Asked by Tyler
Two kids are on a beach pulling an inner tube. One is pulling at a force of 45N[N] and the other one pulling at 60N[SW]. What is the net force? (I tried to do it by taking the 60N and making a right angled triangle with it, i got 42.43N north and west, then I tried to combine those results with the 45N north, i ended up with 42, but I'm not sure if it's right, or if i even did it right.)
Answers
Answered by
MathMate
You basically have it right. Just perhaps not too clear about the direction and how to combine the two x- and y- components. Here's how it can be worked out:
Resolve all forces in the x- (east) and y- (north) directions.
F1=45N due north
=(0,45)
F2=60N due SW
=(-60cos(45°),-60sin(45°))
=(-42.4264,-42.4264) (approx.)
Net force
= F1+F2
= (0,45) + (-42.4264,-42.4264)
=(-42.4264,2.5736)
=√(-42.4264²+2.5736²) in the direction of atan2(2.5736,-42.4264)
=42.5N at S86.53°W
Resolve all forces in the x- (east) and y- (north) directions.
F1=45N due north
=(0,45)
F2=60N due SW
=(-60cos(45°),-60sin(45°))
=(-42.4264,-42.4264) (approx.)
Net force
= F1+F2
= (0,45) + (-42.4264,-42.4264)
=(-42.4264,2.5736)
=√(-42.4264²+2.5736²) in the direction of atan2(2.5736,-42.4264)
=42.5N at S86.53°W
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