Length and Stiffness of an Interatomic Bond - Revision history

Ftariq6 at 05:25, 24 November 2019

2019-11-24T05:25:32Z

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 ===Example 3===
 Forty-five identical springs are placed side by side and connected in parallel to a large massive block. The stiffness of the 45-spring combination is 20,250 N/m. What is the stiffness of one of the individual springs?
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 </div>
-===Example 4===
+===Example 5===
 Five identical springs, each with stiffness 390 N/m, are attached in parallel (that is side by side) to hold up a heavy weight. If these springs were replaced by an equivalent single spring, what should be the stiffness of this single spring?

Ftariq6 at 05:05, 24 November 2019

2019-11-24T05:05:42Z

← Older revision		Revision as of 01:05, 24 November 2019
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	<math> Total\ number\ of\ springs\ = n = 45</math>		<math> Total\ number\ of\ springs\ = n = 45</math>

	<math> Effective\ stiffness\ constant\ = k_{eff} = 20250</math>		<math> Effective\ stiffness\ constant\ = k_{eff} = 20250\ N/m</math>

	<math> Stiffness\ constant\ of\ each\ spring\ = k_{s}</math>		<math> Stiffness\ constant\ of\ each\ spring\ = k_{s}</math>

	<math> k_{eff}=~~45k_~~{s}</math>		<math> k_{eff}=n*k_{s}</math>

	<math>k_{s} = {\frac{k_{eff}}{45}}</math>		<math>k_{s} = {\frac{k_{eff}}{n}}</math>

	<math>k_{s} = {\frac{20250\ N/m}{45}}</math>		<math>k_{s} = {\frac{20250\ N/m}{45}}</math>

	<math>k_{s} = 450\ N/m </math>		<math>k_{s} = 450\ N/m </math>

			</div>

			===Example 4===

			Five identical springs, each with stiffness 390 N/m, are attached in parallel (that is side by side) to hold up a heavy weight. If these springs were replaced by an equivalent single spring, what should be the stiffness of this single spring?

			<div align ="center">

			<math> k_{eff}=n*k_{s}</math>

			<math> k_{s}=390\ N/m</math>

			<math> n = 5</math>

			<math>k_{eff} = {5}*{390\ N/m}</math>

			<math>k_{eff} = {1950\ N/m}</math>

	</div>		</div>

Ftariq6 at 04:55, 24 November 2019

2019-11-24T04:55:01Z

← Older revision		Revision as of 00:55, 24 November 2019
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	'''Claimed by Fehmeen Tariq Fall 2019'''		'''Claimed by Fehmeen Tariq Fall 2019'''
	~~Claimed by Andy Stevens Spring 2017~~
	~~Claimed by Nicole Tansey Fall 2017~~

	This topic covers find the length and stiffness of an Interatomic Bond.		This topic covers find the length and stiffness of an Interatomic Bond.

Ftariq6 at 04:47, 24 November 2019

2019-11-24T04:47:58Z

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-'''Claimed by Andy Stevens Spring 2017'''
+'''Claimed by Fehmeen Tariq Fall 2019'''
 Claimed by Nicole Tansey Fall 2017
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 </div>
 ==Connectedness==

Nicoletansey at 02:04, 29 November 2017

2017-11-29T02:04:25Z

← Older revision		Revision as of 22:04, 28 November 2017
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	[[File:crosssectionalarea.jpg]]		[[File:crosssectionalarea.jpg]]

	~~===Formulas to know===~~

			== Mathematical Model ==
			'''Formulas to Know'''

			In many problems, you will be given the density of the solid that you are being questioned on. It is important to remember that <math>{density} = \frac{mass}{volume}</math>
	Number of bonds in one "chain" of the solid:		Number of bonds in one "chain" of the solid:
	<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>		<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Nicoletansey at 01:45, 29 November 2017

2017-11-29T01:45:14Z

← Older revision		Revision as of 21:45, 28 November 2017
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	Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.		Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

	We can determine the stiffness of an interatomic bond ~~by considering it as a spring~~. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms~~. The diameter~~ of ~~an atom is~~ the space-filling model ~~of a solid~~. To find the radius, we divide the diameter in half. If we can calculate the length of the interatomic bond ~~and~~ the diameter of a single atom, we can use this data to find the stiffness of the interatomic bond, ~~considered~~ as a spring.		With this information, we can determine the stiffness of an interatomic bond. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms, which is just the distance of the two atoms' radii added together since we use the space-filling model. To find the radius, we divide the diameter of a single atom in half. So, the center-to-center distance is essentially equal to the diameter of a single atom since in the space-filling model there is no space between each atom. If we can calculate the length of the interatomic bond (the diameter of a single atom), we can use this data to find the stiffness of the interatomic bond, modeled as a spring.

