Pulley system for block piled on block - Revision history

Vipul: /* Case analysis */

2011-08-13T01:37:43Z

Case analysis

← Older revision		Revision as of 01:37, 13 August 2011
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	<math>\! F \le g[\mu_{s1}(m_A + m_B) + 2\mu_{s2}m_B] = g[\mu_{s1}m_A + (\mu_{s1} + 2\mu_{s2})m_B]</math>		<math>\! F \le g[\mu_{s1}(m_A + m_B) + 2\mu_{s2}m_B] = g[\mu_{s1}m_A + (\mu_{s1} + 2\mu_{s2})m_B]</math>

			===Case that both blocks move===

			Let <math>a</math> denote the magnitude of acceleration of both <math>A</math> and <math>B</math>, with the direction for <math>A</math> being along <math>F</math> (pulling on the string) and the direction for <math>B</math> being opposite to <math>F</math> (being pulled by the string). As before, denote by <math>T</math> the tension in the string, which is the same at both ends of the string due to the assumption of a massless string and massless pulley.

			We have:

			<math>\! F = T + f_{k1} + f_{k2} + m_Aa \qquad (3.1)</math>

			We also have:

			<math>f_{k1} = \mu_{k1}N_A \qquad (3.2)</math>

			and

			<math>f_{k2} = \mu_{k2}N_{AB} \qquad (3.3)</math>

			The force diagram for <math>B</math> gives:

			<math>T = f_{k2} + m_Ba \qquad (3.4)</math>

			Plug in (3.2), (3.3), and (3.4) into (3.1) and get:

			<math>\! F = \mu_{k1}N_A + 2\mu_{k2}N_{AB} + (m_A + m_B)a \qquad (3.5)</math>

			Now plug in (1.1) and (1.3) into this:

			<math>\! F = \mu_{k1}(m_A + m_B)g + 2\mu_{k2}m_Bg + (m_A + m_B)a \qquad (3.6)</math>

			Rearranging, we obtain that:

			<math>\! a = \frac{F}{m_A + m_B} - \mu_{k1}g - \frac{2\mu_{k2}m_B}{m_A + m_B} \qquad (3.7)</math>

			The condition for this case to occur is that <math>F</math> be greater than the maximum value for static friction to apply, i.e.:

			<math>\! F \ge g[\mu_{s1}(m_A + m_B) + 2\mu_{s2}m_B] = g[\mu_{s1}m_A + (\mu_{s1} + 2\mu_{s2})m_B]</math>

Vipul: /* Case analysis */

2011-08-13T01:28:30Z

Case analysis

← Older revision		Revision as of 01:28, 13 August 2011
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	* The accelerations of <math>A</math> and <math>B</math> have equal magnitudes and opposite directions.		* The accelerations of <math>A</math> and <math>B</math> have equal magnitudes and opposite directions.
	* In particular, if either block moves, ''both'' blocks move and they do so in opposite directions, so there is bound to be slipping. It is not possible for the two blocks to move as a system.		* In particular, if either block moves, ''both'' blocks move and they do so in opposite directions, so there is bound to be slipping. It is not possible for the two blocks to move as a system.

			===Case that neither block moves===

			Let <math>T</math> denote the magnitude of tension in the string. There is no slipping of surfaces, so static friction is operative both between <math>A</math> and the floor and between <math>A</math> and <math>B</math>. Denote by <math>f_{s1}</math> the magnitude of static friction between <math>A</math> and the floor and by <math>f_{s2}</math> the magnitude of static friction between <math>A</math> and <math>B</math>.

