Question

Three forces F, root 2F and 2F are acting at 0 ,45 and 120 from x axis on a body.
What force should be applied o it to keep it in equilibrium?

Answer 1

Answer:

Let us denote the resultant of these 3 forces by R. The X- component of the resultant = Rx= F + F√2 •Cos45°+ 2FCos120° =F+F√2 • 1/√2 -2F• 1/2 =F+F-F =F

Ycomponent of the resultant= 0+F√2 • Sin45°+2F Sin120°= F√2• 1/√2 +2F •√3/2= F+F√3

Hence the resultant R= √( F^2+{F+F√3}^2) =√(F^2+F^2{4+2√3}) =F [1+4+2√3]^1/2 = F√{5+2√3)

To keep it in equilibrium, we have to exert a force whose magnitude is F √(5+2√3) & direction opposite to R.

Direction of R is tan theta = Ycomponent/Component = (F+F√3)/F = 1+√3

Hence the direction of the force to be applied= tan^-1 {1+√3} and the magnitude= F√{5+2√3}

Step-by-step explanation: