Question

If the 1 kg standard body is accelerated by only${\overrightarrow{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{j}}$and ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{i}}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{N}\stackrel{\mathbf{⏜}}{\mathbf{j}}$, then what is${\overrightarrow{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}$ (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive x-direction? What are the (d) magnitude and (e) angle of $\overrightarrow{\mathbf{a}}$?

Short answer

a) The net force in the vector form is,${\overrightarrow{F}}_{net}=1\stackrel{⏜}{i}-2\stackrel{⏜}{j}$

b) The magnitude of the net force is 2.24 N

c) The angle made by the net force is$-63.45°$

d) The magnitude of the direction is$2.24\mathrm{m}/{\mathrm{s}}^2$

e) The angle made by the direction is$-63.5°$

Step-by-step solution

Given information

1)${\overrightarrow{F}}_1=3\stackrel{⏜}{i}+4\stackrel{⏜}{j}$

2)$\overrightarrow{F_2}=-2\stackrel{⏜}{i}-6\stackrel{⏜}{j}$

3)$m=1\mathrm{kg}$

Understanding the concept of Newton’s law

Newton’s law states the force acting on the body is the product of mass of the body and the acceleration with which body is moving.

Two forces are given in vector notations. We can use vector algebra to add these two forces and find the resultant and magnitude of them. Using Newton’s second law of motion, we can find the magnitude and acceleration of the object.

Formula

${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{a}$

Here,${\mathbf{F}}_{\mathbf{net}}$ is the force,m is the mass of the object, and a is the acceleration of the object.

(a) Write F→net in unit-vector notation

Add both the forces in unit vector notation as below:

$$\begin{gathered} {\overrightarrow{F}}_{net}=F_1+F_2 \\ =3\stackrel{⏜}{i}+4\stackrel{⏜}{j}-2\stackrel{⏜}{i}-6\stackrel{⏜}{j} \\ =1\stackrel{⏜}{i}-2\stackrel{⏜}{j} \end{gathered}$$

Therefore, the net force is$1\stackrel{⏜}{i}-2\stackrel{⏜}{j}$

(b) Calculate the magnitude of F→net

From part (a), the magnitude of force is calculated as,

$$\begin{gathered} \left|F_{act}\right|=\sqrt{1^2+{\left(-2\right)}^2} \\ =2.24\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the force is 2.24 N.

(c) Define F→net as an angle relative to the positive x-direction 

Using the trigonometry, we can find the angle of force as

$$\begin{gathered} \tan \theta =\frac{y}{x} \\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right) \\ =-\frac{2}{1} \\ =-63.45° \end{gathered}$$

Therefore, the angle is $-63.45°$.

(d) Calculate the magnitude of acceleration

Use Newtons second law to calculate the acceleration from the net force.

$$\begin{gathered} F_{net}=m\times a \\ a=\frac{F_{net}}{m} \\ =\frac{2.24\mathrm{N}}{1\mathrm{kg}} \\ =2.24\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration is $2.24\mathrm{m}/{\mathrm{s}}^2$.

(e) Calculate the direction of acceleration

Direction of acceleration is same as direction of net force that is$-63.45°$