			For block <math>A</math>, we have, by [[Newton's first law]]:

			<math>\! F = T + f_{s1} + f_{s2} \qquad (2.1)</math>

			Also:

			<math>f_{s1} \le \mu_{s1}N_A \qquad (2.2)</math>

			and

			<math>f_{s2} \le \mu_{s2}N_{AB} \qquad (2.3)</math>

			For block <math>B</math>, we have, by Newton's first law:

			<math>\! T = f_{s2} \qquad (2.4)</math>

			Plugging in (2.4) into (2.1), we obtain that:

			<math>\! F = f_{s1} + 2f_{s2} \qquad (2.5)</math>

			Plugging in (2.2), (2.3), (1.1), and (1.3) into (2.5), we get:

			<math>\! F \le g[\mu_{s1}(m_A + m_B) + 2\mu_{s2}m_B] = g[\mu_{s1}m_A + (\mu_{s1} + 2\mu_{s2})m_B]</math>

Vipul: Created page with "==The scenario== Consider one block placed on top of another, with the bottom block being $A$ and the block on top being $B$ . $A$ is on a fixed ..."

2011-08-13T01:15:18Z

Created page with "==The scenario== Consider one block placed on top of another, with the bottom block being <math>A</math> and the block on top being <math>B</math>. <math>A</math> is on a fixed ..."

New page

==The scenario==

Consider one block placed on top of another, with the bottom block being <math>A</math> and the block on top being <math>B</math>. <math>A</math> is on a fixed horizontal floor and the surface of contact between <math>A</math> and <math>B</math> is also horizontal. There is a pulley system connecting <math>A</math> and <math>B</math>, with both sides of the pulley rope being horizontal.

Consider the following conventions:

* <math>\mu_{s1}</math> is the [[limiting coefficient of static friction]] between <math>A</math> and the floor.
* <math>\mu_{k1}</math> is the [[coefficient of kinetic friction]] between <math>A</math> and the floor.
* <math>\mu_{s2}</math> is the [[limiting coefficient of static friction]] between <math>A</math> and <math>B</math>.
* <math>\mu_{k2}</math> is the [[coefficient of kinetic friction]] between <math>A</math> and <math>B</math>.

Assume that <math>\mu_{k1} \le \mu_{s1}</math> and <math>\mu_{k2} \le \mu_{s2}</math>.

We assume the pulley and string to be massless, so that the tension at both ends of the string is equal.

==Normal force and gravitational force==

===On the upper block===

[[File:Upperblockforcediagram.png|thumb|500px|right]]

{{action reaction force concept|The normal force on the upper block and <math>m_Bg</math> (the gravitational force) are equal and opposite but do ''not'' form an action reaction pair.}}

If we denote by <math>N_{AB}</math> the normal force between the blocks, then <math>N_{AB}</math> acts upward on the upper block. The gravitational force <math>m_Bg</math> acts downward. Since there is no net acceleration in the vertical direction, we get:

<math>\! N_{AB} = m_Bg \qquad (1.1)</math>

===On the lower block===

If we denote by <math>N_A</math> the normal force between the lower block and the floor, we obtain, by [[Newton's first law]], that:

<math>N_A = N_{AB} + m_Ag \qquad (1.2)</math>

Plugging in (1) into (2) we get:

<math>N_A = (m_A + m_B)g \qquad (1.3)</math>

==Case analysis==

We consider an external force <math>F</math> applied horizontally, in the direction away from the direction the string is attached, on the lower block. The analysis is exactly the same for the upper block (give explanation). Note the following:

* The tension in the string at both its <math>A</math> and <math>B</math> ends is equal.
* The accelerations of <math>A</math> and <math>B</math> have equal magnitudes and opposite directions.
* In particular, if either block moves, ''both'' blocks move and they do so in opposite directions, so there is bound to be slipping. It is not possible for the two blocks to move as a system.

Pulley system for block piled on block - Revision history

Vipul: /* Case analysis */

Vipul: /* Case analysis */

Vipul: Created page with "==The scenario== Consider one block placed on top of another, with the bottom block being A and the block on top being B. A is on a fixed ..."

Vipul: Created page with "==The scenario== Consider one block placed on top of another, with the bottom block being $A$ and the block on top being $B$ . $A$ is on a fixed ..."