In a football game, a quarterback (G), hit simultaneously by two linebackers, experiences horizontal forces of 412 N [27.0° W of N] and 478 N [36.0° N of E]. Determine the net force of the quarterback.
break the two forces into N, E components, and add.
I don't completely understand
So you want me to split the two forces into North and East then add it
because when I do that I get a different answer than there is
To determine the net force experienced by the quarterback, we need to analyze the horizontal components and vertical components separately. The net force is the vector sum of these components.
Step 1: Separate the given forces into their horizontal and vertical components.
Force 1: 412 N [27.0° W of N]
- Horizontal component (H1): 412 N * cos(27.0°)
- Vertical component (V1): 412 N * sin(27.0°)
Force 2: 478 N [36.0° N of E]
- Horizontal component (H2): 478 N * sin(36.0°)
- Vertical component (V2): 478 N * cos(36.0°)
Step 2: Determine the total horizontal and vertical components.
Horizontal component (H): H1 - H2
Vertical component (V): V1 + V2
Step 3: Calculate the magnitude and direction of the net force using the total horizontal and vertical components.
Magnitude of the net force = √(H^2 + V^2)
Direction of the net force = tan^(-1)(V/H)
Let's calculate the values:
H1 = 412 N * cos(27.0°) ≈ 365.67 N
V1 = 412 N * sin(27.0°) ≈ 190.16 N
H2 = 478 N * sin(36.0°) ≈ 285.31 N
V2 = 478 N * cos(36.0°) ≈ 385.14 N
H = H1 - H2 ≈ 365.67 N - 285.31 N ≈ 80.36 N
V = V1 + V2 ≈ 190.16 N + 385.14 N ≈ 575.3 N
Magnitude of the net force = √(H^2 + V^2) ≈ √(80.36^2 + 575.3^2) ≈ √(6462.77 N^2 + 331268.09 N^2) ≈ √(337730.86 N^2) ≈ 580.88 N
Direction of the net force = tan^(-1)(V/H) ≈ tan^(-1)(575.3 N / 80.36 N) ≈ 82.36°
Therefore, the net force experienced by the quarterback is approximately 580.88 N at an angle of 82.36° relative to the horizontal direction.