Question

The cable of a hoist has a cross-sectional area of 80 mm2. The hoist is used to lift a crate weighing 500 kg. The free length of the cable is 30 m. Assume all deformation is elastic.

(a) What is the stress on the cable?

Answer 1

The stress= 6.13×10^7N/m^2

Step-by-step explanation:

Formular for calculating stress=F/A

Where F is force

A is crossectional area of the specimen

But W= mg

W= weight of crate

m= mass of crate

g= acceleration due to gravity

W= 500kg× 9.81=4905N

Stress = 4905N/ 8.0×10^-7m^2

Stress= 6.13×10^7M/m^2

Answer 2

6.25×10^6m

Step-by-step explanation:

Tensile stress of an elastic materials is defined as the ratio of the force exerted on the cable to its cross sectional area. Mathematically,

Tensile stress = Applied Force/Cross sectional area.

Given the weight of the crate = 500kg

Force = mg = 500×10

Force applied = 5000N

Cross sectional area = 80mm²

Since 1mm² = 1×10^-6m²

80mm² = 80×10^-6m²

Tensile stress = 5000/80×10^-6

Tensile stress = 6.25×10^6m

The stress of the cable is therefore 6.25×10^6m