Problem — Resultant of two forces

Problem

Two forces lie in the same plane: $\vv{F}_1$ has magnitude $5\;\text{N}$ and makes an angle of $0^\circ$ with the $x$ axis; $\vv{F}_2$ has magnitude $4\;\text{N}$ and is directed at $60^\circ$ above the $x$ axis. Calculate the magnitude and direction of the resultant $\vv{R} = \vv{F}_1 + \vv{F}_2$.

Solution

Step 1 — Cartesian components. We decompose each force along the axes. The first is directed along $x$; the second makes $60^\circ$ with $x$:

$\begin{aligned} F_{1x} &= 5\;\text{N}, & F_{1y} &= 0 \\ F_{2x} &= 4\cos 60^\circ = 4\cdot 0{,}5 = 2{,}0\;\text{N}, & F_{2y} &= 4\sin 60^\circ = 4\cdot 0{,}866 = 3{,}46\;\text{N} \end{aligned}$

Step 2 — Components of the resultant. We add component by component:

$R_x = F_{1x} + F_{2x} = 5 + 2{,}0 = 7{,}0\;\text{N}, \qquad R_y = F_{1y} + F_{2y} = 0 + 3{,}46 = 3{,}46\;\text{N}$

Step 3 — Magnitude.

$R = \sqrt{R_x^2 + R_y^2} = \sqrt{7{,}0^2 + 3{,}46^2} = \sqrt{49 + 12{,}0} = \sqrt{61{,}0} \approx 7{,}8\;\text{N}$

Step 4 — Direction. The angle with respect to the $x$ axis is:

$\theta = \arctan\!\frac{R_y}{R_x} = \arctan\!\frac{3{,}46}{7{,}0} = \arctan 0{,}49 \approx 26^\circ$

Result.

$\ev{R \approx 7{,}8\;\text{N}, \qquad \theta \approx 26^\circ}$

Note on the answer key $R \approx 7{,}2\;\text{N}$, $\theta \approx 34^\circ$) is not consistent with the data of the exercise. With $F_1 = 5\;\text{N}$ at $0^\circ$ and $F_2 = 4\;\text{N}$ at $60^\circ$ one inevitably gets $R = \sqrt{61} \approx 7{,}8\;\text{N}$ and $\theta \approx 26^\circ$, as above. The correct value is the one derived here: this is a typo in the answer key.

The answer printed in the margin of the book (

Linked atoms

Argomenti: Vectors Concetti: Components of a vector Competenze: Vector decomposition Metodi: Decomposition into Cartesian components