A body is at equilibrium under the action of three forces. One force is 10N acting due east and one is 5N in the direction 60° north east. What is the magnitude and direction of the third force?​

If the body is in equilibrium the two forces add up and the third is the opposite of the resultant.

F(1x)=F(1)=10 N

F(2x)=F(2)cos60=5•0.5=2.5 N

F(2y) =F(2)sin60 = 5•0.866= 4.33 N

F(3x) =- F(x)=- (10+2.5 )= -12.5 N

F(3y) =- F(2y)= - 4.33 N

F(3) = sqrt{ F(3x)²+F(3y)²} =13.23 N

tan φ = F(3y)/F(3x) =4.33/12.5=0.364

φ = 19.1⁰ (south-west)

Explanation:

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