Posted by **Jacob** on Monday, November 16, 2009 at 3:56pm.

The following forces act on a hockey puck (round, rubber disk) sitting on a frictionless surface: F_1 = 12 N at 15 degrees; F_2 = 28 N at 125 degrees; and F_3 = 33.3 N at 235 degrees. All the forces are in the plane of the ice. Determine the net force on the puck.

- physics -
**bobpursley**, Monday, November 16, 2009 at 4:11pm
Net force= sum of all.

First, net force at 0 degrees

F@0=12cos15+28cos125+33.3cos235

calculate those, and add.

Then, the forces at 90 deg

F@90=12sin15+28sin125+33.3sin235

calculate, and sum.

Net force= sum of those two forces.

I like to do it this way.

1) angle=arctan F@0/F@90

2) magnitude= F@0/cos(angle)

check that logic.

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