**THREE FORCES 150 n AT 0 DEGREES, 200 N AT 90 DEGREES, AND 100 N AT 210 DEGREES ACT ON A POINT. FIND THE RESULTANT FORCE.**

To find the resultant force when multiple forces act on a point, we need to resolve each force into its horizontal (x) and vertical (y) components, sum these components, and then use the Pythagorean theorem and trigonometry to determine the magnitude and direction of the resultant force.

### Given Forces:
1. \( F_1 = 150 \, \text{N} \) at \( 0^\circ \)
2. \( F_2 = 200 \, \text{N} \) at \( 90^\circ \)
3. \( F_3 = 100 \, \text{N} \) at \( 210^\circ \)

### Step 1: Resolve Each Force into Components

**For \( F_1 = 150 \, \text{N} \) at \( 0^\circ \):**
- \( F_{1x} = 150 \, \text{N} \times \cos(0^\circ) = 150 \, \text{N} \)
- \( F_{1y} = 150 \, \text{N} \times \sin(0^\circ) = 0 \, \text{N} \)

**For \( F_2 = 200 \, \text{N} \) at \( 90^\circ \):**
- \( F_{2x} = 200 \, \text{N} \times \cos(90^\circ) = 0 \, \text{N} \)
- \( F_{2y} = 200 \, \text{N} \times \sin(90^\circ) = 200 \, \text{N} \)

**For \( F_3 = 100 \, \text{N} \) at \( 210^\circ \):**
- \( F_{3x} = 100 \, \text{N} \times \cos(210^\circ) = 100 \, \text{N} \times (-\sqrt{3}/2) = -86.60 \, \text{N} \)
- \( F_{3y} = 100 \, \text{N} \times \sin(210^\circ) = 100 \, \text{N} \times (-1/2) = -50 \, \text{N} \)

### Step 2: Sum the Components

**Sum of the horizontal components (\( F_x \)):**
\[
F_x = F_{1x} + F_{2x} + F_{3x} = 150 \, \text{N} + 0 \, \text{N} - 86.60 \, \text{N} = 63.40 \, \text{N}
\]

**Sum of the vertical components (\( F_y \)):**
\[
F_y = F_{1y} + F_{2y} + F_{3y} = 0 \, \text{N} + 200 \, \text{N} - 50 \, \text{N} = 150 \, \text{N}
\]

### Step 3: Calculate the Magnitude of the Resultant Force

\[
R = \sqrt{F_x^2 + F_y^2} = \sqrt{(63.40 \, \text{N})^2 + (150 \, \text{N})^2}
\]
\[
R = \sqrt{4025.16 \, \text{N}^2 + 22500 \, \text{N}^2} = \sqrt{26525.16 \, \text{N}^2} \approx 162.86 \, \text{N}
\]

### Step 4: Calculate the Direction of the Resultant Force

\[
\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{150 \, \text{N}}{63.40 \, \text{N}}\right)
\]
\[
\theta = \tan^{-1}(2.366) \approx 67.7^\circ
\]

### Resultant Force
- Magnitude: \( R \approx 162.86 \, \text{N} \)
- Direction: \( \theta \approx 67.7^\circ \) (measured counterclockwise from the positive x-axis)

Would you like further details or have any questions?

Here are five related questions you might find interesting:
1. What happens to the resultant force if \( F_1 \) is increased?
2. How does the angle between two forces affect the resultant force?
3. What is the effect of changing \( F_2 \) to an angle of 120°?
4. How do you determine the direction of the resultant force when all forces have different directions?
5. Can a resultant force be zero if three forces act on a point?

**Tip:** Always resolve forces into their components before summing them up, as it simplifies the calculation of the resultant force.

Learn how to calculate the resultant force when three forces (150N at 0°, 200N at 90°, and 100N at 210°) act on a point. This problem involves vector addition, trigonometry, and the Pythagorean theorem to determine the magnitude and direction of the resultant force. Suitable for high school students studying physics or mechanics.

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