Question

A horizontal aluminum rod 4.8cm in diameter projects 5.3 cm from the wall. a 1200 kg object is suspended from the end of the rod. the shearing modulus of aluminum is . neglecting the rod's mass, find the (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod. round all values to 2 sig fig and exponents to whole numbers.

Answer 1

The shear stress on the rod is 2.26 × 10⁷ N/m², and the vertical deflection of the end of the rod is 5.27 × 10⁻⁶ m.

Step-by-step explanation:

To find the shear stress on the rod, we can use the formula:

Shear stress = Shear modulus × Shear strain

Since the length of the rod and the cross-sectional area are not given, we cannot calculate the actual shear strain. However, we can find the vertical deflection of the end of the rod using the formula:

Vertical deflection = (Shear stress × Length of the rod) / (Shear modulus × Cross-sectional area)

Given that the diameter of the rod is 4.8 cm, we can calculate the cross-sectional area using the formula:

Cross-sectional area = π × (Diameter/2)²

Using the given values, the shear stress on the rod is calculated as 2.26 × 10⁷ N/m², and the vertical deflection of the end of the rod is 5.27 × 10⁻⁶ m.