Question

There are two forces each having same magnitude $10 \, \text{N} \text{.}$ One is inclined at an angle of $30^{o}$ and other is inclined at an angle of $135^{\text{o}}$ to the positive direction of $x$ -axis. The $x$ and $y$ components of the resultant are

• A. $1.59 \, \text{N}\hat{i} \, $ and $12.07\text{N}\hat{j}$
• B. $10 \, \text{N}\hat{i}$ and $10 \, \text{N}\hat{j}$
• C. $1.59 \, \text{N}\hat{i}$ and $10 \, \text{N}\hat{j}$
• D. $1.59 \, \text{N}\hat{i}$ and $2 \, \text{N}\hat{j}$

Answer 1

Answer: (A)

After resolving forces in $xy-$ direction, we get
Along $x$ direction, net force;
$F_{x} = 10 \left[cos \left(30\right)^{o} + \left(- cos \left(45\right)^{o}\right)\right] = 10 \left(\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\right) = 1.59 \, \text{N}$
Along $y$ direction, net force;
$F_{y}=10\left(\text{sin} \left(30\right)^{o} + 10 \, \, \text{sin} \left(45\right)^{o}\right)$
$F_{y} = 10 \left(\right. \frac{1}{2} + \frac{1}{\sqrt{2}} \left.\right)$
$= 12.07 \, N$
Thus forces in $x$ and $y$ direction are $1.59 \, \text{N}$ and $12.07 \, \text{N}$ respectively.