A two member frame is pin connected at O, Y and X (as shown in figure). The cable is attached to W, passes over smooth peg at Z and is attached to load of 500 N. Determine vertical components of reactions at pin O, Y and X.

This question was previously asked in

UPRVUNL AE ME 2014 Official Paper

Option 1 : 1000 N, -500 N, 1000 N

At joint Z for equilibrium

ΣF_{y }= 0

F_{OZ} cos 56.3 = 500 N

F_{OZ} = 901 N

Now ΣF_{x} = 0

F_{OZ} sin 56.3 = F_{wz}

F_{wz} = 750 N

Now for R_{x} and R_{y}

From Vertical Equilibrium of the whole frame

R_{x} + R_{y} = 500 N

From moment equilibrium ΣM about pin O = Zero

R_{x }× 1 = 500 × 3 + R_{y} × 1 (the moment due to cable will cancel out each other)

R_{x} = 1500 + R_{y}

⇒ R_{x} = 1500 + 500 - R_{x}

⇒ 2 R_{x} = 2000 N.m

**⇒ ****R _{x }= 1000 N**

**∴**** R _{y }= 500 N – 1000 N = - 500 N **

Now for reaction at point O

Force in member OW and OZ = 901 N

At joint O,

Since the members OW and OZ are pushing the joint O

⇒ 901 cos 56.3 + 901 cos 56.3 = 1000 N

**So the reaction at O will be acting vertically upwards 1000 N**.