2d rigid body equilibrium example determine the reactions at a and b on the beam.

Explanation:

Step1: Set up force - equilibrium equations

Let the horizontal reaction at A be $A_x$, the vertical reaction at A be $A_y$, and the force in the link BC be $F_{BC}$. In the x - direction, $\sum F_x=A_x - F_{BC}\cos\theta = 0$. In the y - direction, $\sum F_y=A_y+F_{BC}\sin\theta-600 - 800=0$. First, find $\theta$. $\tan\theta=\frac{1.5}{1}$, so $\theta=\arctan(1.5)\approx56.31^{\circ}$.

Step2: Set up moment - equilibrium equation

Take the moment about point A, $\sum M_A=-600\times1 - 800\times(1 + 2)-900+F_{BC}\sin\theta\times1 = 0$. [ \begin{align*} -600-2400 - 900+F_{BC}\sin(56.31^{\circ})\times1&=0\ -3900+F_{BC}\times0.832&=0\ F_{BC}&=\frac{3900}{0.832}\approx4687.5\text{ N} \end{align*} ]

Step3: Solve for $A_x$ and $A_y$

From $\sum F_x = 0$, $A_x=F_{BC}\cos\theta=4687.5\times\cos(56.31^{\circ})\approx2600\text{ N}$. From $\sum F_y = 0$, $A_y=600 + 800 - F_{BC}\sin\theta=1400-4687.5\times0.832=1400 - 3900=-2500\text{ N}$.

The reaction at B is the force in the link BC. The magnitude of the reaction at B is $F_{BC}\approx4687.5\text{ N}$ along the link BC.

Answer:

$A_x = 2600\text{ N}$, $A_y=-2500\text{ N}$, the reaction at B has a magnitude of approximately $4687.5\text{ N}$ along the link BC.