To determine the resultant of the three forces, we must first break down the forces into their vector components along the $x$-axis and $y$-axis. Then, we'll sum the components along both axes to find the resultant vector.

Step 1: Determine the direction of the forces

Given the description, each force acts from the origin toward the opposite corner of a rectangle in different quadrants:

• $P_1$ in the first quadrant: Its components are both positive.
• $P_2$ in the second quadrant: The $x$-component is negative, and the $y$-component is positive.
• $P_3$ in the fourth quadrant: The $x$-component is positive, and the $y$-component is negative.

Step 2: Break each force into components

Assuming that the forces act at 45° angles (which is typical for forces directed along diagonals of rectangles in quadrants):

1. Force $P_1 = 802 \, \text{N}$ in the first quadrant:

   • $P_{1x} = 802 \cos 45^\circ = \frac{802}{\sqrt{2}} = 567 \, \text{N}$
   • $P_{1y} = 802 \sin 45^\circ = \frac{802}{\sqrt{2}} = 567 \, \text{N}$

2. Force $P_2 = 421 \, \text{N}$ in the second quadrant:

   • $P_{2x} = 421 \cos 135^\circ = -\frac{421}{\sqrt{2}} = -298 \, \text{N}$
   • $P_{2y} = 421 \sin 135^\circ = \frac{421}{\sqrt{2}} = 298 \, \text{N}$

3. Force $P_3 = 390 \, \text{N}$ in the fourth quadrant:

   • $P_{3x} = 390 \cos 315^\circ = \frac{390}{\sqrt{2}} = 276 \, \text{N}$
   • $P_{3y} = 390 \sin 315^\circ = -\frac{390}{\sqrt{2}} = -276 \, \text{N}$

Step 3: Sum the components along each axis

• Resultant in the $x$-direction: $R_x = P_{1x} + P_{2x} + P_{3x} = 567 + (-298) + 276 = 545 \, \text{N}$

• Resultant in the $y$-direction: $R_y = P_{1y} + P_{2y} + P_{3y} = 567 + 298 + (-276) = 589 \, \text{N}$

Step 4: Calculate the magnitude of the resultant force

The magnitude of the resultant force is given by: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{545^2 + 589^2} = \sqrt{297025 + 346921} = \sqrt{643946} \approx 802 \, \text{N}$

Step 5: Calculate the direction of the resultant force

The direction (angle $\theta$) of the resultant force relative to the positive $x$-axis is: $\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{589}{545}\right) \approx 47.8^\circ$

Final Result:

• Magnitude of the resultant force: $802 \, \text{N}$
• Direction of the resultant force: $47.8^\circ$

Would you like further clarification on any step?

5 Related Questions:

1. How do force vectors change if forces act at angles other than 45°?
2. What happens if one or more forces are not in the plane but instead in 3D space?
3. How do you resolve forces acting in different directions using the law of cosines?
4. How would the resultant change if the magnitudes of the forces were different?
5. How can you calculate the resultant of more than three forces acting on an object?

Tip:

Always make sure to decompose vectors into components accurately, especially when working with non-standard angles or dimensions!