The image contains a physics problem asking for the magnitude of the resultant vector for a system of forces. Here’s what I can infer from the image:

The forces shown in the diagram are:

• 50 N at 0° (along the positive x-axis),
• 60 N at 90° (along the positive y-axis),
• 70 N at 120° (measured counterclockwise from the positive x-axis).

To determine the magnitude of the resultant vector, we use vector addition:

1. Resolve each force into its x and y components:

   • $F_1 = 50 \, \text{N}$ at $0^\circ$:
     $F_{1x} = 50 \, \cos(0^\circ) = 50 \, \text{N}, \, F_{1y} = 50 \, \sin(0^\circ) = 0 \, \text{N}$
   • $F_2 = 60 \, \text{N}$ at $90^\circ$:
     $F_{2x} = 60 \, \cos(90^\circ) = 0 \, \text{N}, \, F_{2y} = 60 \, \sin(90^\circ) = 60 \, \text{N}$
   • $F_3 = 70 \, \text{N}$ at $120^\circ$:
     $F_{3x} = 70 \, \cos(120^\circ) = -35 \, \text{N}, \, F_{3y} = 70 \, \sin(120^\circ) = 60.62 \, \text{N}$

2. Sum up the components:

   • Total $F_x = F_{1x} + F_{2x} + F_{3x} = 50 + 0 - 35 = 15 \, \text{N}$
   • Total $F_y = F_{1y} + F_{2y} + F_{3y} = 0 + 60 + 60.62 = 120.62 \, \text{N}$

3. Calculate the magnitude of the resultant force using the Pythagorean theorem: $F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{(15)^2 + (120.62)^2} = \sqrt{225 + 14549.18} \approx 121.54 \, \text{N}$

Thus, the magnitude of the resultant force is approximately 121.54 N.

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Related Questions:

1. What is the direction of the resultant force?
2. How does vector resolution help in finding the resultant of multiple forces?
3. Can we calculate the resultant force using graphical methods?
4. How would the result change if one of the forces had a different magnitude or direction?
5. What is the significance of angles in vector addition?

Tip: When solving vector addition problems, always resolve each vector into x and y components to simplify the process.