Question

In the overhead view of Fig.5-65, five forces pull on a box of mass m=4.0 kg. The force magnitudes are${\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{11}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}{\mathbf{F}}_{\mathbf{2}}\mathbf{=}\mathbf{17}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}{\mathbf{F}}_{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}{\mathbf{F}}_{\mathbf{4}}\mathbf{=}\mathbf{14}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{and}\mathbf{}{\mathbf{F}}_{\mathbf{5}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{angle}\mathbf{}{\mathbf{\theta }}_{\mathbf{4}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{30}\mathbf{°}$Find the box’s acceleration (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the xaxis.

Short answer

(a) Acceleration in unit vector notation is$\left(1.0\right)\hat{i}-\left(1.3\right)\hat{j}$.

(b) Magnitude of acceleration is$1.6\mathrm{m}/{\mathrm{s}}^2$.

(c) Angle relative to positive x axis is$-50°$_.

Step-by-step solution

Given information

$$\begin{gathered} 1)\mathrm{m}=4.0\mathrm{kg}, \\ 2){\mathrm{F}}_1=11\mathrm{N}, \\ 3){\mathrm{F}}_2=17\mathrm{N}, \\ 4){\mathrm{F}}_3=3.0\mathrm{N}, \\ 5){\mathrm{F}}_4=14\mathrm{N}, \\ 6){\mathrm{F}}_5=5.0\mathrm{N}\mathrm{and} \\ 7){\mathrm{\theta }}_4=30° \end{gathered}$$

Understanding the concept

The vector quantity can be resolved into components along x and y axis. Newton’s second law states that the product of mass and acceleration of the object is equal to the force acting on the object.

Use i and j to write x and y components of vector. First, find net forces using given forces, then using Newton’s second law we can find acceleration.

Once the acceleration vector is found, find magnitude and angle.

Formulae:

$$\begin{gathered} \mathbf{1}\mathbf{)}\mathbf{}\mathbf{}{\overline{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{F}\hat{\mathbf{i}}\mathbf{+}\mathbf{F}\hat{\mathbf{j}} \\ \mathbf{2}\mathbf{)}\mathbf{}\mathbf{}\mathbf{F}\mathbf{=}\mathbf{ma} \\ \mathbf{3}\mathbf{)}\mathbf{}\mathbf{}\mathbf{\theta }\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{y}{x}\right) \end{gathered}$$

Calculate the acceleration in unit vector notation

(a)

Write the components of force to find component of resultant force.

$$\begin{gathered} F_x=F_3+F_4\cos \theta -F_1\mathrm{and} \\ {\mathrm{F}}_{\mathrm{y}}={\mathrm{F}}_5+{\mathrm{F}}_4\sin \theta -F_2 \end{gathered}$$

Substitute the values to calculate x and y components.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{x}}=\left(3.0+14\cos 30°-11\right) \\ =4.1\mathrm{N} \\ {\mathrm{F}}_{\mathrm{y}}=\left(14\sin 30°\right)+5-17 \\ =-5\mathrm{N} \end{gathered}$$

Now, calculate the net force on the box.

${\mathrm{F}}_{\mathrm{net}}=\left(4.1\right)\hat{\mathrm{i}}-\left(5\right)\hat{\mathrm{j}}\mathrm{N}$

Apply Newton’s second law to find acceleration:

role="math" localid="1660904306598" $$\begin{gathered} a=\frac{F}{m} \\ =\frac{\left(4.1\right)\hat{i}-\left(5\right)\hat{j}}{4.0} \\ =\left(0.1\right)\hat{i}-\left(1.3\right)\hat{j} \end{gathered}$$

Therefore, the acceleration vector of the box is $\left(0.1\right)\hat{i}-\left(1.3\right)\hat{j}$.

Calculate the magnitude of acceleration

(b)

To calculate the magnitude, use the acceleration vector.

$$\begin{gathered} \mathrm{a}=\sqrt{{\left(0.1\right)}^2+{\left(1.3\right)}^2} \\ =1.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Calculate the angle made by acceleration vector with x axis

(c)

Using the trigonometry, find the angle as below:

$$\begin{gathered} \theta =\tan \left(\frac{1.3}{1.0}\right) \\ =-50° \end{gathered}$$

Angle is $50°$clockwise with positive x axis.