**A horizontal aluminum rod 4.6 cm in diameter projects 7.1 cm from a wall. A 820 kg object is suspended from the end of the rod. The shear modulus of aluminum is 3.0·1010 N/m2. Neglecting the rod's mass, find (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod.**

To solve this problem, we need to calculate the shear stress on the rod and the vertical deflection of the end of the rod when a mass is suspended from it. The shear modulus and the geometry of the rod will be used to determine these values.

### Given Data:
- Diameter of the rod, \( d = 4.6 \, \text{cm} = 0.046 \, \text{m} \)
- Length of the rod, \( L = 7.1 \, \text{cm} = 0.071 \, \text{m} \)
- Mass of the object, \( m = 820 \, \text{kg} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)
- Force due to the object, \( F = mg = 820 \times 9.8 = 8036 \, \text{N} \)
- Shear modulus of aluminum, \( G = 3.0 \times 10^{10} \, \text{N/m}^2 \)

### Part (a): Shear Stress on the Rod

Shear stress \( \tau \) is defined as the force applied per unit area:

\[
\tau = \frac{F}{A}
\]

where \( A \) is the cross-sectional area of the rod. The cross-sectional area \( A \) of a circular rod is given by:

\[
A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.046}{2}\right)^2 = \pi \times (0.023)^2 \approx 1.661 \times 10^{-3} \, \text{m}^2
\]

Thus, the shear stress \( \tau \) is:

\[
\tau = \frac{8036 \, \text{N}}{1.661 \times 10^{-3} \, \text{m}^2} \approx 4.837 \times 10^6 \, \text{N/m}^2 = 4.837 \, \text{MPa}
\]

### Part (b): Vertical Deflection of the End of the Rod

The vertical deflection \( \Delta y \) at the end of the rod due to the shear force is given by:

\[
\Delta y = \frac{F \times L}{A \times G}
\]

Substituting the values:

\[
\Delta y = \frac{8036 \times 0.071}{1.661 \times 10^{-3} \times 3.0 \times 10^{10}} \, \text{m}
\]

\[
\Delta y \approx \frac{570.556}{4.983 \times 10^7} \, \text{m} \approx 1.145 \times 10^{-5} \, \text{m} \approx 11.45 \, \mu\text{m}
\]

### Final Answers:
(a) The shear stress on the rod is approximately **4.837 MPa**.

(b) The vertical deflection of the end of the rod is approximately **11.45 µm**.

Would you like further details on this solution or have any questions? Here are some related questions:

1. How would the shear stress change if the rod were made of a different material with a different shear modulus?
2. What would happen to the deflection if the rod's length were doubled?
3. How does the diameter of the rod affect the shear stress and deflection?
4. What is the significance of the shear modulus in determining the deflection of the rod?
5. How would the results differ if the rod had a rectangular cross-section instead of a circular one?

**Tip:** When dealing with shear stress and deflection, remember that the geometry of the object plays a crucial role in determining the final values.

Learn how to calculate shear stress and vertical deflection in an aluminum rod problem. Solve for shear stress and vertical deflection using the shear modulus and dimensions of the rod. Suitable for college-level mechanics and solid mechanics courses.

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