∙∙Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m

∙∙Two circular rods, one steel and the other copper, are...

Key Concepts

Tensile Force

Tensile force is an external force that pulls on a material, causing it to stretch. It is fundamental in problems involving deformation because it initiates the stress and subsequent strain within the material.

Stress

Stress is defined as the force applied per unit area of a material. In tensile problems, it quantifies the internal forces resisting the applied load and is calculated by dividing the tensile force by the cross?sectional area of the rod.

Strain

Strain measures the relative deformation of a material under stress; it is the ratio of the change in length to the original length. As a dimensionless quantity, strain provides a normalized measure of deformation regardless of the material’s size.

Elongation

Elongation is the actual change in length experienced by a material when subjected to a tensile force. It is obtained by multiplying the strain by the original length, giving a direct measure of how much the material has stretched.

Young's Modulus

Young's modulus is a material property that indicates its stiffness by relating tensile stress to strain through a linear relationship. In the elastic regime, it allows for the calculation of elongation when the stress is known.

Hooke's Law

Hooke's law states that, within the elastic limit of a material, the strain is directly proportional to the applied stress. This principle underpins the calculation of deformation and elongation in materials subjected to tensile forces, making it essential for understanding elastic behavior.

Transcript

00:01 Hi there, troy g here for numerate, solving a problem having to do with the length, elasticity, and our constant of proportionality here, young's modulus, for a combination situation of copper and steel.

00:14 Young's modulus relates the stress of material undergoes and the deformation, the strain that it experiences.

00:22 Stress, we think of force per unit of cross -sectional area, and strain is the amount that the length changes.

00:31 Okay, in proportion to the original length of the object.

00:36 So we're told here both materials have a length of 0 .75 meters.

00:42 We're also given in the problem that the diameter of the wire system is 150 centimeters.

00:48 Let's communicate that in meters, 1 .5 times 10 to the negative second meters, divided by 100.

00:55 Of course.

00:57 The tensile force is 4 ,000 newtons on this combination metal wire.

01:03 And not given right in the problem, but easily findable in your textbook or online, young's modulus for steel is 2 times 10 to the 11th pascales, and for copper is 1 .1 times 10 to the 11th pascales.

01:15 This fits with our experience, wherein we are familiar with the fact that it is easier to stretch and deform copper than it is steel, hence the lower young's modulus.

01:28 So let's go here.

01:29 We're asked to solve for the strain on each wire in this situation.

01:37 So let's use our relationship for young's modulus to figure that out.

01:43 If we take, do some quick algebra and move the strain, multiply both sides by the strain and divide both sides by young's modulus, we can, come up with an expression where we can say the strain, in other words, the change in length over length is equal to, why don't we write it out like this, one over young's modulus multiplied by force divided by area.

02:14 So hopefully you can see the algebra happening there.

02:17 Of course, we're given the diameter of the wire, the cross -sectional area that we're concerned with is going to be pi times half of our diameter squared.

02:30 All right, so that's gonna go in there for a.

02:35 And again, if we can put it all into one expression, i think it's the best way to do both calculations.

02:43 We can say the strain is then one over young's modulus multiplied by the force.

02:51 Let's bring a factor of four from the bottom of our area, equation there.

02:57 Let's bring that up to the top and that's going to leave pi times the diameter squared there at the bottom...

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