Question

A $2\text{kg}$ object is subjected to three forces that give it an acceleration $\overrightarrow{a}=-\left(8\hat{i}\right){\text{m/s}}^2$. If two of the three forces are ${\overrightarrow{F}}_1=(30\hat{i}+16\hat{j})\text{N}$ and ${\overrightarrow{F}}_2=(-12\hat{i}+8\hat{j})\text{N}$, find the third force ($\text{in N}$).

• A. $34\hat{i}+24\hat{j}$
• B. $-34\hat{i}-24\hat{j}$
• C. $18\hat{i}+24\hat{j}$
• D. $-18\hat{i}-24\hat{j}$

Answer 1

Answer: (B)

Step 1: Given that:

Mass($m$) of the object = $2kg$

Net acceleration acting on the body $\overrightarrow{a}=(-8\hat{i})m{s}^{-2}$

First force acting on the body is $\overrightarrow{F_1}=(30\hat{i}+16\hat{j})N$

Second force acting on the body that is $\overrightarrow{F_2}=(-12\hat{i}+8\hat{j})N$

Step 2: Calculation of the net force acting on the body:

According to the second law of motion;

$Force\left(F\right)=mass\left(m\right)\times Acceleration\left(a\right)$

Thus,

$\overrightarrow{F_{net}}=2kg\times (-8\hat{i})N$

$\overrightarrow{F_{net}}=(-16\hat{i})N$

Step 3: Calculation of the third force acting on the body:

Let the third force acting on the body is $\overrightarrow{F_3}=(x\hat{i}+y\hat{j})N$

Now, the net force acting on the body is given by;

$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=\overrightarrow{F_{net}}$

Putting the values, we get

$(30\hat{i}+16\hat{j})N+(-12\hat{i}+8\hat{j})N+(x\hat{i}+y\hat{j})N=(-16\hat{i})N$

$30\hat{i}-12\hat{i}+x\hat{i}+16\hat{j}+8\hat{j}+y\hat{j}=-16\hat{i}$

$(18+x)\hat{i}+(24+y)\hat{j}=-16\hat{i}+0\hat{j}$

$(18+x)\hat{i}+(24+y)\hat{j}=-16\hat{i}+0\hat{j}$

Comparing the coefficients of

$\hat{i}$ and $\hat{j}$ on both sides, we get

$18+x=-16$

$x=-16-18$

$x=-34$

And,

$24+y=0$

$y=-24$

Thus, the third force acting on the body will be

$\overrightarrow{F_3}=\{-34\hat{i}+(-24)\hat{j}\}$

$\overrightarrow{F_3}=(-34\hat{i}-24\hat{j})N$

Thus,

Option A) $(-34\hat{i}-24\hat{j})N$ is the correct option.