problem 3.9 - find the reactions at supports a and b. the weight of the rod is negligible.

Explanation:

Step1: Take moment about A

Let the reaction at B be $R_B$. The moment - equilibrium equation $\sum M_A = 0$. The 3 - kip load creates a clock - wise moment and the reaction at B creates a counter - clockwise moment. $3\times4=R_B\times(4 + 4)$

Step2: Solve for $R_B$

From the equation $12 = 8R_B$, we can find $R_B$ by dividing both sides of the equation by 8. $R_B=\frac{12}{8}=1.5$ kip

Step3: Use vertical force equilibrium

Let the vertical reaction at A be $R_{Ay}$. The sum of vertical forces $\sum F_y=0$. $R_{Ay}+R_B - 3=0$

Step4: Solve for $R_{Ay}$

Substitute $R_B = 1.5$ kip into the equation $R_{Ay}+1.5 - 3=0$. Then $R_{Ay}=3 - 1.5 = 1.5$ kip. Since there are no horizontal forces acting on the structure, the horizontal reaction at A, $R_{Ax}=0$.

Answer:

Reaction at A: $R_{Ax}=0$ kip, $R_{Ay}=1.5$ kip; Reaction at B: $R_B = 1.5$ kip