Questions: Two forces, F1 and F2, act at a point, as shown in ( Figure 1). F1 has a magnitude of 9.80 N and is directed at an angle of α=65.0° above the negative x axis in the second quadrant. F2 has a magnitude of 5.60 N and is directed at an angle of β′=53.7° below the negative x axis in the third quadrant. What is the y component Fy of the resultant force? Express your answer in newtons. Fy=4.4 N What is the magnitude F of the resultant force? Express your answer in newtons. F=8.7 N What is the angle y that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis Express your answer in degrees. γ= degrees

Two forces, F1 and F2, act at a point, as shown in ( Figure 1). F1 has a magnitude of 9.80 N and is directed at an angle of α=65.0° above the negative x axis in the second quadrant. F2 has a magnitude of 5.60 N and is directed at an angle of β′=53.7° below the negative x axis in the third quadrant. What is the y component Fy of the resultant force? Express your answer in newtons. Fy=4.4 N What is the magnitude F of the resultant force? Express your answer in newtons. F=8.7 N What is the angle y that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis Express your answer in degrees. γ= degrees
Transcript text: Two forces, $\vec{F}_{1}$ and $\vec{F}_{2}$, act at a point, as shown in ( Figure 1). $\vec{F}_{1}$ has a magnitude of 9.80 N and is directed at an angle of $\alpha=65.0^{\circ}$ above the negative $x$ axis in the second quadrant. $\vec{F}_{2}$ has a magnitude of 5.60 N and is directed at an angle of $\beta^{\prime}=53.7^{\circ}$ below the negative $x$ axis in the third quadrant. What is the $y$ component $F_{y}$ of the resultant force? Express your answer in newtons. \[ F_{y}=4.4 \mathrm{~N} \] What is the magnitude $F$ of the resultant force? Express your answer in newtons. \[ F=8.7 \mathrm{~N} \] What is the angle $y$ that the resultant force forms with the negative $x$ axis? In this problem, assume that positive angles are measured clockwise from the negative $x$ axis Express your answer in degrees. $\gamma=$ $\square$ degrees
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Solution

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Solution Steps

Step 1: Identify the components of the forces
  • Force \( F_1 \): Magnitude = 80.0 N, Angle = 65.0° above the negative x-axis in the second quadrant.
    • \( F_{1x} = -80.0 \cos(65.0°) \)
    • \( F_{1y} = 80.0 \sin(65.0°) \)
  • Force \( F_2 \): Magnitude = 65.0 N, Angle = 53.7° below the negative x-axis in the third quadrant.
    • \( F_{2x} = -65.0 \cos(53.7°) \)
    • \( F_{2y} = -65.0 \sin(53.7°) \)
Step 2: Calculate the components of each force
  • Force \( F_1 \):
    • \( F_{1x} = -80.0 \cos(65.0°) = -33.8 \) N
    • \( F_{1y} = 80.0 \sin(65.0°) = 72.5 \) N
  • Force \( F_2 \):
    • \( F_{2x} = -65.0 \cos(53.7°) = -38.6 \) N
    • \( F_{2y} = -65.0 \sin(53.7°) = -51.9 \) N
Step 3: Sum the components to find the resultant force
  • Resultant Force \( F_R \):
    • \( F_{Rx} = F_{1x} + F_{2x} = -33.8 + (-38.6) = -72.4 \) N
    • \( F_{Ry} = F_{1y} + F_{2y} = 72.5 + (-51.9) = 20.6 \) N

Final Answer

  • Magnitude of the resultant force:
    • \( F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2} = \sqrt{(-72.4)^2 + (20.6)^2} = 75.3 \) N
  • Angle of the resultant force:
    • \( \theta = \tan^{-1}\left(\frac{F_{Ry}}{F_{Rx}}\right) = \tan^{-1}\left(\frac{20.6}{-72.4}\right) = -15.9° \) (below the negative x-axis)
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