Adding Vectors by Components — Question 5

Question 5

Make the resultant point exactly north

Vector A is 10 m at $30^\circ$ north of east. Vector B has unknown nonnegative magnitude $b$ and points $60^\circ$ north of west. Their resultant must point exactly north.

1. Use the horizontal components to find b.

2. Find the resultant magnitude and verify that its vertical component is positive.

3. Can the same direction choices produce a zero resultant or a due-south resultant for any $b\geq0$?

Question 5 - Solution

Make the resultant point exactly north

(a) Cancel the horizontal projections.

$$R_x=10\cos30^\circ-b\cos60^\circ=0\Rightarrow\boxed{b=10\sqrt3\,\mathrm m}$$ $A_x=5\sqrt3$ m and $B_x=-b\cos60^\circ=-b/2$. Requiring $R_x=0$ gives $b=10\sqrt3\ \mathrm m\approx17.321$ m. A zero horizontal component alone does not yet select north rather than south.

(b) Check the remaining component.

$$R_y=10\sin30^\circ+10\sqrt3\sin60^\circ=5+15=20\,\mathrm m$$ $R_y=10\sin30^\circ+b\sin60^\circ=5+(10\sqrt3)(\sqrt3/2)=20$ m. Thus $\vec R=20\hat\jmath$ m, genuinely due north.

(c) Use signs to exclude alternatives.

$$R_y=5\,\mathrm m+\tfrac{\sqrt3}{2}b>0\quad(b\ge0)\Rightarrow\vec R\ne\vec0\text{ and cannot point south}$$ For every $b\geq0$, $R_y=5+(\sqrt3/2)b>0$. The resultant can never vanish or point south. Although vectors with other directions might cancel A, this restricted B direction always contributes an additional northward component.