The image presents three statics problems involving forces and angled supports. The task is to analyze each free body diagram and determine the components of the forces acting on the supports A and B. The first problem has a weight of 400N supported by two angled members, A and B, with B inclined at 30 degrees to the vertical. The second problem shows a weight W supported by members A and B, with A at an angle of 40 degrees to the vertical and B horizontal. The third problem depicts a weight W of 160N supported by members A and B, with B at an angle of 50 degrees to the horizontal, and A acting vertically.

1. Problem Description

The image presents three statics problems involving forces and angled supports. The task is to analyze each free body diagram and determine the components of the forces acting on the supports A and B. The first problem has a weight of 400N supported by two angled members, A and B, with B inclined at 30 degrees to the vertical. The second problem shows a weight W supported by members A and B, with A at an angle of 40 degrees to the vertical and B horizontal. The third problem depicts a weight W of 160N supported by members A and B, with B at an angle of 50 degrees to the horizontal, and A acting vertically.

2. Solution Steps

Problem 1:

* Force $W = 400 N$ acts downwards.

* Force B is at 30 degrees to the vertical. The components of B are: $B_x = B \cos(60^\circ) = B \sin(30^\circ)$ and $B_y = B \cos(30^\circ)$.

* Force A is acting horizontally. So $A_x = A$ and $A_y = 0$.

* Equilibrium in the x direction: $A = B_x = B \sin(30^\circ)$

* Equilibrium in the y direction: $B_y = W = B \cos(30^\circ)$

* Therefore, $B = \frac{W}{\cos(30^\circ)} = \frac{400}{\cos(30^\circ)} = \frac{400}{\sqrt{3}/2} = \frac{800}{\sqrt{3}} N$

* And, $A = B \sin(30^\circ) = \frac{800}{\sqrt{3}} \cdot \frac{1}{2} = \frac{400}{\sqrt{3}} N$

Problem 2:

* Force $W$ acts downwards.

* Force A is at 40 degrees to the vertical. The components of A are: $A_x = A \sin(40^\circ)$ and $A_y = A \cos(40^\circ)$.

* Force B is horizontal and acting towards the right. So $B_x = B$ and $B_y = 0$.

* Equilibrium in the x direction: $A_x = B$, i.e., $A \sin(40^\circ) = B$.

* Equilibrium in the y direction: $A_y = W$, i.e., $A \cos(40^\circ) = W$

* Therefore, $A = \frac{W}{\cos(40^\circ)}$

* And $B = A \sin(40^\circ) = \frac{W \sin(40^\circ)}{\cos(40^\circ)} = W \tan(40^\circ)$. If $W$ is provided, A and B can be calculated.

Problem 3:

* Force $W = 160N$ acts downwards.

* Force B is at 50 degrees to the horizontal. The components of B are: $B_x = B \cos(50^\circ)$ and $B_y = B \sin(50^\circ)$.

* Force A is acting vertically upwards. So $A_x = 0$ and $A_y = A$.

* Equilibrium in the x direction: $B_x = 0$, i.e., $B \cos(50^\circ) = 0$. Because $cos(50^\circ) \neq 0$, this implies $B=0$.

* Equilibrium in the y direction: $A + B_y = W$, i.e., $A + B \sin(50^\circ) = W$

* Since $B=0$, we have $A=W=160N$

3. Final Answer

Problem 1: $A = \frac{400}{\sqrt{3}} N$, $B = \frac{800}{\sqrt{3}} N$

Problem 2: $A = \frac{W}{\cos(40^\circ)}$, $B = W \tan(40^\circ)$ where W is the magnitude of the hanging weight.

Problem 3: $A = 160 N$, $B = 0 N$