Question

Three astronauts, propelledby jet backpacks, push and guide a 120 kg asteroid toward a processing dock, exerting the forces shown inFig. 5-29, with ${\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{32}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{F}}_{\mathbf{2}}\mathbf{=}\mathbf{55}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{F}}_{\mathbf{3}}\mathbf{=}\mathbf{41}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}\mathbf{}{\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{°}\mathbf{,}$and .What is the asteroid’s acceleration(a) in unit-vector notation and as (b) a magnitude and (c) a direction relative to the positive direction of the x axis?

Short answer

1. $\overrightarrow{\mathrm{a}}=\left(0.86\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{i}}-\left(0.16\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{j}}$
   
2. magnitude is$0.88\mathrm{m}/{\mathrm{s}}^2$
3. Direction is$-11°$

Step-by-step solution

Identification of given data

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

Concept of vector addition and subtraction

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:

$F_{net}=ma$ …(i)

$\left|\overrightarrow{a}\right|=\sqrt{a_x^2+a_y^2}$ …(ii)

$\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$ …(iii)

(a) Determining the acceleration vector in unit-vector notation.

The net force is given by,

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

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

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

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

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

Thus, acceleration is given by,

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

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

(b) Determining the magnitude of acceleration vector

The magnitude is given by,

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

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

(c) Determining the direction of acceleration vector

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

Thus, the direction is$-11°$