To determine the reaction components at the fixed support $A$ (neglecting the weight of the beam), we need to resolve the forces into their components and apply equilibrium equations. Here's how we'll proceed:

Step 1: Resolve the 600 N Force

The 600 N force is applied at an angle of $45^\circ$. Break it into horizontal and vertical components:

$F_x = 600 \cos(45^\circ) = 600 \cdot 0.707 = 424.2 \, \text{N}$ $F_y = 600 \sin(45^\circ) = 600 \cdot 0.707 = 424.2 \, \text{N}$

Step 2: Equilibrium Conditions

At support $A$, there are three unknowns:

1. $A_x$: Horizontal reaction force
2. $A_y$: Vertical reaction force
3. $M_A$: Moment reaction at $A$

The beam is in static equilibrium, so the following conditions hold:

1. $\Sigma F_x = 0$ (Sum of horizontal forces is zero)
2. $\Sigma F_y = 0$ (Sum of vertical forces is zero)
3. $\Sigma M_A = 0$ (Sum of moments about point $A$ is zero)

Step 3: Apply $\Sigma F_x = 0$

The horizontal forces are:

• $A_x$: Reaction at $A$ in the horizontal direction
• $424.2 \, \text{N}$: Horizontal component of the 600 N force

$A_x - 424.2 = 0$ $A_x = 424.2 \, \text{N}$

Step 4: Apply $\Sigma F_y = 0$

The vertical forces are:

• $A_y$: Reaction at $A$ in the vertical direction
• $1200 \, \text{N}$: Downward force
• $424.2 \, \text{N}$: Vertical component of the 600 N force

$A_y - 1200 - 424.2 = 0$ $A_y = 1624.2 \, \text{N}$

Step 5: Apply $\Sigma M_A = 0$

Take moments about point $A$. Clockwise moments are positive. The moment contributions are:

1. The 1200 N force at 2 m: $M_1 = 1200 \cdot 2 = 2400 \, \text{N·m (clockwise)}$
2. The vertical component of the 600 N force ($424.2 \, \text{N}$) at 6 m: $M_2 = 424.2 \cdot 6 = 2545.2 \, \text{N·m (clockwise)}$
3. The horizontal component of the 600 N force ($424.2 \, \text{N}$) does not create a moment about $A$ because its line of action passes through $A$.

The total moment about $A$ must be zero: $M_A - M_1 - M_2 = 0$ $M_A - 2400 - 2545.2 = 0$ $M_A = 4945.2 \, \text{N·m}$

Final Results

1. $A_x = 424.2 \, \text{N}$ (horizontal reaction)
2. $A_y = 1624.2 \, \text{N}$ (vertical reaction)
3. $M_A = 4945.2 \, \text{N·m}$ (moment reaction)

Would you like further clarification or more detailed steps? Let me know if you'd like to explore additional concepts!

Related Questions

1. How do we calculate reactions if the beam has a distributed load instead of point forces?
2. How would the results change if the beam's weight was included in the problem?
3. Can you explain how to analyze beams with multiple supports instead of a fixed support?
4. What are the differences between pinned, roller, and fixed supports in terms of reactions?
5. How do we determine the center of gravity for non-uniform beams?

Tip:

Always check the direction and location of forces carefully when breaking them into components—small mistakes here can lead to incorrect results!