The direction of the pull is -70.93°, or approximately 71° South of West.
Given the vector as F= (−2980.0i^ + 8200.0j^)N
The magnitude of the pull is given by the formula below:
Magnitude of pull, |F| = √(F_x^2 + F_y^2)
Here, F_x is the force in the x-direction and F_y is the force in the y-direction. We know that:
F_x = -2980.0 N and F_y = 8200.0 N
Therefore, the magnitude of the pull is:
|F| = √(F_x^2 + F_y^2)|F| = √((-2980.0)^2 + (8200.0)^2)|F| = √(88840000) |F| = 9425.89 N
The direction of the pull can be found as follows:
θ = tan⁻¹(F_y/F_x)θ = tan⁻¹(8200.0/-2980.0)θ = -70.93°
The direction of the pull is -70.93°, or approximately 71° South of West.
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