91Ó°ÊÓ

$$\begin{gathered} {\mathrm{F}}_1=32\mathrm{N} \\ {\mathrm{F}}_2=55\mathrm{N} \\ \mathrm{F}3=41\mathrm{N} \\ {\mathrm{θ}}_1=30° \\ {\mathrm{θ}}_3=60° \\ \mathrm{m}=120\mathrm{kg} \end{gathered}$$

The problem is based on the simple vector addition and subtraction. Straight addition of numbers is refers to scalar addition. The addition of vectors is not as simple as the addition of scalars. Vectors have both magnitude and direction, and it can be added geometrically.

Formula:

$F_{net}=ma$ …(¾±)

$\left|\stackrel{→}{a}\right|=\sqrt{a_x^2+a_y^2}$ …(¾±¾±)

$\mathrm{θ}={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$ …(¾±¾±¾±)

The net force is given by,

${\stackrel{→}{F}}_{net}={\stackrel{→}{F}}_1+{\stackrel{→}{F}}_2+{\stackrel{→}{F}}_3$

Thus, using equation (i) the acceleration can be written as,

$$\begin{gathered} \stackrel{→}{a}={\stackrel{→}{F}}_{net}/m \\ =\left({\stackrel{→}{F}}_1+{\stackrel{→}{F}}_2+{\stackrel{→}{F}}_3\right)/m \end{gathered}$$

In unit vector notation, the forces exerted on the astronauts are given as,

role="math" localid="1660894011636" $$\begin{gathered} {\stackrel{→}{\mathrm{F}}}_1=32\mathrm{N}\left(\cos 30°\hat{\mathrm{i}}+\sin 30°\hat{\mathrm{j}}\right) \\ =\left(27.7\mathrm{N}\right)\hat{\mathrm{i}}\left(16\mathrm{N}\right)\hat{\mathrm{j}} \\ {\stackrel{→}{\mathrm{F}}}_2=55\mathrm{N}\left(\cos 0°\hat{\mathrm{i}}+\sin 0°\hat{\mathrm{j}}\right) \\ =\left(55\mathrm{N}\right)\hat{\mathrm{i}} \\ {\stackrel{→}{\mathrm{F}}}_1=41\mathrm{N}\left(\cos \left(-60°\right)\hat{\mathrm{i}}+\sin \left(-60°\right)\hat{\mathrm{j}}\right) \\ =\left(20.5\mathrm{N}\right)\hat{\mathrm{i}}-\left(35.5\mathrm{N}\right)\hat{\mathrm{j}} \end{gathered}$$

Thus, acceleration is given by,

$$\begin{gathered} \stackrel{→}{\mathrm{a}}=\frac{\left(27.7\hat{\mathrm{i}}+16\hat{\mathrm{j}}\right)\mathrm{N}+\left(55\hat{\mathrm{i}}\right)\mathrm{N}+\left(20.5\hat{\mathrm{i}}-35.5\hat{\mathrm{j}}\right)\mathrm{N}}{120} \\ =\left(0.86\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}-\left(0.16\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{j}} \end{gathered}$$

Thus, acceleration in unit-vector notation is,$\stackrel{→}{a}=\left(0.86\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}-\left(0.16\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{j}}$

The magnitude is given by,

$$\begin{gathered} \left|\stackrel{→}{a}\right|=\sqrt{a_x^2+a_y^2} \\ =\sqrt{\left(0.86\mathrm{m}/{\mathrm{s}}^2\right)+{\left(-0.16\mathrm{m}/{\mathrm{s}}^2\right)}^2} \\ =0.88\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the magnitude is$0.88\mathrm{m}/{\mathrm{s}}^2$

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{a_y}{a_x}\right) \\ ={\tan }^{-1}\left(\frac{-0.16\mathrm{m}/{\mathrm{s}}^2}{0.86\mathrm{m}/{\mathrm{s}}^2}\right) \\ =-11° \end{gathered}$$

Thus, the direction is$-11°$