From point A(2,1,-1) acts a force F with magnitude 7 and…
From point $A(2,1,-1)$ acts a force $\vec{F}$ with magnitude $7$ and projections: on the $x$-axis $X=2$, on the $y$-axis $Y=-3$. Determine the direction and sense of force $\vec{F}$.
Explanation of Method
Direction and sense of a force from its magnitude and two projections.
Step 1: Find the $z$-projection
\[ |\vec{F}|^2 = X^2+Y^2+Z^2 \;\Longrightarrow\; 49 = 4 + 9 + Z^2 \;\Longrightarrow\; Z^2 = 36 \;\Longrightarrow\; Z = \pm 6 \]
Both signs are possible: the data do not decide whether the force points up or down.
Step 2: Direction cosines
\[ \cos\alpha = \frac{2}{7} \;\Rightarrow\; \alpha \approx 73.4^\circ, \qquad \cos\beta = -\frac{3}{7} \;\Rightarrow\; \beta \approx 115.4^\circ \]
\[ Z=+6:\ \cos\gamma = \frac{6}{7},\ \gamma \approx 31.0^\circ; \qquad Z=-6:\ \cos\gamma = -\frac{6}{7},\ \gamma \approx 149.0^\circ \]
Step 3: The two possible forces
\[\vec{F}_1 = 2\,\vec{i} - 3\,\vec{j} + 6\,\vec{k}, \qquad \vec{F}_2 = 2\,\vec{i} - 3\,\vec{j} - 6\,\vec{k}\]
They have the same $x$- and $y$-projections and differ only in the sign of the $z$-projection ($\vec F_1-\vec F_2=12\vec k$); they are mirror images of each other in the horizontal plane through $A$. Applied at $A$, their tips are $B_1(4,-2,5)$ and $B_2(4,-2,-7)$.
Verification
\[ |\vec{F}_{1,2}| = \sqrt{4+9+36} = 7 \;\checkmark, \qquad \frac{4+9+36}{49} = 1 \;\checkmark \]
- $\vec F_1=2\vec i-3\vec j+6\vec k$
- $\vec F_2=2\vec i-3\vec j-6\vec k$
Result
| $F_x$ | $F_y$ | $F_z$ | $\alpha$ | $\beta$ | $\gamma$ | |
|---|---|---|---|---|---|---|
| $\vec F_1$ | $2$ | $-3$ | $6$ | $73.4^\circ$ | $115.4^\circ$ | $31.0^\circ$ |
| $\vec F_2$ | $2$ | $-3$ | $-6$ | $73.4^\circ$ | $115.4^\circ$ | $149.0^\circ$ |