Inclined Plane - Revision history

Aspektor3: Added an explanation of why the force of gravity affects objects on an inclined plane in the horizontal directions.

2023-12-03T21:40:13Z

Added an explanation of why the force of gravity affects objects on an inclined plane in the horizontal directions.

← Older revision		Revision as of 17:40, 3 December 2023
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	[[File:IPMathMod.jpeg\|center\|600px\|A Representation for the Mathematical Model]]		[[File:IPMathMod.jpeg\|center\|600px\|A Representation for the Mathematical Model]]

			It is important to understand the forces that act on an object traveling on an inclined plane. For example, the three forces acting here are the force of gravity, the normal force, and the force of friction. However, imagine if an object was moving up a frictionless inclined plane. Clearly, it would eventually slow down, stop, and begin moving back down the plane. This is because of the force of gravity. While the force of gravity only acts in the -y direction, it can be broken down into its x and y components to show that it has some sort of effect on the object moving up the inclined plane in the x-direction opposite to its movement. If it did not, then the object would keep rolling upwards forever.

	'''Variables:'''		'''Variables:'''

Laurence12799: /* Middling */

2019-07-18T02:53:44Z

Middling

← Older revision		Revision as of 22:53, 17 July 2019
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	:[[File:IPMiddlingFBD1.jpeg\|200px\|A Free Body Diagram for the Car]]		[[File:IPMiddlingFBD1.jpeg\|200px\|A Free Body Diagram for the Car]]
	[[File:IPMiddlingFBD2.jpeg\|213px\|A Free Body Diagram for the Counterweight]]		[[File:IPMiddlingFBD2.jpeg\|213px\|A Free Body Diagram for the Counterweight]]

Laurence12799: /* Middling */

2019-07-18T02:53:31Z

Middling

← Older revision		Revision as of 22:53, 17 July 2019
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	[[File:IPMiddlingFBD1.jpeg\|200px\|A Free Body Diagram for the Car]]		:[[File:IPMiddlingFBD1.jpeg\|200px\|A Free Body Diagram for the Car]]
	[[File:IPMiddlingFBD2.jpeg\|213px\|A Free Body Diagram for the Counterweight]]		[[File:IPMiddlingFBD2.jpeg\|213px\|A Free Body Diagram for the Counterweight]]

Laurence12799 at 20:13, 17 June 2019

2019-06-17T20:13:18Z

← Older revision		Revision as of 16:13, 17 June 2019
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	~~Claimed by Catherine Grey Spring 2017<br>~~
	~~Edited by Laurence Leon Summer 2019~~
	==The Main Idea==		==The Main Idea==

Laurence12799: /* A Mathematical Model */

2019-06-13T13:36:01Z

A Mathematical Model

← Older revision		Revision as of 09:36, 13 June 2019
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	::Note that <math>F_y</math> tends to be <math>0</math>, so <math>\mathbf{F_{net}} = \sqrt{{F_x}^2} = F_x</math>		::Note that <math>F_y</math> tends to be <math>0</math>, so <math>\mathbf{F_{net}} = \sqrt{{F_x}^2} = F_x</math>

	:<math>F_x = F_f - F_{g_x} = F_f - mg \ \text{sin}θ</math>		:<math>F_x = F_f - F_{g_x} = F_f - mg \ \text{sin}\theta</math>

	:<math>F_y = F_N - F_{g_y} = F_N - mg \ \text{cos}θ </math>		:<math>F_y = F_N - F_{g_y} = F_N - mg \ \text{cos}\theta </math>

	===A Computational Model===		===A Computational Model===

Laurence12799: /* A Mathematical Model */

2019-06-13T13:35:24Z

A Mathematical Model

← Older revision		Revision as of 09:35, 13 June 2019
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	::Note that <math>F_y</math> tends to be <math>0</math>, so <math>\mathbf{F_{net}} = \sqrt{{F_x}^2} = F_x</math>		::Note that <math>F_y</math> tends to be <math>0</math>, so <math>\mathbf{F_{net}} = \sqrt{{F_x}^2} = F_x</math>

	:<math>F_x = F_f - F_{g_x} = F_f - mg \ \text[sin]θ</math>		:<math>F_x = F_f - F_{g_x} = F_f - mg \ \text{sin}θ</math>

	:<math>F_y = F_N - F_{g_y} = F_N - mg \ \text{cos}θ </math>		:<math>F_y = F_N - F_{g_y} = F_N - mg \ \text{cos}θ </math>

Laurence12799: /* A Mathematical Model */

2019-06-13T13:35:06Z

A Mathematical Model

← Older revision		Revision as of 09:35, 13 June 2019
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	::Note that <math>F_y</math> tends to be <math>0</math>, so <math>\mathbf{F_{net}} = \sqrt{{F_x}^2} = F_x</math>		::Note that <math>F_y</math> tends to be <math>0</math>, so <math>\mathbf{F_{net}} = \sqrt{{F_x}^2} = F_x</math>

	:<math>F_x = F_f - F_{g_x} = F_f - mg \ ~~sinθ~~</math>		:<math>F_x = F_f - F_{g_x} = F_f - mg \ \text[sin]θ</math>

	:<math>F_y = F_N - F_{g_y} = F_N - mg \ ~~cosθ~~ </math>		:<math>F_y = F_N - F_{g_y} = F_N - mg \ \text{cos}θ </math>

