Math Problem Statement

Determine the magnitude of the resultant force and its direction, measured clockwise from the negative x-axis, for forces F1 = 200 N at 45 degrees and F2 = 150 N at 30 degrees.

Solution

The problem asks to determine the magnitude of the resultant force and its direction (measured clockwise from the negative x-axis) for two given forces, F1=200NF_1 = 200 \, \text{N} and F2=150NF_2 = 150 \, \text{N}, applied at angles of 4545^\circ and 3030^\circ, respectively.

Step 1: Resolve the forces into their components

Force F1=200NF_1 = 200 \, \text{N} acts at 00^\circ from the positive x-axis:

  • F1x=200Ncos(0)=200NF_{1x} = 200 \, \text{N} \cdot \cos(0^\circ) = 200 \, \text{N}
  • F1y=200Nsin(0)=0NF_{1y} = 200 \, \text{N} \cdot \sin(0^\circ) = 0 \, \text{N}

Force F2=150NF_2 = 150 \, \text{N} acts at 3030^\circ below the negative x-axis (angle measured clockwise):

  • F2x=150Ncos(30)=150×0.866=129.9NF_{2x} = 150 \, \text{N} \cdot \cos(30^\circ) = 150 \times 0.866 = 129.9 \, \text{N} in the negative x direction.
  • F2y=150Nsin(30)=150×(0.5)=75NF_{2y} = 150 \, \text{N} \cdot \sin(30^\circ) = 150 \times (-0.5) = -75 \, \text{N}

Step 2: Calculate the total components of the resultant force

  • FRx=F1x+F2x=200N129.9N=70.1NF_{\text{Rx}} = F_{1x} + F_{2x} = 200 \, \text{N} - 129.9 \, \text{N} = 70.1 \, \text{N}
  • FRy=F1y+F2y=0N75N=75NF_{\text{Ry}} = F_{1y} + F_{2y} = 0 \, \text{N} - 75 \, \text{N} = -75 \, \text{N}

Step 3: Calculate the magnitude of the resultant force

The magnitude of the resultant force FRF_R is found using the Pythagorean theorem: FR=(FRx)2+(FRy)2F_R = \sqrt{(F_{\text{Rx}})^2 + (F_{\text{Ry}})^2} Substituting the values: FR=(70.1)2+(75)2=4914.01+5625=10539.01102.66NF_R = \sqrt{(70.1)^2 + (-75)^2} = \sqrt{4914.01 + 5625} = \sqrt{10539.01} \approx 102.66 \, \text{N}

Step 4: Determine the direction of the resultant force

The direction θR\theta_R is given by: θR=tan1(FRyFRx)\theta_R = \tan^{-1} \left( \frac{F_{\text{Ry}}}{F_{\text{Rx}}} \right) θR=tan1(7570.1)46.07\theta_R = \tan^{-1} \left( \frac{-75}{70.1} \right) \approx -46.07^\circ Since the angle is negative, it lies in the fourth quadrant (below the x-axis). To express the angle measured clockwise from the negative x-axis, we adjust: θR=180+(46.07)=133.93\theta_R = 180^\circ + (-46.07^\circ) = 133.93^\circ

Final Answer:

  • The magnitude of the resultant force is approximately 102.66N102.66 \, \text{N}.
  • The direction of the resultant force is approximately 133.93133.93^\circ, measured clockwise from the negative x-axis.

Would you like more details or clarifications on this? Here are five related questions you might explore:

  1. How would the solution change if F1F_1 had a different angle?
  2. What happens to the resultant force if both F1F_1 and F2F_2 are in the same direction?
  3. How can vector addition be graphically represented for this problem?
  4. What is the significance of resolving vectors into their components?
  5. How would the result change if the forces were in three dimensions?

Tip: Always sketch the problem and resolve vectors into components for easier analysis.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Trigonometry
Force Components
Pythagorean Theorem

Formulas

F_R = sqrt((F_Rx)^2 + (F_Ry)^2)
F_Rx = F1x + F2x
F_Ry = F1y + F2y
θ_R = tan^(-1)(F_Ry / F_Rx)

Theorems

Pythagorean Theorem
Trigonometric Functions

Suitable Grade Level

College-level Physics or Advanced High School Physics