Part 1 (Problem 005):

We are tasked with finding the tension force $F_T$ in the cable. Since the beam is in equilibrium, we will use the conditions of static equilibrium:

1. Sum of forces in the horizontal and vertical directions must be zero.
2. Sum of torques (moments) about any point must be zero.

Let's choose the pivot point at the wall where the beam is attached (pin connection). This will eliminate the reaction forces at the pin in the torque equation.

Torque Calculation:

We need to ensure the beam is in rotational equilibrium. Let the length of the beam be $L = 5.09 \, \text{m}$.

• Weight of the beam = $435 \, \text{N}$, acting at its center (at $2.545 \, \text{m}$ from the pin).
• The 481 N person stands at a distance of $2.01 \, \text{m}$ from the pin.
• The tension in the cable creates a torque about the pin. The cable is at an angle of $58^\circ$, and the moment arm of the tension is $L = 5.09 \, \text{m}$.

The torque equation around the pin: $\text{Sum of torques} = 0$ Clockwise torque due to the weight of the beam and the person: $\tau_{\text{weight of beam}} = 435 \, \text{N} \times 2.545 \, \text{m}$ $\tau_{\text{weight of person}} = 481 \, \text{N} \times 2.01 \, \text{m}$ Counterclockwise torque due to the tension in the cable (only the vertical component contributes to torque): $\tau_{T} = F_T \times \sin(58^\circ) \times 5.09 \, \text{m}$

Setting the total torque equal to zero: $F_T \times \sin(58^\circ) \times 5.09 = (435 \times 2.545) + (481 \times 2.01)$ Solving for $F_T$: $F_T = \frac{(435 \times 2.545) + (481 \times 2.01)}{\sin(58^\circ) \times 5.09}$

Let me calculate this value.The tension force $F_T$ in the cable is approximately $480.45 \, \text{N}$.

Part 2 (Problem 006):

Now, we need to find the magnitude of the reaction force $R$ exerted on the beam by the wall. Since the beam is in equilibrium, the sum of forces in both the horizontal and vertical directions must also be zero.

• Horizontal forces: The horizontal component of the tension $F_T$ is balanced by the horizontal reaction force at the wall $R_x$.

  $R_x = F_T \cos(58^\circ)$

• Vertical forces: The vertical reaction force $R_y$ at the wall balances the vertical component of the tension and the total weight (beam + person).

  $R_y + F_T \sin(58^\circ) = 435 \, \text{N} + 481 \, \text{N}$

After solving for $R_x$ and $R_y$, we can combine them to find the total reaction force $R$ using:

$R = \sqrt{R_x^2 + R_y^2}$

Let me calculate this for you.The reaction force $R$ exerted on the beam by the wall is approximately $568.73 \, \text{N}$.

Would you like any further details or explanations about these calculations? Here are a few related questions for deeper understanding:

1. What happens if the angle of the cable changes? How does it affect the tension $F_T$?
2. How would the problem change if the person stood closer to the pin?
3. What if the beam was longer or shorter? How would it impact the forces?
4. Can you explain why we neglect the horizontal forces in the torque calculation?
5. What would happen if there was no pin at the wall? How would the system behave?

Tip: In static equilibrium problems, breaking down forces into their horizontal and vertical components can help simplify the problem-solving process.