	===A Computational Model===		===A Computational Model===

Laurence12799: /* Part II */

2019-06-13T13:33:52Z

Part II

← Older revision		Revision as of 09:33, 13 June 2019
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	:::<math>\int (6.534 - t^3 + 0.5t^2) \ dt = 6.534t - \frac{t^4}{4} + \frac{t^3}{6} + C</math>		:::<math>\int (6.534 - t^3 + 0.5t^2) \ dt = 6.534t - \frac{t^4}{4} + \frac{t^3}{6} + C</math>

	::Since <math>v(10)</math> must equal <math>297.94 \ \frac{m}{s}</math>, we have:		::Since <math>v_x(10)</math> must equal <math>297.94 \ \frac{m}{s}</math>, we have:

	:::<math>297.94 = 6.534 \times 10 - \frac{10^4}{4} + \frac{10^3}{6} + C</math>		:::<math>297.94 = 6.534 \times 10 - \frac{10^4}{4} + \frac{10^3}{6} + C</math>
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	::Now we set <math>v_x(t_\text{o})</math> equal to <math>0</math>, so that we may solve for <math>t_\text{o}</math>:		::Now we set <math>v_x(t_\text{o})</math> equal to <math>0</math>, so that we may solve for <math>t_\text{o}</math>:

	:::<math>v_x(t_\text{o}) = 6.534t_\text{o} - \frac{{t_\text{o}}^4}{4} + \frac{{t_\text{o}}^3}{6} = 0</math>		:::<math>v_x(t_\text{o}) = 6.534t_\text{o} - \frac{{t_\text{o}}^4}{4} + \frac{{t_\text{o}}^3}{6} + 2565.93 = 0</math>

	::Using a graphing calculator, we can solve the polynomial. The only solutions are <math>t_\text{o} = -10.~~4863~~ \ \text{and} \ t_\text{o} = 10.~~2743~~</math>. We reason that only the second solution makes sense and that at <math>t_\text{o} = 10.27</math> seconds the block comes to rest.		::Using a graphing calculator, we can solve the polynomial. The only solutions are <math>t_\text{o} = -9.84 \ \text{and} \ t_\text{o} = 10.30</math>. We reason that only the second solution makes sense and that at <math>t_\text{o} = 10.30</math> seconds the block comes to rest.

	==Technical Usage==		==Technical Usage==

Laurence12799: /* Part II */

2019-06-13T13:30:15Z

Part II

← Older revision		Revision as of 09:30, 13 June 2019
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	::Therefore, <math>C</math> must equal:		::Therefore, <math>C</math> must equal:

	:::<math>C = 297.94 - 65.34 + 2500 - \frac{1000}{6} = ~~2899~~.27</math>		:::<math>C = 297.94 - 65.34 + 2500 - \frac{1000}{6} = 2565.93</math>

	::Now we set <math>v_x(t_\text{o})</math> equal to <math>0</math>, so that we may solve for <math>t_\text{o}</math>:		::Now we set <math>v_x(t_\text{o})</math> equal to <math>0</math>, so that we may solve for <math>t_\text{o}</math>:

Laurence12799: /* Part II */

2019-06-13T13:22:06Z

Part II

← Older revision		Revision as of 09:22, 13 June 2019
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	::It is easy to see that the antiderivative is:		::It is easy to see that the antiderivative is:

	:::<math>\int (6.534 - t^3 + 0.5t^2) \ dt = 6.534t - \frac{t^4}{4} - \frac{t^3}{6} + C</math>		:::<math>\int (6.534 - t^3 + 0.5t^2) \ dt = 6.534t - \frac{t^4}{4} + \frac{t^3}{6} + C</math>

	::Since <math>v(10)</math> must equal <math>297.94 \ \frac{m}{s}</math>, we have:		::Since <math>v(10)</math> must equal <math>297.94 \ \frac{m}{s}</math>, we have:

	:::<math>297.94 = 6.534 \times 10 - \frac{10^4}{4} - \frac{10^3}{6} + C</math>		:::<math>297.94 = 6.534 \times 10 - \frac{10^4}{4} + \frac{10^3}{6} + C</math>

	::Therefore, <math>C</math> must equal:		::Therefore, <math>C</math> must equal:

	:::<math>C = 297.94 - 65.34 + 2500 + \frac{1000}{6} = 2899.27</math>		:::<math>C = 297.94 - 65.34 + 2500 - \frac{1000}{6} = 2899.27</math>

	::Now we set <math>v_x(t_\text{o})</math> equal to <math>0</math>, so that we may solve for <math>t_\text{o}</math>:		::Now we set <math>v_x(t_\text{o})</math> equal to <math>0</math>, so that we may solve for <math>t_\text{o}</math>:

	:::<math>v_x(t_\text{o}) = 6.534t_\text{o} - \frac{{t_\text{o}}^4}{4} - \frac{{t_\text{o}}^3}{6} = 0</math>		:::<math>v_x(t_\text{o}) = 6.534t_\text{o} - \frac{{t_\text{o}}^4}{4} + \frac{{t_\text{o}}^3}{6} = 0</math>

	::Using a graphing calculator, we can solve the polynomial. The only solutions are <math>t_\text{o} = -10.4863 \ \text{and} \ t_\text{o} = 10.2743</math>. We reason that only the second solution makes sense and that at <math>t_\text{o} = 10.27</math> seconds the block comes to rest.		::Using a graphing calculator, we can solve the polynomial. The only solutions are <math>t_\text{o} = -10.4863 \ \text{and} \ t_\text{o} = 10.2743</math>. We reason that only the second solution makes sense and that at <math>t_\text{o} = 10.27</math> seconds the block comes to rest.