We are given the problem of hanging a beam and sign using a cable with a breaking strength of 417 N. To find the minimum angle $\theta$ between the beam and the cable, we need to consider the forces and torques acting on the system. Here's a breakdown of the problem:

Known Values:

• Breaking strength of cable $T_{\text{max}} = 417 \, \text{N}$
• Weight of the sign $W_{\text{sign}} = 237 \, \text{N}$
• Weight of the beam $W_{\text{beam}} = 79.8 \, \text{N}$
• Length of the beam $L_{\text{beam}} = 1.50 \, \text{m}$
• Sign's horizontal dimension $d_{\text{sign}} = 1.00 \, \text{m}$
• Sign's vertical dimension $h_{\text{sign}} = 0.800 \, \text{m}$

Step 1: Forces Acting on the Beam

The forces acting on the system are:

• The tension $T$ in the cable at some angle $\theta$.
• The weight of the sign $W_{\text{sign}} = 237 \, \text{N}$, applied at the end of the beam.
• The weight of the beam $W_{\text{beam}} = 79.8 \, \text{N}$, applied at its center (since the beam is uniform).
• The reaction force at the wall $F_{\text{wall}}$, which has both horizontal and vertical components, $F_{\text{wall,x}}$ and $F_{\text{wall,y}}$, respectively.

Step 2: Torque Calculation

To maintain equilibrium, the sum of the torques about the pivot point (where the beam is attached to the wall) must be zero. We'll assume counterclockwise torques as positive and clockwise torques as negative.

The torque due to the weight of the beam is:

= 79.8 \, \text{N} \times \frac{1.50 \, \text{m}}{2} = 59.85 \, \text{N} \cdot \text{m}$$ The torque due to the weight of the sign is: $$\tau_{\text{sign}} = W_{\text{sign}} \times L_{\text{beam}} = 237 \, \text{N} \times 1.50 \, \text{m} = 355.5 \, \text{N} \cdot \text{m}$$ The torque due to the tension in the cable must balance these torques. If the cable is attached at the end of the beam, the perpendicular distance from the pivot to the cable is $$L_{\text{beam}} \sin \theta$$. The torque due to the tension is: $$\tau_{\text{tension}} = T \times L_{\text{beam}} \sin \theta = T \times 1.50 \, \text{m} \times \sin \theta$$ ### Step 3: Setting Up the Torque Equilibrium For rotational equilibrium, the sum of the torques must be zero: $$T \times 1.50 \, \text{m} \times \sin \theta = 59.85 \, \text{N} \cdot \text{m} + 355.5 \, \text{N} \cdot \text{m}$$ $$T \times 1.50 \, \text{m} \times \sin \theta = 415.35 \, \text{N} \cdot \text{m}$$ $$T \times \sin \theta = \frac{415.35 \, \text{N} \cdot \text{m}}{1.50 \, \text{m}} = 276.9 \, \text{N}$$ ### Step 4: Solving for $$\theta$$ The tension in the cable cannot exceed the breaking strength, so: $$T \leq T_{\text{max}} = 417 \, \text{N}$$ Substitute $$T = 417 \, \text{N}$$ into the equation: $$417 \times \sin \theta = 276.9 \, \text{N}$$ $$\sin \theta = \frac{276.9}{417} = 0.664$$ $$\theta = \arcsin(0.664) \approx 41.66^\circ$$ ### Final Answer: The minimum angle $$\theta$$ between the beam and the cable is approximately $$41.66^\circ$$. --- Would you like further clarification or details on any part of the solution? ### Related Questions: 1. What would happen if the cable's breaking strength was lower than 417 N? 2. How would the angle $$\theta$$ change if the sign's weight was increased? 3. What forces act on the wall in this setup? 4. How would the problem change if the beam was not uniform? 5. How does the position of the sign affect the tension in the cable? ### Tip: Always check the breaking strength of materials in tension-related problems to avoid failures due to overloading!