Determine the distance at which the lead $Q = 150N$ should be placed from point B to keep the bar horizontal

The bar is supported by two plane surfaces. Plane surfaces can give only normal reaction (reaction perpendicular to the surface.)

Hence, the FBD can be drawn as follows

Given: the bar should be horizontal. It means that net moment of the system should be zero. This condition will help us determine X. But before that we will need to find values of reactions at A and B using COE. Let reactions at A and B be Ra and Rb.

Since, Ra and Rb are perpendicular to planes their angles will be (90 – 30) and (90-50) respectively.

$∑Fx =0 \\ Ra \cos60 – Rb \cos40 = 0 … i \\ and\space \space ∑Fy = 0 \\ Ra \sin60 –P – Q + Rb \sin40 = 0 \\ Ra \sin60 + Rb \sin40 = 400 .. ii \\ $

Solving (i) and (ii)

$Ra = 311N $ and $Rb = 203N $

Taking third COE, $∑M = 0$ (anticlockwise positive) about point B (because X is distance of Q from B)

$$Q \times X + P \times 2.5 – Ra \sin60 \times 3 = 0 $$

$150 \times X + 250 \times 2.5 – 311 \sin60 \times 3 = 0\\ X = 1.22m$