Question

Force $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ acting on body in such a way that their resultant $\overrightarrow{F_3}$ is $80\text{N}$ at ${30}^{\circ }$ East of North. If $\overrightarrow{F_2}$ has magnitude of $20\text{N}$ at ${45}^{\circ }$ North of West. Find the magnitude of $\overrightarrow{F_1}$
[Take $\cos {30}^{\circ }=0.866,\cos {45}^{\circ }=\sin {45}^{\circ }=0.707$]

• A. $70.27\text{N}$
• B. $80.27\text{N}$
• C. $87.34\text{N}$
• D. $77.27\text{N}$

Answer 1

The correct option is D $77.27\text{N}$

Given, $|\overrightarrow{F_3}|=80\text{N}$ and $|\overrightarrow{F_2}|=20\text{N}$
The given situation can be shown as

As we know,
$\overrightarrow{F_3}=\overrightarrow{F_1}$ +$\overrightarrow{F_2}$
or, $\overrightarrow{F_1}=\overrightarrow{F_3}-$ $\overrightarrow{F_2}$ ,
on resolving vectors along the respective axes we get
$\overrightarrow{F_3}=\left(80\sin {30}^{\circ }\right)\hat{i}+\left(80\cos {30}^{\circ }\right)\hat{j}$
$\overrightarrow{F_3}=\left(40\right)\hat{i}+\left(69.28\right)\hat{j}$
$\overrightarrow{F_2}=(-20\sin {45}^{\circ })\hat{i}+\left(20\cos {45}^{\circ }\right)\hat{j}$
$\overrightarrow{F_2}=(-14.14)\hat{i}+\left(14.14\right)\hat{j}$
$\overrightarrow{F_1}=(40+14.14)\hat{i}+(69.28-14.14)\hat{j}$
$\overrightarrow{F_1}=\left(54.14\right)\hat{i}+\left(55.14\right)\hat{j}$
$|\overrightarrow{F_1}|=\sqrt{{54.14}^2+{55.14}^2}$
$|\overrightarrow{F_1}|=77.27\text{N}$