Question

There are two forces on the$2.00\mathrm{kg}$box in the overhead view of Figure, but only one is shown. For $F_1=20.0\mathrm{N},\mathrm{a}=12.0\mathrm{m}/{\mathrm{s}}^2$ , and $\theta =30.0°$ , find the second force

(a) in unit-vector notation and as

(b) a magnitude of second force and

(c) an angle relative to the positive direction of the xaxis

Short answer

1. Second force in unit-vector notation is $\left(-32.0\mathrm{N}\right)\hat{i}-\left(20.76\mathrm{N}\right)\hat{j}$.
2. The magnitude of second force is $38.14\mathrm{N}$.
3. An angle relative to the positive direction of the x axis is$-147°$

Step-by-step solution

The given data

• Mass of the box,$m=2\mathrm{kg}$.
• Magnitude of first force,$F_1=20\mathrm{N}$.
• Acceleration in the direction of second force, $a=12\mathrm{m}/{\mathrm{s}}^2$.

• Angle of second force with y-axis, $\theta =30.0°$.

Understanding the concept of Newton’s law of motion

As per Newton’s second law, the force acting on the body is product of mass and acceleration of the body. If there are more than one force acting on the body, the net force is equal to the vector sum of all the forces. The expression of force in terms of mass m and acceleration$\overrightarrow{\mathbf{a}}$is given by,

${\overrightarrow{\mathit{F}}}_{Net\mathbf{}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{a}}$

Using the formula for Newton’s second law, we can find the second force in unit-vector notation and its magnitude.

Formula:

The net force on a particle according to Newton’s second law,

$$\begin{gathered} \overrightarrow{F}net=\overrightarrow{F_1}+\overrightarrow{F_2} \\ =M\overrightarrow{a} \end{gathered}$$ (i)

a) Calculation of second force in unit-vector notation

From equation (i), we can get the second force as follows.

$\overrightarrow{F_2}=M\overrightarrow{a}-\overrightarrow{F_1}$

Find the component of the acceleration along both the forces. The value of acceleration and angle are$\overrightarrow{a}=12\mathrm{m}/{\mathrm{s}}^2,\mathrm{\theta }=30.0°$ respectively.

$\overrightarrow{a}=\left(12\sin \left(30°\right)\mathrm{m}/{\mathrm{s}}^2\right)\hat{i}-\left(12\cos \left(30°\right)\mathrm{m}/{\mathrm{s}}^2\right)\hat{j}$

$=\left(-6.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{i}-\left(10.38\mathrm{m}/{\mathrm{s}}^2\right)\hat{j}$

Substitute these values in the above equation for force $\overrightarrow{F_2}$.

$$\begin{gathered} \overrightarrow{F_2}=\left(2.00\mathrm{kg}\right)\left[\left(-6.00\mathrm{m}/{\mathrm{s}}^2\right)\hat{i}-\left(10.38\mathrm{m}/{\mathrm{s}}^2\right)\hat{j}\right]-\left(20.0\mathrm{N}\right)\hat{i} \\ =\left(-32.0\mathrm{N}\right)\hat{i}-\left(20.76\mathrm{N}\right)\hat{j} \end{gathered}$$

Hence, the second force is$\left(-32.0\mathrm{N}\right)\hat{i}-\left(20.76\mathrm{N}\right)\hat{j}$

b) Calculations for the magnitude of the second force

From the x and y component of the forces, we can find the magnitude of the second force. Substitute the values of x and y components to find the magnitude.

$$\begin{gathered} \left|\overrightarrow{F_2}\right|=\sqrt{{\left(-32.0\mathrm{N}\right)}^2+{\left(-20.76\mathrm{N}\right)}^2} \\ =38.14\mathrm{N} \end{gathered}$$

Hence, the value of magnitude of the force is $38.14\mathrm{N}$

c) Calculation for the angle relative to the positive direction of the x-axis

The angle that the second force makes can be given as:

$\varphi ={\tan }^{-1}\left(\frac{-20.76\mathrm{N}}{-32.0\mathrm{N}}\right)$

$$\begin{gathered} 32.97 \\ \approx 33° \end{gathered}$$

As both x and y components are negative, the force is in third quadrant. So, the angle $33°$is measured in counter clockwise direction from negative x-axis. The angle with the positive x-axis measured in counterclockwise direction would be,

$$\begin{gathered} \phi =33°-180° \\ =-147° \end{gathered}$$

Hence, the value of the angle is $-147°$.