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

The correct option is B $-34\hat{i}-24\hat{j}$

Given: Acceleration of the body, $\overrightarrow{a}=-8\hat{i}{\text{m/s}}^2$
${\overrightarrow{F}}_1=(30\hat{i}+16\hat{j})\text{N}$
${\overrightarrow{F}}_2=(-12\hat{i}+8\hat{j})\text{N}$
Net force acting on the object $\overrightarrow{F}=m\overrightarrow{a}=2\times (-8)\hat{i}=-\left(16\hat{i}\right)$
Let the third force be ${\overrightarrow{F}}_3=x\hat{i}+y\hat{j}\Rightarrow {\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=F=-\left(16\text{N}\right)\hat{i}$
$\Rightarrow 30\hat{i}+16\hat{j}-12\hat{i}+8\hat{j}+x\hat{i}+y\hat{j}=-16\hat{i}$
$\Rightarrow (30-12+x)\hat{i}+(16+8+y)\hat{j}=-16\hat{i}$
$\Rightarrow 18+x=-16$ and $24+y=0$
$\Rightarrow x=-34$ and $y=-24$
$\therefore$ The third force will be ${\overrightarrow{F}}_3=(-34\hat{i}-24\hat{j})\text{N}$