Problem

$\bullet$ $\bullet$ A steel cable with cross-sect…

03:15

Problem 20 Medium Difficulty

$\bullet$ A steel wire has the following properties:
$$\begin{array}{l}{\text { Length }=5.00 \mathrm{m}} \\ {\text { Cross-sectional area }=0.040 \mathrm{cm}^{2}} \\ {\text { Young's modulus = } 2.0 \times 10^{11} \mathrm{Pa}} \\ {\text { Shear modulus = } 0.84 \times 10^{11} \mathrm{Pa}} \\ {\text { Proportional limit }=3.60 \times 10^{8} \mathrm{Pa}} \\ {\text { Breaking stress }=11.0 \times 10^{8} \mathrm{Pa}}\end{array}$$
The wire is fastened at its upper end and hangs vertically.
(a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

Answer

a) 147 $\mathrm{kg}$
b) 0.9 $\mathrm{cm}$
c) 449 $\mathrm{kg}$

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Physics 101 Mechanics

College Physics

Chapter 11

Elasticity and Periodic Motion

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Video Transcript

to do part a recall that the stress of an object is equal to the perpendicular force divided by the cross sectional area. So in this case, were given that we want the stress to be the proportional limit, which is 3.6 times 10 to the eighth. So we know F perp over a is equal to 3.6 times 10 to the eighth Pascal's, which is the proportional limit. In this case, the perpendicular forces just equal to the weight of the object. So wait, being W W over is now equal to 3.6 times 10 to the eighth best gals. And then we can solve this for W. Since we know the cross sectional area and so W is equal to 3.6 times 10 to the eighth. Pascal's Time's a cross sectional area, which is of being syrup 0.4 times 10 to the minus four, and this is a meter squared. So since everything is in S I u nits, whenever we plug this into the calculator, we're going to get an answer or Newton's. And that answer is one point for Times 10 cubed Nunes. And that's the maximum weight if you want to stay under the proportional limit here. And so the search party and Barbie, we want to figure out the change in length under this proportional limit. And so were called that young module asses Eagle to L. Not. Time's a perpendicular force over a telltale. So solving this for Delta L. We get Delta l is equal to the original length times the perpendicular force over Young's mom AngelesTimes cross sectional area And we know everything in the right hand side here. So he was plug in. Were you five for the originally time is one point for four times 10. Cute. This was what we found in party. We're going to divide by the Mongols. They give us 2.0 times 10 to the 11 and then the area which were used in part as 110.4 time stenting lives for and that's a meter square. When we do this out, I get nine times 10 to the minus three and this is in meters since everything used tears in this eye. And so that's the answer to part me. So for Parsi, we want to do that using the exact same setup as part, eh? Except in this case, F per over eh is equal to 11 times 10 to the eighth Pascal's. So this number here is different because this is now the breaking stress, but the exact process will be the same. And so we solve this out. We get that the wait. Is he going to four point for times 10? Cute. And so this is a little higher because we're allowing it to stress a little more. And so this is the final answer.

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