om he pin reactions at all of the pins 250 lbs 100 lbs 6 6 S

om he pin reactions at all of the pins. 250 lbs 100 lbs 6\' 6

Solution

Let us consider member BC. Let the force at C on member BC be Cx and Cy in x and y direction respectively

From moment equilibrium about B,we get

250*8+4*Cx=0

Cx=-250*8/4=-500 lb=500 lb(towards left)

Therefore, if we consider member AC,force Cx acts towards right

let us consider moment equilibrium of member AC about A

100*6+500*4-12*Cy=0

Cy=216.667 lbs(upwaqrds)

From vertical equilibrium of AC,A_y=216.667-100=116.667 lbs(upwards)

from horizontal equilibrium of AC, Ax=500 lbs(towards left)

From vertical equilibrium of BC, By=216.667 lb(vertically upwards)

From horizontal equilibrium of BC, Bx=250 lbs(towards right)

Therefore, to summarise

Ax=500 lbs; Ay=116.667 lbs; Resultant = sqrt(500²+116.667²)=513.43 lbs

Bx=250 lbs; By=216.667 lbs; Resultant = sqrt(250²+216.667²)=330.82lbs

Cx=500 lbs; Cy=216.667 lbs; Resultant = sqrt(500²+216.667²)=544.93 lbs