Posted by **MAdason** on Tuesday, October 6, 2009 at 9:58am.

Five forces act on an object: 1: 60 N at 90 degrees, 2: 40 N at 0 degrees, 3: 80 N at 270 degrees, 4: 40 N at 180 degrees, and 5: 50 N at 60 degrees. What is the magnitude and direction of the 6th froce that would produce equilibrium?

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**drwls**, Tuesday, October 6, 2009 at 10:12am
The vector sum of the six forces must be zero. Write that in the form of two equations (one each for the x and y components). Then solve for the unknown components of the sixth force.

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**Anonymous**, Monday, October 4, 2010 at 9:07pm
hkjh

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**Teklu**, Thursday, December 30, 2010 at 2:18am
At a point forces of 60N and 90N are acting and the angle between them is 60 degree. Determine the magnitude and the direction of their Resultant by graphical method

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**Anonymous**, Thursday, November 17, 2011 at 9:18pm
-30 N

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**Henry**, Saturday, August 8, 2015 at 12:20pm
Fr = 60[90o] + 40[0o] + 80[270o] + 40[180] + 50[60o].

X=60*Cos90+40*Cos0+80*Cos270+40*Cos180+50*Cos60 = 0 + 40 + 0 - 40 + 25 = 25 N.

Y=60*sin90+40*sin0+80*sin270+50*sin60 =

60 + 0 - 80 + 0 + 43.3 = 23.3 N.

Fr = X+Yi = 25 + 23.3i.

Fe = -25 - 23.3i = 34.2N[43.2o+180] =

34.2N[223.2o].

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