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?

6.25*10^6m

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

The stress = 6.13*10^7N/m^2

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