An 8.00 m long beam with a mass of 8.70 kg is suspended horizontally from a post, as shown. The left end of the beam is attached to a frictionless hinge on the vertical post that allows the beam to rotate up and down freely. The right end of the beam is supported by a wire that runs to the top of the post. The wire makes a 50° angle with the vertical post. The beam's shape is symmetric and it is made of a homogeneous material so that its center of mass is located at its geometric center. What tension in the wire is required to support the beam horizontally?

To determine the tension in the wire required to support the beam horizontally, we can analyze the beam's forces and torques in equilibrium.

### Given:
- Length of the beam, \( L = 8.00 \, \text{m} \)
- Mass of the beam, \( m = 8.70 \, \text{kg} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)
- Angle of the wire with the vertical, \( \theta = 50^\circ \)

### Solution Steps:

1. **Calculate the weight of the beam**:
   \[
   W = mg = 8.70 \times 9.8 = 85.26 \, \text{N}
   \]

2. **Identify the point of torque rotation**:
   We'll take torques about the hinge at the left end of the beam. Since the beam is in equilibrium, the sum of torques about this point must be zero.

3. **Determine the torque due to the beam’s weight**:
   The center of mass of the beam is at its midpoint (4.00 m from the hinge). The torque due to the beam's weight is:
   \[
   \tau_{\text{weight}} = W \cdot \text{distance} = 85.26 \times 4.00 = 341.04 \, \text{Nm}
   \]
   This torque is directed clockwise.

4. **Determine the torque due to the tension in the wire**:
   Let \( T \) be the tension in the wire. The wire makes a 50° angle with the vertical, so the horizontal component of the tension provides the torque:
   \[
   \tau_{\text{tension}} = T \sin(50^\circ) \times L
   \]
   Since the length \( L = 8.00 \, \text{m} \), the torque becomes:
   \[
   \tau_{\text{tension}} = T \sin(50^\circ) \times 8.00
   \]

5. **Set up the torque equilibrium equation**:
   For the beam to be in equilibrium, the counterclockwise torque due to the tension must equal the clockwise torque due to the weight of the beam:
   \[
   T \sin(50^\circ) \times 8.00 = 341.04
   \]

6. **Solve for \( T \)**:
   \[
   T = \frac{341.04}{8.00 \times \sin(50^\circ)}
   \]
   Calculating \( \sin(50^\circ) \approx 0.7660 \),
   \[
   T = \frac{341.04}{8.00 \times 0.7660} = \frac{341.04}{6.128} \approx 55.67 \, \text{N}
   \]

### Answer:
The required tension in the wire to support the beam horizontally is approximately:
\[
T \approx 55.67 \, \text{N}
\]

Would you like more details, or do you have any questions?

---

1. How would the tension change if the angle of the wire were increased?
2. What would happen to the tension if the mass of the beam doubled?
3. How does the location of the center of mass affect the calculation?
4. Can you calculate the reaction forces at the hinge?
5. What role does the symmetry of the beam play in this problem?

**Tip:** Always check the balance of forces and torques in equilibrium problems; starting from one pivot point simplifies the analysis.

Learn how to calculate the tension in a wire supporting a horizontal beam with a detailed step-by-step solution. This statics problem involves torque and equilibrium concepts, suitable for high school physics students in Grades 11-12.

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