	==Length of an Interatomic Bond==		==Length of an Interatomic Bond==

Nicoletansey at 01:39, 29 November 2017

2017-11-29T01:39:24Z

← Older revision		Revision as of 21:39, 28 November 2017
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	'''Claimed by Andy Stevens Spring 2017'''		'''Claimed by Andy Stevens Spring 2017'''
	Claimed by Nicole Tansey Fall 2017		Claimed by Nicole Tansey Fall 2017

	This topic covers find the length and stiffness of an Interatomic Bond.		This topic covers find the length and stiffness of an Interatomic Bond.

	==The Main Idea==		==The Main Idea==

	A solid ~~objects~~ is made up of a ~~bunch~~ of ~~tiny balls (~~atoms) that are held together by ~~springs (~~chemical bonds). The relaxed ~~legth~~ of the ~~little~~ spring between two atoms ~~(the interatomic bond)~~ is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.		A solid object is made up of a large amount of atoms that are held together by chemical bonds in a lattice-like structure as seen in ''Figure 1''. Each atom is connected to its neighboring atoms by these chemical bonds. One way to approximate this solid object model is to imagine the atoms as balls and the chemical bonds as springs-- a ball and spring model! The relaxed length of the microscopic spring between two atoms is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.


			''Figure 1''
	[[File:ballcube.jpg\|center]]		[[File:ballcube.jpg\|center]]

Nicoletansey at 01:17, 29 November 2017

2017-11-29T01:17:24Z

← Older revision		Revision as of 21:17, 28 November 2017
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	'''Claimed by Andy Stevens Spring 2017'''		'''Claimed by Andy Stevens Spring 2017'''
			Claimed by Nicole Tansey Fall 2017
	This topic covers find the length and stiffness of an Interatomic Bond.		This topic covers find the length and stiffness of an Interatomic Bond.

Astevens37: /* Example 2 */

2017-04-10T03:45:00Z

Example 2

← Older revision		Revision as of 23:45, 9 April 2017
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	'''Step 1'''		'''Step 1'''
	Spring Stiffness of Entire Wire		Spring Stiffness of Entire Wire
			<div align ="center">

	<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>		<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

	<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>		<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

			</div>


	'''Step 2'''		'''Step 2'''
	Cross-Sectional Area of Wire		Cross-Sectional Area of Wire

			<div align ="center">

	<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>		<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

			</div>


	'''Step 3'''		'''Step 3'''
	Cross-Sectional Area of One Atom		Cross-Sectional Area of One Atom

			<div align ="center">

	<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>		<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

			</div>


	'''Step 4'''		'''Step 4'''
	Number of Chains		Number of Chains

			<div align ="center">

	<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>		<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

			</div>


	'''Step 5'''		'''Step 5'''
	Number of Bonds in One Chain		Number of Bonds in One Chain

			<div align ="center">

	<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>		<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

			</div>


	'''Step 6'''		'''Step 6'''
	Stiffness of One Interatomic Spring		Stiffness of One Interatomic Spring

			<div align ="center">

	<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>		<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

	<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>		<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

			</div>

	==Connectedness==		==Connectedness==

Astevens37: /* Example 1 */

2017-04-10T03:44:07Z

Example 1

← Older revision		Revision as of 23:44, 9 April 2017
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	''b. What is the diameter of a single zinc atom?''		''b. What is the diameter of a single zinc atom?''

			<div align ="center">

	<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>		<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
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	<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>		<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

			</div>

	''c. How many zinc atoms make up one side of the penny?''		''c. How many zinc atoms make up one side of the penny?''

			<div align ="center">

	<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>		<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

			</div>

	''d. Calculate the interatomic spring stiffness for zinc.''		''d. Calculate the interatomic spring stiffness for zinc.''

			<div align ="center">

	<math>y = {\frac{k_{si}}{d_{atom}}}</math>		<math>y = {\frac{k_{si}}{d_{atom}}}</math>

	<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>		<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

			</div>

	===Example 2===		===Example 2===