Asked by allan
What is the direction of the net external force
on the crate (as an angle between −180◦
and
180◦
, measured from the positive x axis with
counterclockwise positive)?
Answer in units of ◦
Ive solved for the net external force in the x and y direction.
Fx=-23.3N
Fy=409.5N
I know i have to use the arc Tangent but i tried all the ways with these two numbers and nothing worked.
on the crate (as an angle between −180◦
and
180◦
, measured from the positive x axis with
counterclockwise positive)?
Answer in units of ◦
Ive solved for the net external force in the x and y direction.
Fx=-23.3N
Fy=409.5N
I know i have to use the arc Tangent but i tried all the ways with these two numbers and nothing worked.
Answers
Answered by
Henry
Fx = -23.3 N.
Fy = 409.5 N.
2nd Quadrant.
Tan Ar = Fy/Fx = 409.5/-23.3 = -17.57511
Ar = -86.7o = Reference angle(Q4).
A = -86.7 + 180 = 93.3o, CCW(Q2) = Direction.
Fy = 409.5 N.
2nd Quadrant.
Tan Ar = Fy/Fx = 409.5/-23.3 = -17.57511
Ar = -86.7o = Reference angle(Q4).
A = -86.7 + 180 = 93.3o, CCW(Q2) = Direction.
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