I only have 2 hours left to submit my answer Problem 623 and

I only have 2 hours left to submit my answer.

Problem 6.23 and pinal B Part A Expres your answer ing three significant figures. Continue

Let x direction be horizontal pointing towards right and y be vertically upwards.

Balancing vertical direction forces,

Ray + Rby = w*4

Balancing horizontal drection forces,

Rax + Rbx = 0

Balancing moments about B,

Rax*(3*sin30) + Ray*(3*cos30 + 4) = (w*4)*(4/2)

Also, by geometry, Rax / Ray = Tan30

Solving these 4 eqns we get,

Rax*(3*sin30) + (Rax / Tan30)*(3*cos30 + 4) = (w*4)*(4/2)

Rax = 0.6188*w

Rbx = -0.6188*w

Ray = 1.072*w

Rby = 2.928*w

Load at A, Ra = sqrt (Rax² + Ray²)

= sqrt [(0.6188*w)² + (1.072*w)²]

Ra = 1.2378*w

Load at B, Rb = sqrt (Rbx² + Rby²)

= sqrt [(-0.6188*w)² + (2.928*w)²]

Rb = 2.993*w

For A, 4 = 1.2378*w or w = 3.2315 kN/m

For B, 8 = 2.993*w or w = 2.672 kN/m

We need to select lower of the two loads. Hence, w = 2.672 kN/m