection, then use the circle.

$$B_x=3-6=-3\,\mathrm m,\quad B_y^2=25-9=16\,\mathrm{m^2}\Rightarrow\boxed{\vec B=(-3,\pm4)\,\mathrm m}$$ $R_x=6+B_x=3$ implies $B_x=-3$ m. The length condition gives $B_y^2=25-9=16$, so $\vec B=(-3,4)$ m or $(-3,-4)$ m. Both corrections satisfy the required 5 m magnitude.

(b) Evaluate the two surviving sums.

$$\vec R_+=(3,6),\ R_+=3\sqrt5,\ \theta_+=63.435^\circ;\quad\vec R_-=(3,-2),\ R_-=\sqrt{13},\ \theta_-=326.310^\circ\quad\text{(m)}$$ They give $\vec R_+=(3,6)$ m, magnitude $3\sqrt5\approx6.708$ m, angle $63.435^\circ$; and $\vec R_-=(3,-2)$ m, magnitude $\sqrt{13}\approx3.606$ m, angle $326.310^\circ$. The same horizontal target does not constrain the vertical sign.

(c) Apply the final directional observation.

$$R_y>0\Rightarrow2+B_y>0\Rightarrow B_y=+4\,\mathrm m$$ Above the horizontal requires $R_y>0$, selecting $\vec B=(-3,4)$ m. Geometrically the fixed $B_x$ line intersects the circle $|\vec B|=5$ m at two points. The added sign information chooses one intersection; it does not alter the earlier constraints.