Components of Vectors — Question 4

Question 4

One component leaves two possible directions

A straight displacement has magnitude 13 m and horizontal component $A_x=5$ m. Nothing is initially known about its vertical direction.

1. Find every possible value of $A_y$ and the corresponding standard direction angles.

2. What extra observation selects a unique answer? Use the observation that the endpoint is below the starting point.

3. Classify the possibilities if a second report instead states $A_x=13$ m or $A_x=14$ m, with the same magnitude.

Question 4 - Solution

One component leaves two possible directions

(a) Retain both square-root branches.

$$A_y^2=13^2-5^2=144\Rightarrow\boxed{A_y=\pm12\,\mathrm m},\quad\theta=67.380^\circ\text{ or }292.620^\circ$$ $A_y^2=13^2-5^2=144\ \mathrm{m^2}$, so $A_y=\pm12$ m. The directions are $\tan^{-1}(12/5)\approx67.38^\circ$ and $360^\circ-67.38^\circ=292.62^\circ$. A magnitude is not enough to select the sign.

(b) Use a directional observation.

$$A_y<0\Rightarrow(A_x,A_y)=(5,-12)\,\mathrm m$$ An endpoint below the start means $A_y<0$. It selects $A_y=-12$ m and $292.62^\circ$, or $67.38^\circ$ below positive $x$. Merely knowing the vertical separation is 12 m would not resolve the ambiguity.

(c) Check feasibility before taking a root.

$$A_x=13\Rightarrow A_y=0;\quad A_x=14\Rightarrow A_y^2=-27\,\mathrm{m^2}<0$$ For $A_x=13$ m, $A_y=0$ and the only direction is positive $x$. For $A_x=14$ m, $A_y^2=-27\ \mathrm{m^2}$, impossible for a real vector. In orthogonal axes, every component satisfies $|A_i|\leq A$.