Question
Two ropes are attached to a skater as sketched in the figure below and exert forces on her as shown. Find the magnitude and direction of the total force exerted by the ropes on the skater if θ1 = 32.0° and θ2 = 53.0°.
what is the magnitude and direction (horizontal) ?
what is the magnitude and direction (horizontal) ?
Answers
You have not provided enough information for us to help you. The magnitudes of the two forces, F1 and F2, are needed. The total force component along the axis from which the angles are measured will be
Fx = F1 cos 32 + F2 cos 53.
Along the perpendicular axis the net force component is
Fy = F1 sin 32 + F2 sin 53.
The magnitude of the resultant force is
F = sqrt(Fx^2 + Fy^2)
and the direction is tan^1 Fy/Fx
Fx = F1 cos 32 + F2 cos 53.
Along the perpendicular axis the net force component is
Fy = F1 sin 32 + F2 sin 53.
The magnitude of the resultant force is
F = sqrt(Fx^2 + Fy^2)
and the direction is tan^1 Fy/Fx
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