What is the net force acting on the ring in the figure below?

Question

Clink on the link to see the graph

http://i277.photobucket.com/albums/kk80/mitsuki_chan_pf_fullmoon/graph.jpg?t=1218425705

Answer 1

Consider horizontal as x axis with towards right as positive and vertical as y axis with towards up as positive.

Force along x axis Fx = 64 + 128 cos 30 deg - 128 = 64 + 128 * sqrt(3/2) - 128

= 46.85 N

Force along y axis Fy = 128 sin 30 deg = 128/2 = 64 N

Total force F = sqrt(Fx^2 + Fy^2) = sqrt(46.84^2 + 64^2) = sqrt(6291) = 79.3 N

Angle made by with x axis = tan-1(Fy/Fx) = tan-1(64/46.85) = 54 deg

Ans: 79.3 N at 54 deg with x axis.

Answer 2

in X:

64 - 128 + 128cos(30°) =

64 - 128 + 110.85         = 46.85 N   EASTwards

in Y:

128sin(30°) = 64 N  NORTHwards

result:

a² + b²  = c²                   sqrt(c²) = c

64² + 46.85² =

4096 + 2194.92 = 6290.92

sqrt(6290.92) = 79.31 N      module

orientation:  Arctan(64/46.85)= 53.79471 degrees

Answer 3

just figure out north/south and east/west. its 64N north and 46.85N east, so that would be 79.31N at 53.79degrees north of east.

make it a good day.