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 3.0 x 10¹⁰ N/m². 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. vaia topper

Answer 1

The shear stress on the rod is 7.95 x 10^8 N/m^2 and the vertical deflection of the end of the rod is 1.04 x 10^-3 cm.

Step-by-step explanation:

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

Shear Stress = (Shearing Modulus) * (Shearing Strain)

The shearing strain can be calculated by dividing the horizontal projection of the rod from the wall by the length of the rod:

Shearing Strain = (Horizontal Projection) / (Length of Rod)

Substituting the given values, we have:

Shearing Strain = 5.3 cm / 200 cm = 0.0265

Now we can calculate the shear stress:

Shear Stress = (3.0 x 10^10 N/m^2) * (0.0265) = 7.95 x 10^8 N/m^2

(b) To find the vertical deflection of the end of the rod, we can use the formula:

Vertical Deflection = (Load) * (Length of Rod) / (Shearing Modulus * (Cross-sectional area of Rod))

Substituting the given values, we have:

Vertical Deflection = (1200 kg * 9.8 m/s^2) * (200 cm) / (3.0 x 10^10 N/m^2 * (pi * (2.4 cm)^2)) = 1.04 x 10^-3 cm

Therefore, the shear stress on the rod is 7.95 x 10^8 N/m^2 and the vertical deflection of the end of the rod is 1.04 x 10^-3 cm.