To determine the magnitude of the force exerted on the beam due to the weight of the sign, we can analyze the forces in play.

Step 1: Given Information

• The mass of the sign $m = 10.8 \, \text{kg}$.
• The angle between the cable and the beam $\theta = 38.2^\circ$.
• Gravitational acceleration $g = 9.81 \, \text{m/s}^2$.

Step 2: Calculating the weight of the sign

The force due to gravity (weight) on the sign can be calculated as: $W = m \cdot g = 10.8 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 105.948 \, \text{N}$

This weight is supported entirely by the tension in the cable.

Step 3: Resolving the tension force in the cable

The tension in the cable can be resolved into two components:

1. The horizontal component $T_x$.
2. The vertical component $T_y$.

Since the vertical component of the tension in the cable must balance the weight of the sign, we have: $T_y = W = 105.948 \, \text{N}$ From the angle, the vertical component of the tension is related to the total tension $T$ by: $T_y = T \cdot \sin(\theta)$ Thus, we can solve for the total tension $T$: $T = \frac{T_y}{\sin(\theta)} = \frac{105.948 \, \text{N}}{\sin(38.2^\circ)}$ $T = \frac{105.948 \, \text{N}}{0.6188} \approx 171.23 \, \text{N}$

Step 4: Horizontal component of the force on the beam

The horizontal component of the tension $T_x$ can be calculated using: $T_x = T \cdot \cos(\theta)$ Substitute the value of $T$: $T_x = 171.23 \, \text{N} \cdot \cos(38.2^\circ) = 171.23 \, \text{N} \cdot 0.7855 \approx 134.48 \, \text{N}$

Step 5: Conclusion

The magnitude of the force exerted on the beam is 134.48 N in the horizontal direction.

Would you like more details or have any questions about this?

Here are five follow-up questions that expand on the topic:

1. How would the force on the beam change if the angle of the cable were increased?
2. What would happen to the tension if the mass of the sign increased?
3. Can we determine the horizontal force exerted by the wall on the beam?
4. How would the system be affected if the beam were inclined instead of horizontal?
5. How would we calculate the forces if the sign were accelerating downward?

Tip: Always resolve forces into horizontal and vertical components when dealing with angles and tensions!