Animated Solution for Physics - Kinematics: The resultant of these forces OP, OQ, OR, OS and OT is approximately ...... N. [Take, $\sqrt{3}=1.7$, $\sqrt{2}=1.4$ and given $\hat{i}$ and $\hat{j}$ unit vectors along X, Y axis]

Imagine you are standing at the origin of a coordinate system, and five different ropes are pulling you in various directions. This is exactly what our problem presents: a system of five concurrent forces acting at the origin $O$. Our mission is to find the single resultant force ${\vec{F}}_{net}$ that can replace all of these individual pulls.

Adding vectors geometrically by drawing them tip-to-tail would be a nightmare here. Instead, we use the elegant method of resolution of vectors. By breaking down each force into its horizontal ($x$) and vertical ($y$) components, we transform a complex geometry problem into simple arithmetic.

Resolving Vectors

The core idea is to express every force in the form $\vec{F}=F_x\hat{i}+F_y\hat{j}$. We must pay close attention to two things: the angle given (is it with the X-axis or Y-axis?) and the quadrant the vector lies in (which determines the positive or negative signs).

Let's tackle them one by one, strictly adhering to the problem's instruction to use $\sqrt{3}=1.7$ and $\sqrt{2}=1.4$.

1. Resolving $\overrightarrow{OP}$: Vector $\overrightarrow{OP}$ has a magnitude of $20\text{ N}$ and makes an angle of $30^{\circ }$ with the Y-axis. Because the angle is with the Y-axis, the $x$-component uses sine and the $y$-component uses cosine.

Using $\sqrt{3}=1.7$, we get exactly:

2. Resolving $\overrightarrow{OQ}$: Vector $\overrightarrow{OQ}$ is $10\text{ N}$ at $30^{\circ }$ with the X-axis. This is standard.

3. Resolving $\overrightarrow{OR}$: Vector $\overrightarrow{OR}$ is in the fourth quadrant ($x$ is positive, $y$ is negative). It's $20\text{ N}$ at $45^{\circ }$ with the X-axis.

Since $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$, and we are given $\sqrt{2}=1.4$, the value is $0.7$.

4. Resolving $\overrightarrow{OS}$: Vector $\overrightarrow{OS}$ is in the third quadrant (both $x$ and $y$ are negative). It's $15\text{ N}$ at $45^{\circ }$ with the negative X-axis.

5. Resolving $\overrightarrow{OT}$: Vector $\overrightarrow{OT}$ is in the second quadrant ($x$ is negative, $y$ is positive). Notice the angle is $60^{\circ }$ with the Y-axis.

The Magic of Following Instructions

Now, we simply sum all the $x$-components and all the $y$-components.

Combining these, our final resultant vector is:

Notice how perfectly the numbers aligned to give us an exact match with Option (a). Many students ignore the given approximations and use $1.732$ or $1.414$, ending up with messy decimals like $9.21\hat{i}+5.08\hat{j}$ and having to guess the closest option. By trusting the problem setter and using the exact values provided, we arrived at the pristine answer smoothly. Always read the instructions carefully!