The force vec F=5vec i+4vec j+3vec k (in kg, i.e.…

The force $\vec F=5\vec i+4\vec j+3\vec k$ (in kg, i.e. kilogram-force) moves its point of application along the displacement $\vec S=2\vec i+\vec j-2\vec k$ (in m). Find the work done and the angle between $\vec F$ and $\vec S$.

Explanation of Method

The work of a force as a scalar product F·S.

Step 1: Work

\[ A=\vec F\cdot\vec S=5\cdot2+4\cdot1+3\cdot(-2)=10+4-6=8. \]

\[A=8\ \text{kgf}\cdot\text{m}\approx78.5\ \text{J}\]

($1\ \text{kgf}=9.81\ \text{N}$, so $8\ \text{kgf}\cdot\text{m}=78.48\ \text{J}$.)

Step 2: Angle

\[ |\vec F|=\sqrt{25+16+9}=5\sqrt2,\qquad |\vec S|=\sqrt{4+1+4}=3, \]

\[ \cos\theta=\frac{8}{5\sqrt2\cdot3}=\frac{8}{15\sqrt2}=\frac{4\sqrt2}{15}\approx0.3771. \]

\[\theta\approx67.8^\circ\]

Interpretation. Only the component of $\vec S$ along $\vec F$ does work:

\[ \mathrm{Pr}_{\vec F}\vec S=\frac{8}{5\sqrt2}=\frac{4\sqrt2}{5}\ \text{m}, \]

\[ A=|\vec F|\cdot\mathrm{Pr}_{\vec F}\vec S=5\sqrt2\cdot\frac{4\sqrt2}{5}=8. \]

$\theta$ $O$ $\vec S$ $\vec F$

$\vec S=2\vec i+\vec j-2\vec k$
$\vec F=5\vec i+4\vec j+3\vec k\ (\text{half scale})$
$\mathrm{Pr}_{\vec F}\vec S=\tfrac{4\sqrt2}{5}\,\mathrm m$

$\vec F$ is drawn at half scale. The angle $\theta\approx67.8^\circ$ is acute, so the work is positive.

Result

\[\boxed{\,A=8\ \text{kgf}\cdot\text{m}\approx78.5\ \text{J}\,}\]

\[\boxed{\,\theta\approx67.8^\circ\,}\]

Remark

Correction. $\arccos(0.3771)=67.84^\circ$, so it rounds to $67.8^\circ$, not $67.9^\circ$.