Name Harry Miles SHORT ANSWER. Write the word or phrase that...

Name Harry Miles SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A 7.0-kg pipe is hung from a steel wire that is 2.5 m long and has a diameter of 0.75 mm. Young's modulus for steel is 2.0 x 10^11 N/m^2. (a) By what distance does the wire stretch? (b) If the wire diameter were doubled, what would be the stretch? 2) An aluminum wire and a steel wire, each of length 2.0 m, are hung from the ceiling. A 5.0-kg mass is suspended from the lower end of each wire. The aluminum wire has a diameter of 2.2 mm. What must be the diameter of the steel wire if it is to stretch the same distance as the aluminum wire, so that the two wires maintain equal lengths after the masses are attached? Young's modulus for aluminum is 0.70 x 10^11 Pa and for steel it is 2.0 x 10^11 Pa. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 3) When a 125-kg piece of equipment is hung from a steel wire of length 75 cm, it stretches the wire by 4.00 mm. If instead the wire had a length of 150 cm, but all else were the same, by what distance would the same equipment stretch the wire? A) 16.0 mm B) 1.00 mm C) 8.00 mm D) 2.00 mm E) 4.00 mm 4) A solid steel column is 4.0 m long and 0.20 m in diameter. Young's modulus for this steel is 2.0 x 10^11 N/m^2. By what distance does the column shrink when a 5000-kg truck is supported by it? A) 8.0 x 10^-7 m B) 3.2 x 10^-6 m C) 3.1 x 10^-5 m D) 7.8 x 10^-6 m 5) A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young's modulus for steel is 2.0 x 10^11 Pa. A) 4.5 cm B) 2.5 cm C) 3.5 cm D) 1.5 cm

Transcript

00:01 Hi there, we have five questions here under youngs modulus.

00:06 Number one, okay, let's go to number one.

00:09 We have a steel wire and suspended by it is a 7 kilogram pipe.

00:16 The initial length and the diameter of the steel wire are given here, and we are to determine the amount of stretching that's caused by the 7 kilogram object.

00:27 So to get delta l, or the change in the...

00:31 Length we'll be using the definition of youngs modulus so we see that young's modulus is equivalent to perpendicular force whether it's tensile or compressive but in this case this perpendicular force is tensile because it causes the steel wire to be stretched and then times the original length of the wire divided by the area and since we're given the diameter here we'll express area in terms of pi d squared over four and then we have the change in the length so by algebra we can simplify this as so we isolate the delta l on the left hand side so that this would still be f l not pi d squared over four times yungs modulus for steel.

01:29 So we have here the tensile force here that is exerted on the steel is of course equivalent to the weight of the suspended object.

01:42 So for f we just multiply the mass times acceleration due to gravity l0, pi d squared over 4 times l.

01:53 So we put down the values now.

01:55 We have 7 kilograms acceleration due to gravity magnitude 9 .8 and the regional length is 2 .5 we have pi the diameter has to be in meter so this is 0 .75 to the minus 3 all over 4 times yungs modulus for steel that's 10 to the 11th okay so our change in length here is we have 1 .94 times 10 to the minus 3.

02:31 So in two significant figures, let's make it 1 .9 raised to negative 3 meter.

02:38 Or that's 1 .9 millimeters.

02:43 Okay.

02:45 Let's box our final answer.

02:47 For question b, what if the diameter of the steel wire were doubled? what's going to happen to the change in the length? so to help us answer this, we just need to determine the relationship between the diameter and the change in length.

03:06 So we can use this equation here.

03:10 All the other factors are constant except for delta l and d.

03:14 So we have delta l in the numerator and in the denominator, you have the square of the diameter.

03:20 So we can say that delta l is inversely proportional to the square of the diameter, which, means that if we, sorry, let's write it again, delta l is inversely proportional to the square.

03:35 So if the diameter is doubled, then the inverse of two is one half, but we need to square it.

03:42 So the factor of change would be one -fourth.

03:45 So one -fourth of the delta -l, delta -l is 1 .9mm.

03:51 So in this case, the delta -l will just be one -fourth of this, and that's about, um, 0 .475 millimeter.

04:02 Okay.

04:04 So our work is made easier by determining the relationship using the corresponding equation where the two compared quantities are present.

04:17 Let's go to number two.

04:19 We have two wires here.

04:20 One is made of aluminum and the other is steel.

04:23 These are the yungs modulus of each steel.

04:27 Now the question is, if both wires are exposed to the weight of a 5 kilogram object, given this diameter of aluminum, 2 .2 millimeters, we need to determine the diameter of the steel such that when these two wires are exposed to this tensile force here, they would have the same amount of stretching.

04:53 So we'll start with that relationship, that change in aluminum equals change in steel wire.

05:04 Okay, so we've already determined the equation for delta l earlier, so we can directly use that.

05:12 We derive that delta l is just f, regional length over pi d squared over four times yang's modulus.

05:22 So we will write this expression on the left and say that this is f.

05:30 Okay.

05:31 So since the two wire is exposed to the same amount of mass, then we can say that the tensile force here in the aluminum is just equivalent to the tensile force on the steel.

05:48 So we can just call it as f7t for both.

05:52 And then same here.

05:54 They both have the same initial length divided by, okay, what color am i re -rosing now? it's stick to green.

06:03 Okay.

06:04 So we have pi.

06:06 This should be the diameter of the aluminum over four times diangus modulus for aluminum.

06:12 So this is now the left -hand side of our equation.

06:15 For the right -hand side, we still have tension, length l -sab -0, and then pi, diameter of steel, squared over 4 times the youngest modulus for steel.

06:28 So let's cancel out all the variables that are identical.

06:33 We have tensile force here equivalent to that that cancels out.

06:37 Same length.

06:38 So we have one in the numerator here and then pi over 4 is out.

06:44 So further simplifying this, if we cross multiply, we have diameter of steel squared times yungs modulus of steel equals diameter of aluminum squared yunges modulus for aluminum.

07:01 Isolating diameter of steel on the left hand side...

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