Properties of solid materials. Hardness Resistant to cutting, indentation and abrasion. The Mohs scale of hardness grades ten minerals from the softest (talc) rated 1, to the hardest (diamond) rated 10
The elastic limit can be seen on the graph.
This is where it stops obeying Hookes law.
F = k ∆ x
k is measured in units of newtons per metre (Nm -1).
Extension, ∆x = Stretched length – Original length = . 0.42m – 0.38m = 0.04 m
F = k ∆ x
So, k = 2.0 N
= 50 N m-1
2.0N = k x 0.04m
1.A metal guitar string stretches 4mm when a 10N force is applied.
2.A rubber band is 6cm long. When it is loaded with 2.5N, its length becomes 10.4cm. Further loading increases the length to 16.2cm when the force is 5N.
Does the rubber band obey Hooke’s law when the force on it is 5N? Justify your answer with a suitable calculation.
3.A vertical spring stretches 10cm from its original position under 200g load, determine the spring load.
A stretching force is also called a tensile force.
Tensile stress = tensile force (N)
cross-section area (m2)
σ = F / A
unit – Pa (pascal) or Nm-2
Note: 1 Pa = 1 Nm-2
Nylon has an ultimate tensile stress of 7x107Pa. The cross sectional area of the nylon rope in the diagram is 3x10-5m2. If the Bassem weighs 600N. Will it hold him?
This is less than the ultimate tensile stress, he’ll live to see another physics example! Dammit! : )
This is the stress required to cause a material to break.
Sometimes known as the ultimate tensile stress.
Answer: 4.0x108 Pa
Tensile strain = extension
ε = ΔL / L
unit – none (it’s a ratio like pi)
Qu: A steel wire of length 2m and diameter 0.4mm is extended by 4.0mm when a stretching force of 50N is applied. Calculate
b) The strain on the wire
Answer: 2x10-3 = 20%
tensile strain, ε
A ΔLWhat are the units?
A metal wire of original length 1.6m, cross sectional area 0.8 mm2 extends by 4mm when stretched by a tensile force of 200N.
Calculate the wire’s (a) strain, (b) stress & (c) Young Modulus
b) stress: 200/8x107=2.5x108Pa